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On May 10 2012 06:18 frogrubdown wrote:Show nested quote +On May 10 2012 05:11 Misder wrote:Honestly, I'm really surprised that the ping-pong ball problem is still in discussion, and I'm glad actually. Here is the problem again: + Show Spoiler +On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradoxThe question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng. Well, I guess this is what I'm using my 500th post on. + Show Spoiler +
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random
P(B)=0.5 since there are equal number of balls of each color and they are drawn at random.
P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason.
+ Show Spoiler +
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
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On May 10 2012 20:36 TanGeng wrote:
Mountaineers and Sherpas
A mountaineer is attempting a climb through a mountain gap from base camp A to base camp B. The route takes six days. At base camp A, the mountaineer finds sherpas for hire. He can get them to carry his supplies and equipment but must feed them and must not allow the sherpas to starve. The mountaineer and the sherpas can carry a maximum of four days supply while climbing. How does the mountaineer get from base camp to base camp in six days? How many days of supply does he buy?
The next year, the mountaineer returns to make the climb but no sherpas are available. How does the mountaineer make the climb on his own?
+ Show Spoiler +16 days' worth of supplies, hires 3 sherpas. After the first day, one sherpa gives 1 days' worth of supplies to each of the other two sherpas, then returns home. After the second day, each of the remaining sherpas gives 1 days' worth of supplies to the mountaineer, then return. At this point, there's four days left to go, and the mountaineer has four days' worth of food.
The next year, assuming the supplies don't spoil and he can find them after he stores them along the road, he takes four days' worth of supplies, advances for 1 day, plants 2 days' worth of supplies, then returns. Then he takes four more days' worth of supplies, advances 1 day, retrieves 1 day of supplies, advances another day, plants 2 days' worth of supplies, and returns, retrieving the rest of the cache of supplies left at the day 1 stop. Finally, he takes 4 days' worth of supplies, advances 2 days, retrieves the cache with 2 days' supplies, and a-moves for the win.
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On May 11 2012 00:09 Zaphod Beeblebrox wrote:Show nested quote +On May 10 2012 06:18 frogrubdown wrote:On May 10 2012 05:11 Misder wrote:Honestly, I'm really surprised that the ping-pong ball problem is still in discussion, and I'm glad actually. Here is the problem again: + Show Spoiler +On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradoxThe question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng. Well, I guess this is what I'm using my 500th post on. + Show Spoiler +
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random
P(B)=0.5 since there are equal number of balls of each color and they are drawn at random.
P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason. + Show Spoiler +
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
I agree with your explanation...
+ Show Spoiler +
But I've also seen a lot of arguing based on the wording of the riddle, rather than the actual mathematics or process behind solving it. Clearly, if we all agreed on what the question was specifically asking, there would probably be no debate 
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On May 11 2012 01:10 Zato-1 wrote:Show nested quote +On May 10 2012 20:36 TanGeng wrote:
Mountaineers and Sherpas
A mountaineer is attempting a climb through a mountain gap from base camp A to base camp B. The route takes six days. At base camp A, the mountaineer finds sherpas for hire. He can get them to carry his supplies and equipment but must feed them and must not allow the sherpas to starve. The mountaineer and the sherpas can carry a maximum of four days supply while climbing. How does the mountaineer get from base camp to base camp in six days? How many days of supply does he buy?
The next year, the mountaineer returns to make the climb but no sherpas are available. How does the mountaineer make the climb on his own? + Show Spoiler +16 days' worth of supplies, hires 3 sherpas. After the first day, one sherpa gives 1 days' worth of supplies to each of the other two sherpas, then returns home. After the second day, each of the remaining sherpas gives 1 days' worth of supplies to the mountaineer, then return. At this point, there's four days left to go, and the mountaineer has four days' worth of food.
The next year, assuming the supplies don't spoil and he can find them after he stores them along the road, he takes four days' worth of supplies, advances for 1 day, plants 2 days' worth of supplies, then returns. Then he takes four more days' worth of supplies, advances 1 day, retrieves 1 day of supplies, advances another day, plants 2 days' worth of supplies, and returns, retrieving the rest of the cache of supplies left at the day 1 stop. Finally, he takes 4 days' worth of supplies, advances 2 days, retrieves the cache with 2 days' supplies, and a-moves for the win.
+ Show Spoiler +I think you can do it with 2 sherpas and 12 days worth of supply. The three of them use 3 days worth of supply from sherpa 1 on first day, and send him home with his remaining day worth. The mountaineer and the other sherpa still have full supply, and they use up 2 days worth of supply from sherpa 2 on the second day, and sent him home with his remaining 2 days of supply, while the mountaineer still has his 4 days of supply to get him all the way to camp B.
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On May 10 2012 05:50 Aelfric wrote:Assassination Attempt (This one is a bit harder):+ Show Spoiler +
The King of a small country invites 1000 senators to his annual party. As gifts, each senator brings the King a bottle of wine, for a grand total of 1000 bottles of wine. Each bottle is signed by the senator who gave it.
At the end of the party, the Queen tells the King that one of the senators is trying to assassinate him, and has put deadly poison in the bottle of wine he gave as a gift. Unfortunately, the Queen doesn't know which senator is the traitor (and thus doesn't know which bottle of wine has the poison in it).
The King has 10 servants. He views them as expendable, and does not care if they live or die. He decides to use them to figure out which bottle is poisoned, which will then indicate which senator is trying to assassinate him.
His plan is to make each servant drink from zero or more of the bottles of wine. The King knows that the poison is such that if a servant drinks it, he will feel fine until noon on the next day, at which point he will instantly drop dead.
The King must know for sure who the traitor is by noon on the day after the party, or else the traitor will try to find another way to assassinate him. This essentially means that he has one shot to make his servants drink the wine in order to figure out which is the poison wine.
Note that the King can make any of the servants drink from any of the wine bottles. He does not need to make all of the servants drink wine if he doesn't want to. Any servant who drinks from the poisoned bottle will die the next day at noon.
How can the King figure out for sure who the traitor is by noon on the following day?
+ Show Spoiler +1: 1-512
2: 1-256, 513-768
3: 1-128, 257-384, 513-640, 769-896
4: 1-64, 129-192, 257-320, 385-448, 513-576, 641-704, 769-832, 897-960
5: 1-32, 65-96, 129-160, 193-224, 257-288, 321-352, 385-416, 449-480, 513-544, 577-608, 641-672, 705-736, 769-800, 833-864, 897-928, 961-992
6: 1-16, 33-48, 65-80, 97-112, 129-144, 161-176, 193-208, 225-240, 257-272, 289-304, 321-336, 353-368, 385-400, 417-432, 449-464, 481-496, 513-528, 545-560, 577-592, 609-624, 641-656, 673-688, 705-720, 737-752, 769-784, 801-816, 833-848, 865-880, 897-912, 929-944, 961-976, 993-1000
7: 1-8, 17-24, 33-40, 49-56, 65-72, 81-88, 97-104, 113-120, 129-136, 145-152, 161-168, 177-184, 193-200, 209-216, 225-232, 241-248, 257-264, 273-280, 289-296, 305-312, 321-328, 337-344, 353-360, 369-376, 385-392, 401-408, 417-424, 433-440, 449-456, 465-472, 481-488, 497-504, 513-520, 529-536, 545-552, 561-568, 577-584, 593-600, 609-616, 625-632, 641-648, 657-664, 673-680, 689-696, 705-712, 721-728, 737-744, 753-760, 769-776, 785-792, 801-808, 817-824, 833-840, 849-856, 865-872, 881-888, 897-904, 913-920, 929-936, 945-952, 961-968, 977-984, 993-1000
8: 1-4, 9-12, 17-20, 25-28, 33-36, 41-44, 49-52,57-60, 65-68, 73-76, 81-84, 89-92, 97-100, 105-108, 113-116, 121-124, 129-132, 137-140, 145-148, 153-156, 161-164, 169-172, 177-180, 185-188, 193-196, 201-204, 209-212, 217-220, 225-228, 233-236, 241-244, 249-252,257-260, 265-268, 273-276, 281-284, 289-292, 297-300, 305-308, 313-316, 321-324, 329-332, 337-340, 345-348, 353-356, 361-364, 369-372, 377-380, 385-388, 393-396, 401-404, 409-412, 417-420, 425-428, 433-436, 441-444, 449-452,457-460, 465-468, 473-476, 481-484, 489-492, 497-500, 505-508, 513-516, 521-524, 529-532, 537-540, 545-548, 553-556, 561-564, 569-572, 577-580, 585-588, 593-596, 601-604, 609-612, 617-620, 625-628, 633-636, 641-644, 649-652,657-660, 665-668, 673-676, 681-684, 689-692, 697-700, 705-708, 713-716, 721-724, 729-732, 737-740, 745-748, 753-756, 761-764, 769-772, 777-780, 785-788, 793-796, 801-804, 809-812, 817-820, 825-828, 833-836, 841-844, 849-852,857-860, 865-868, 873-876, 881-884, 889-892, 897-900, 905-908, 913-916, 921-924, 929-932, 937-940, 945-948, 953-956, 961-964, 969-972, 977-980, 985-988, 993-996
9: X01-X02, X05-X06, X09-X10, X13-X14, X17-X18, X21-X22, X25-X26, X29-X30, X33-X34, X37-X38, X41-X42, X45-X46, X49-X50, X53-X54, X57-X58, X61-X62, X65-X66, X69-X70, X73-X74, X77-X78, X81-X82, X85-X86, X89-X90, X93-X94, X97-X98, where X is valid for all whole numbers from 0 through 9
10: All odd numbers between 1 and 999
If each servant drinks from the bottles listed above, then they will all drink from between 500 and 512 bottles, between 2 and 10 servants will die, and there's a unique combination of dead servants for every possible number of the poison bottle, which allows the poisoned bottle to be pinpointed precisely.
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On May 11 2012 00:09 Zaphod Beeblebrox wrote:Show nested quote +On May 10 2012 06:18 frogrubdown wrote:On May 10 2012 05:11 Misder wrote:Honestly, I'm really surprised that the ping-pong ball problem is still in discussion, and I'm glad actually. Here is the problem again: + Show Spoiler +On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradoxThe question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng. Well, I guess this is what I'm using my 500th post on. + Show Spoiler +
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random
P(B)=0.5 since there are equal number of balls of each color and they are drawn at random.
P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason. + Show Spoiler +
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
Your argument has been presented and refuted repeatedly in this thread. I don't really know what else to say. It matters that the pocket was originally chosen at random, and the fact that we now know with certainty that the drawn ball is orange doesn't change that. If you can express your complaint in the probability calculus, showing my false assumption and demonstrating your result, you should.
Your reasoning leads to absurdity in a number of cases. Suppose there are two pockets, one with a billion orange balls and one with a billion white balls and one orange ball and a third pocket with a billion white balls. You reach into a pocket at random and draw an orange ball. How confident should you be that you drew from the all orange pocket? By your reasoning, 50%. This is because your reasoning has it that the only information you learn is that you drew from one of the two not-all white pockets, and that the previous chances are rendered irrelevant. Don't you understand how absurd this is? If you repeated the trial thousands of times, the answer would always be the all orange pocket, but you claim that it is 50% the mostly white pocket.
Please, just construct the 6 ball version as an experiment and do 30 trials. I guarantee you that the majority of times you have drawn an orange ball first will be times that you drew from the 2 ball pocket.
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On May 11 2012 01:19 DarkPlasmaBall wrote:Show nested quote +On May 11 2012 00:09 Zaphod Beeblebrox wrote:On May 10 2012 06:18 frogrubdown wrote:On May 10 2012 05:11 Misder wrote:Honestly, I'm really surprised that the ping-pong ball problem is still in discussion, and I'm glad actually. Here is the problem again: + Show Spoiler +On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradoxThe question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng. Well, I guess this is what I'm using my 500th post on. + Show Spoiler +
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random
P(B)=0.5 since there are equal number of balls of each color and they are drawn at random.
P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason. + Show Spoiler +
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
I agree with your explanation... + Show Spoiler +But I've also seen a lot of arguing based on the wording of the riddle, rather than the actual mathematics or process behind solving it. Clearly, if we all agreed on what the question was specifically asking, there would probably be no debate
There variants of this type of puzzle that do involve crucial ambiguities, such as the boy or girl paradox, but this simply isn't one of them. Sorry, but your side simply is wrong about how to evaluate the appropriate conditional probabilities and an experiment would demonstrate that after not too many trials.
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On May 11 2012 02:10 frogrubdown wrote:Show nested quote +On May 11 2012 01:19 DarkPlasmaBall wrote:On May 11 2012 00:09 Zaphod Beeblebrox wrote:On May 10 2012 06:18 frogrubdown wrote:On May 10 2012 05:11 Misder wrote:Honestly, I'm really surprised that the ping-pong ball problem is still in discussion, and I'm glad actually. Here is the problem again: + Show Spoiler +On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradoxThe question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng. Well, I guess this is what I'm using my 500th post on. + Show Spoiler +
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random
P(B)=0.5 since there are equal number of balls of each color and they are drawn at random.
P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason. + Show Spoiler +
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
I agree with your explanation... + Show Spoiler +But I've also seen a lot of arguing based on the wording of the riddle, rather than the actual mathematics or process behind solving it. Clearly, if we all agreed on what the question was specifically asking, there would probably be no debate There variants of this type of puzzle that do involve crucial ambiguities, such as the boy or girl paradox, but this simply isn't one of them. Sorry, but your side simply is wrong about how to evaluate the appropriate conditional probabilities and an experiment would demonstrate that after not too many trials.
Nuh-huh, you're wrong! 
But I'll actually defend why I say that 
+ Show Spoiler +Because there aren't any conditional probabilities that you need to work out the math for, and so you don't need to invoke Bayes' theorem at all. You're already holding an orange ball (which technically means the probability of selecting an orange ball on the first try is 1, since it already occurred). Regardless, that event is now in the past.
What that event does tell you, however, is that all pockets with zero orange balls are now out of play (as you can't select an orange ball from a pocket that has two white balls). This leaves the pocket with two orange balls, and the pocket with one white and one orange. The probabilities of selecting a particular orange ball is not an issue, as you are already holding an orange ball. The point is now that you have two pockets, one with the second ball being necessarily white, and the other with the second ball being necessarily orange.
The question is not how you got to the first orange ball, but how you get to the second orange ball.
Therefore, the probability of the second ball being orange is 1/2.
EDIT: It's important to note that...
+ Show Spoiler +There's a difference between saying "What's the probability of pulling 2 orange balls from the same pocket" (given the same pocket set up) and saying "You're already holding an orange ball, what's the probability that the second one from the same pocket is also orange" (given the same pocket set up). The former invokes necessary conditional probability; the latter doesn't require it, and is what the actual question is asking.
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On May 11 2012 02:00 frogrubdown wrote:Show nested quote +On May 11 2012 00:09 Zaphod Beeblebrox wrote:On May 10 2012 06:18 frogrubdown wrote:On May 10 2012 05:11 Misder wrote:Honestly, I'm really surprised that the ping-pong ball problem is still in discussion, and I'm glad actually. Here is the problem again: + Show Spoiler +On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradoxThe question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng. Well, I guess this is what I'm using my 500th post on. + Show Spoiler +
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random
P(B)=0.5 since there are equal number of balls of each color and they are drawn at random.
P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason. + Show Spoiler +
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
Your argument has been presented and refuted repeatedly in this thread. I don't really know what else to say. It matters that the pocket was originally chosen at random, and the fact that we now know with certainty that the drawn ball is orange doesn't change that. If you can express your complaint in the probability calculus, showing my false assumption and demonstrating your result, you should. Your reasoning leads to absurdity in a number of cases. Suppose there are two pockets, one with a billion orange balls and one with a billion white balls and one orange ball and a third pocket with a billion white balls. You reach into a pocket at random and draw an orange ball. How confident should you be that you drew from the all orange pocket? By your reasoning, 50%. This is because your reasoning has it that the only information you learn is that you drew from one of the two not-all white pockets, and that the previous chances are rendered irrelevant. Don't you understand how absurd this is? If you repeated the trial thousands of times, the answer would always be the all orange pocket, but you claim that it is 50% the mostly white pocket. Please, just construct the 6 ball version as an experiment and do 30 trials. I guarantee you that the majority of times you have drawn an orange ball first will be times that you drew from the 2 ball pocket.
It really depends on how you interpret the question. The first way is "What is the probability that the second ball is orange, given that the first ball is orange?" In this scenario, you're right. The second way is "I'm holding an orange ball. What is the probability that the next ball is orange?" In this scenario, it's no longer a case of conditional probability. You're already holding an orange ball. In your exaggerated example, it doesn't matter if there's only a 1/3 probability that you're holding an orange ball on the first try. You're holding it.
The second case that I highlighted in another way. I draw a single ball. If it's white, then fuck this world I'm outta here. If I draw a orange ball, then I ask the question. Now does it feel like it's 1/2?
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On May 11 2012 03:08 Heh_ wrote:Show nested quote +On May 11 2012 02:00 frogrubdown wrote:On May 11 2012 00:09 Zaphod Beeblebrox wrote:On May 10 2012 06:18 frogrubdown wrote:On May 10 2012 05:11 Misder wrote:Honestly, I'm really surprised that the ping-pong ball problem is still in discussion, and I'm glad actually. Here is the problem again: + Show Spoiler +On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradoxThe question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng. Well, I guess this is what I'm using my 500th post on. + Show Spoiler +
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random
P(B)=0.5 since there are equal number of balls of each color and they are drawn at random.
P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
Well, I think you are quite worng here for one particular reason. + Show Spoiler +
The first draw is NOT random. You are given that the draw has already been decided, and that it turned out orange. Therefore the probability of drawing an orange ball at random is not part of this probability calculation and should be ignored.
The information you are given is that you have drawn an orange ball, so you know that the comination of that pocket is either Orange-White or Orange-Orange. Therefore the probability of the other ball being orange is actually 50%, as it is known that you could not possibly have drawn from the White-White pocket.
The riddle boils down to you being told: "This pocket is NOT White-White" - then finding the probability of it being Orange-Orange. When stated like this it should be obvious that the probability is 50%
Your argument has been presented and refuted repeatedly in this thread. I don't really know what else to say. It matters that the pocket was originally chosen at random, and the fact that we now know with certainty that the drawn ball is orange doesn't change that. If you can express your complaint in the probability calculus, showing my false assumption and demonstrating your result, you should. Your reasoning leads to absurdity in a number of cases. Suppose there are two pockets, one with a billion orange balls and one with a billion white balls and one orange ball and a third pocket with a billion white balls. You reach into a pocket at random and draw an orange ball. How confident should you be that you drew from the all orange pocket? By your reasoning, 50%. This is because your reasoning has it that the only information you learn is that you drew from one of the two not-all white pockets, and that the previous chances are rendered irrelevant. Don't you understand how absurd this is? If you repeated the trial thousands of times, the answer would always be the all orange pocket, but you claim that it is 50% the mostly white pocket. Please, just construct the 6 ball version as an experiment and do 30 trials. I guarantee you that the majority of times you have drawn an orange ball first will be times that you drew from the 2 ball pocket. It really depends on how you interpret the question. The first way is "What is the probability that the second ball is orange, given that the first ball is orange?" In this scenario, you're right. The second way is "I'm holding an orange ball. What is the probability that the next ball is orange?" In this scenario, it's no longer a case of conditional probability. You're already holding an orange ball. In your exaggerated example, it doesn't matter if there's only a 1/3 probability that you're holding an orange ball on the first try. You're holding it. The second case that I highlighted in another way. I draw a single ball. If it's white, then fuck this world I'm outta here. If I draw a orange ball, then I ask the question. Now does it feel like it's 1/2?
Just to be clear, you are accepting the ludicrous result that if you draw an orange ball in the billion-ball-pockets case you think that there's 50% chance it came from the pocket with only one orange ball, right? Same question to you darkplasma.
What on earth does it mean to say "I'm holding an orange ball?" Well, where did you get that orange ball? Did you go out to 7-11 and buy it? Well in that case it tells us nothing. Oh, it turns out that you drew it at random from one of the three pockets. Thanks, that gives me added information about which pocket it likely came from because one of them delivers orange balls more frequently than the other and is hence more likely to be the source. Is that so hard to understand?
The only question that 1/2 is the right answer to is the question of what the probability of the other ball being white is given that you are in a pocket with at least one orange ball (with 0 added information!). This is explicitly not the question. The question tells us that the drawing procedure was random, and that tells us additional information about likelihoods that cannot be ignored.
I'll try one more time to turn this into a more common sense example. You're on the other side of an outfield wall and know that your three friends are taking turns batting but don't know which is up. One of them is incapable of hitting the ball over the fence, another hits the ball over the fence every time, and another hits it over the fence once every 100 swings. You see the ball come over the fence (i.e., you are given that it was in fact a home run). Isn't it obvious that this is fantastic evidence that the middle friend was at the plate. You two think it is only evidence that one of the latter two is at the plate. Don't you see how silly that is?
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Sanya12364 Posts
On May 11 2012 03:08 Heh_ wrote:
The second case that I highlighted in another way. I draw a single ball. If it's white, then fuck this world I'm outta here. If I draw a orange ball, then I ask the question. Now does it feel like it's 1/2?
Yikes, this is still going on. The possibility of drawing the white ball from the mixed pocket is precisely why the answer is 2/3. Because you are drawing a ball and could have drawn the white ball but given the fact that you drew an orange ball invokes conditional probability.
If the question did not involve chances of drawing a white ball. For example:
Your friend picks a pocket with at least one orange ball. He then proves it to you by showing a orange ball. What are the chances that the other ball in the pocket is orange?
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On May 10 2012 05:11 Misder wrote:Honestly, I'm really surprised that the ping-pong ball problem is still in discussion, and I'm glad actually. Here is the problem again: + Show Spoiler +On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradoxThe question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Wow looks like a lot of people are still arguing over this problem
Given the wording of this problem though,
The answer is:
+ Show Spoiler +Definitely 1/2
Step-by-step breakdown of the possibilities (O=Orange, W=White):
Step 1:
OO
OW
WW
Step 2 (eliminate WW, since it is IMPOSSIBLE to pick O from WW):
OO
OW
Step 3 (You pick O from one of these bags, so eliminate it):
O
W
Step 4 (Evaluate probability that the other ball is Orange)
p = 0.5 or 1/2
That's about a clear an answer I can provide given the wording of that problem.
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On May 11 2012 03:43 Akari Takai wrote:Time for science. Computer science that is. Show nested quote +On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
My method for testing was this: 1) Make the bag and pocket and balls, and set two counters (trials && success) at 0. 2) Pick a pocket at random. 3) Pick a ball from that pocket at random. 4) If the ball is white, then we can't include this as part of our trials, so don't change the value of trials or successes. 5) If the ball is orange, then this is a valid trial so increase the trial counter by 1. 6) If the second ball in the pocket is also orange, increase the success counter by 1. 7) Do steps 2-6 1,000,000 times 8) Print out the probability as success/trials CODE! + Show Spoiler +
public class Ball {
public String color;
public Ball (String color) { this.color = color; }
public String getColor() { return color; }
}
public class Pocket {
private Ball[] balls;
private java.util.Random generator = new java.util.Random();
private int first;
public Pocket(Ball a, Ball b) {
balls = new Ball[2];
balls[0] = a;
balls[1] = b;
}
public Ball getFirst() {
first = generator.nextInt(2);
return balls[first];
}
public Ball getLast() {
if (first == 0) return balls[1];
return balls[0];
}
}
public class Bag {
private Pocket[] pockets;
private java.util.Random generator = new java.util.Random();
public Bag() {
pockets = new Pocket[3];
pockets[0] = new Pocket(new Ball("white"), new Ball("white"));
pockets[1] = new Pocket(new Ball("white"), new Ball("orange"));
pockets[2] = new Pocket(new Ball("orange"), new Ball("orange"));
}
public int runTrial() {
// returns 0 if no orange ball was drawn first (can't be counted)
// returns 1 if an orange ball was drawn first but a white ball was drawn second
// returns 2 if an orange ball was drawn first and an orange ball was drawn second
int pocketSelect = generator.nextInt(3);
Ball first = pockets[pocketSelect].getFirst();
if (!first.getColor().equals("orange")) return 0;
Ball last = pockets[pocketSelect].getLast();
if (!last.getColor().equals("orange")) return 1;
return 2;
}
}
public class Tester {
public static void main(String[] args) {
Bag bag = new Bag();
int trials = 0;
int orangeOrange = 0;
for(int i=0; i < 1000000; i++) {
int result = bag.runTrial();
if (result == 0) continue;
else if (results == 1) trials++;
else {
trials++; orangeOrange++;
}
}
System.out.println("There were " + trials + " trials and " + orangeOrange + " orange-orange combinations");
System.out.println("The probability is: " + ((double)orangeOrange/(double)trials));
}
}
and RESULTS! + Show Spoiler +
There were 500170 trials and 332801 orange-orange combinations
The probability is: 0.6653757722374393
There were 500559 trials and 333707 orange-orange combinations
The probability is: 0.6666686644331637
There were 500068 trials and 333559 orange-orange combinations
The probability is: 0.6670272842893367
There were 500562 trials and 333987 orange-orange combinations
The probability is: 0.667224040178839
There were 500334 trials and 333134 orange-orange combinations
The probability is: 0.665823230082305
And expanding to 100,000,000 times instead of 1,000,000 gave + Show Spoiler +
There were 49988348 trials and 33328647 orange-orange combinations
The probability is: 0.6667283143663799
I think this speaks for itself.
The flaw in these steps is that our picker is GUARANTEED to pick an orange ball.
In your program, you set it such that there are three distinct orange balls to pick from.
In reality, if you are guaranteed an orange ball, it's the same thing as having a second person look inside the pockets, and give you an orange ball, in which case it doesn't matter which ball you took from the "orange-orange" pocket
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On May 11 2012 04:05 Akari Takai wrote:Show nested quote +On May 11 2012 03:53 jaerak wrote:On May 11 2012 03:43 Akari Takai wrote:Time for science. Computer science that is. On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
My method for testing was this: 1) Make the bag and pocket and balls, and set two counters (trials && success) at 0. 2) Pick a pocket at random. 3) Pick a ball from that pocket at random. 4) If the ball is white, then we can't include this as part of our trials, so don't change the value of trials or successes. 5) If the ball is orange, then this is a valid trial so increase the trial counter by 1. 6) If the second ball in the pocket is also orange, increase the success counter by 1. 7) Do steps 2-6 1,000,000 times 8) Print out the probability as success/trials CODE! + Show Spoiler +
public class Ball {
public String color;
public Ball (String color) { this.color = color; }
public String getColor() { return color; }
}
public class Pocket {
private Ball[] balls;
private java.util.Random generator = new java.util.Random();
private int first;
public Pocket(Ball a, Ball b) {
balls = new Ball[2];
balls[0] = a;
balls[1] = b;
}
public Ball getFirst() {
first = generator.nextInt(2);
return balls[first];
}
public Ball getLast() {
if (first == 0) return balls[1];
return balls[0];
}
}
public class Bag {
private Pocket[] pockets;
private java.util.Random generator = new java.util.Random();
public Bag() {
pockets = new Pocket[3];
pockets[0] = new Pocket(new Ball("white"), new Ball("white"));
pockets[1] = new Pocket(new Ball("white"), new Ball("orange"));
pockets[2] = new Pocket(new Ball("orange"), new Ball("orange"));
}
public int runTrial() {
// returns 0 if no orange ball was drawn first (can't be counted)
// returns 1 if an orange ball was drawn first but a white ball was drawn second
// returns 2 if an orange ball was drawn first and an orange ball was drawn second
int pocketSelect = generator.nextInt(3);
Ball first = pockets[pocketSelect].getFirst();
if (!first.getColor().equals("orange")) return 0;
Ball last = pockets[pocketSelect].getLast();
if (!last.getColor().equals("orange")) return 1;
return 2;
}
}
public class Tester {
public static void main(String[] args) {
Bag bag = new Bag();
int trials = 0;
int orangeOrange = 0;
for(int i=0; i < 1000000; i++) {
int result = bag.runTrial();
if (result == 0) continue;
else if (results == 1) trials++;
else {
trials++; orangeOrange++;
}
}
System.out.println("There were " + trials + " trials and " + orangeOrange + " orange-orange combinations");
System.out.println("The probability is: " + ((double)orangeOrange/(double)trials));
}
}
and RESULTS! + Show Spoiler +
There were 500170 trials and 332801 orange-orange combinations
The probability is: 0.6653757722374393
There were 500559 trials and 333707 orange-orange combinations
The probability is: 0.6666686644331637
There were 500068 trials and 333559 orange-orange combinations
The probability is: 0.6670272842893367
There were 500562 trials and 333987 orange-orange combinations
The probability is: 0.667224040178839
There were 500334 trials and 333134 orange-orange combinations
The probability is: 0.665823230082305
And expanding to 100,000,000 times instead of 1,000,000 gave + Show Spoiler +
There were 49988348 trials and 33328647 orange-orange combinations
The probability is: 0.6667283143663799
I think this speaks for itself. The flaw in these steps is that our picker is GUARANTEED to pick an orange ball. In your program, you set it such that there are three distinct orange balls to pick from. In reality, if you are guaranteed an orange ball, it's the same thing as having a second person look inside the pockets, and give you an orange ball, in which case it doesn't matter which ball you took from the "orange-orange" pocket You don't actually understand the problem. + Show Spoiler +The problem says that we pick a RANDOM pocket (1/3). You can't just throw out a pocket because it includes no orange ball. It's that we picked a random pocket and the first ball we get turns out to be orange. If THAT has happened, what is the probability that our second ball is also orange. The picture below should help. Picture![[image loading]](http://i.imgur.com/DhiBS.png) Also, the problem is written in precisely the same way here: Bertran's box paradox. There are a lot of scholarly papers about this problem exactly as it is written. A quick googling should help you find them.
...How is it random if we are guaranteeing a result -.-
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On May 11 2012 04:09 jaerak wrote:Show nested quote +On May 11 2012 04:05 Akari Takai wrote:On May 11 2012 03:53 jaerak wrote:On May 11 2012 03:43 Akari Takai wrote:Time for science. Computer science that is. On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
My method for testing was this: 1) Make the bag and pocket and balls, and set two counters (trials && success) at 0. 2) Pick a pocket at random. 3) Pick a ball from that pocket at random. 4) If the ball is white, then we can't include this as part of our trials, so don't change the value of trials or successes. 5) If the ball is orange, then this is a valid trial so increase the trial counter by 1. 6) If the second ball in the pocket is also orange, increase the success counter by 1. 7) Do steps 2-6 1,000,000 times 8) Print out the probability as success/trials CODE! + Show Spoiler +
public class Ball {
public String color;
public Ball (String color) { this.color = color; }
public String getColor() { return color; }
}
public class Pocket {
private Ball[] balls;
private java.util.Random generator = new java.util.Random();
private int first;
public Pocket(Ball a, Ball b) {
balls = new Ball[2];
balls[0] = a;
balls[1] = b;
}
public Ball getFirst() {
first = generator.nextInt(2);
return balls[first];
}
public Ball getLast() {
if (first == 0) return balls[1];
return balls[0];
}
}
public class Bag {
private Pocket[] pockets;
private java.util.Random generator = new java.util.Random();
public Bag() {
pockets = new Pocket[3];
pockets[0] = new Pocket(new Ball("white"), new Ball("white"));
pockets[1] = new Pocket(new Ball("white"), new Ball("orange"));
pockets[2] = new Pocket(new Ball("orange"), new Ball("orange"));
}
public int runTrial() {
// returns 0 if no orange ball was drawn first (can't be counted)
// returns 1 if an orange ball was drawn first but a white ball was drawn second
// returns 2 if an orange ball was drawn first and an orange ball was drawn second
int pocketSelect = generator.nextInt(3);
Ball first = pockets[pocketSelect].getFirst();
if (!first.getColor().equals("orange")) return 0;
Ball last = pockets[pocketSelect].getLast();
if (!last.getColor().equals("orange")) return 1;
return 2;
}
}
public class Tester {
public static void main(String[] args) {
Bag bag = new Bag();
int trials = 0;
int orangeOrange = 0;
for(int i=0; i < 1000000; i++) {
int result = bag.runTrial();
if (result == 0) continue;
else if (results == 1) trials++;
else {
trials++; orangeOrange++;
}
}
System.out.println("There were " + trials + " trials and " + orangeOrange + " orange-orange combinations");
System.out.println("The probability is: " + ((double)orangeOrange/(double)trials));
}
}
and RESULTS! + Show Spoiler +
There were 500170 trials and 332801 orange-orange combinations
The probability is: 0.6653757722374393
There were 500559 trials and 333707 orange-orange combinations
The probability is: 0.6666686644331637
There were 500068 trials and 333559 orange-orange combinations
The probability is: 0.6670272842893367
There were 500562 trials and 333987 orange-orange combinations
The probability is: 0.667224040178839
There were 500334 trials and 333134 orange-orange combinations
The probability is: 0.665823230082305
And expanding to 100,000,000 times instead of 1,000,000 gave + Show Spoiler +
There were 49988348 trials and 33328647 orange-orange combinations
The probability is: 0.6667283143663799
I think this speaks for itself. The flaw in these steps is that our picker is GUARANTEED to pick an orange ball. In your program, you set it such that there are three distinct orange balls to pick from. In reality, if you are guaranteed an orange ball, it's the same thing as having a second person look inside the pockets, and give you an orange ball, in which case it doesn't matter which ball you took from the "orange-orange" pocket You don't actually understand the problem. + Show Spoiler +The problem says that we pick a RANDOM pocket (1/3). You can't just throw out a pocket because it includes no orange ball. It's that we picked a random pocket and the first ball we get turns out to be orange. If THAT has happened, what is the probability that our second ball is also orange. The picture below should help. PictureAlso, the problem is written in precisely the same way here: Bertran's box paradox. There are a lot of scholarly papers about this problem exactly as it is written. A quick googling should help you find them. ...How is it random if we are guaranteeing a result -.-
he picked a pocket at random to begin with, looking at the wording of the original problem. You can think of the problem as you have 3 pockets OO, OW, and WW, you randomly pick out an O, what is the probability that the other one is O. This is equivalent to asking you what is the probability of randomly picking the OO pocket (that is the only choice that results in you picking O and the one left over is also O. It can be seen that this probability is obviously 1/3 since there is only one OO pocket out of 3.
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On May 11 2012 04:14 Akari Takai wrote:Show nested quote +On May 11 2012 04:09 jaerak wrote:On May 11 2012 04:05 Akari Takai wrote:On May 11 2012 03:53 jaerak wrote:On May 11 2012 03:43 Akari Takai wrote:Time for science. Computer science that is. On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
My method for testing was this: 1) Make the bag and pocket and balls, and set two counters (trials && success) at 0. 2) Pick a pocket at random. 3) Pick a ball from that pocket at random. 4) If the ball is white, then we can't include this as part of our trials, so don't change the value of trials or successes. 5) If the ball is orange, then this is a valid trial so increase the trial counter by 1. 6) If the second ball in the pocket is also orange, increase the success counter by 1. 7) Do steps 2-6 1,000,000 times 8) Print out the probability as success/trials CODE! + Show Spoiler +
public class Ball {
public String color;
public Ball (String color) { this.color = color; }
public String getColor() { return color; }
}
public class Pocket {
private Ball[] balls;
private java.util.Random generator = new java.util.Random();
private int first;
public Pocket(Ball a, Ball b) {
balls = new Ball[2];
balls[0] = a;
balls[1] = b;
}
public Ball getFirst() {
first = generator.nextInt(2);
return balls[first];
}
public Ball getLast() {
if (first == 0) return balls[1];
return balls[0];
}
}
public class Bag {
private Pocket[] pockets;
private java.util.Random generator = new java.util.Random();
public Bag() {
pockets = new Pocket[3];
pockets[0] = new Pocket(new Ball("white"), new Ball("white"));
pockets[1] = new Pocket(new Ball("white"), new Ball("orange"));
pockets[2] = new Pocket(new Ball("orange"), new Ball("orange"));
}
public int runTrial() {
// returns 0 if no orange ball was drawn first (can't be counted)
// returns 1 if an orange ball was drawn first but a white ball was drawn second
// returns 2 if an orange ball was drawn first and an orange ball was drawn second
int pocketSelect = generator.nextInt(3);
Ball first = pockets[pocketSelect].getFirst();
if (!first.getColor().equals("orange")) return 0;
Ball last = pockets[pocketSelect].getLast();
if (!last.getColor().equals("orange")) return 1;
return 2;
}
}
public class Tester {
public static void main(String[] args) {
Bag bag = new Bag();
int trials = 0;
int orangeOrange = 0;
for(int i=0; i < 1000000; i++) {
int result = bag.runTrial();
if (result == 0) continue;
else if (results == 1) trials++;
else {
trials++; orangeOrange++;
}
}
System.out.println("There were " + trials + " trials and " + orangeOrange + " orange-orange combinations");
System.out.println("The probability is: " + ((double)orangeOrange/(double)trials));
}
}
and RESULTS! + Show Spoiler +
There were 500170 trials and 332801 orange-orange combinations
The probability is: 0.6653757722374393
There were 500559 trials and 333707 orange-orange combinations
The probability is: 0.6666686644331637
There were 500068 trials and 333559 orange-orange combinations
The probability is: 0.6670272842893367
There were 500562 trials and 333987 orange-orange combinations
The probability is: 0.667224040178839
There were 500334 trials and 333134 orange-orange combinations
The probability is: 0.665823230082305
And expanding to 100,000,000 times instead of 1,000,000 gave + Show Spoiler +
There were 49988348 trials and 33328647 orange-orange combinations
The probability is: 0.6667283143663799
I think this speaks for itself. The flaw in these steps is that our picker is GUARANTEED to pick an orange ball. In your program, you set it such that there are three distinct orange balls to pick from. In reality, if you are guaranteed an orange ball, it's the same thing as having a second person look inside the pockets, and give you an orange ball, in which case it doesn't matter which ball you took from the "orange-orange" pocket You don't actually understand the problem. + Show Spoiler +The problem says that we pick a RANDOM pocket (1/3). You can't just throw out a pocket because it includes no orange ball. It's that we picked a random pocket and the first ball we get turns out to be orange. If THAT has happened, what is the probability that our second ball is also orange. The picture below should help. PictureAlso, the problem is written in precisely the same way here: Bertran's box paradox. There are a lot of scholarly papers about this problem exactly as it is written. A quick googling should help you find them. ...How is it random if we are guaranteeing a result -.- It doesn't say guaranteed anywhere. You reach into one random pocket and pull out an orange ping pong ball. You reach into one RANDOM pocket and [happen to] pull out an orange ping pong ball. [Now that you have done that,] what's the probability of the other ball in the pocket being orange?
It's all in the grammar. As I've already said, there's two ways to interpret it. The first way is the conditional probability thing, which accounts for the probability that the first ball is orange, which is 2/3.
There's another way to visualize the second argument. The three bags remain the same. One person LOOKS at the contents of the bags, and pulls out one orange ball. Now what is the probability that the other ball is orange? 1/2.
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It's really not all in the grammar. you pull and orange first so that is given. there are three orange balls. 2 of the 3 are in the same pocket so the odds that you picked the OO pocket are 2/3. if you picked that pocket the other ball will be orange. The answer to this question is 2/3 regardless of how crazily you read the problem
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EPT
