EPTI think this is a poorly worded riddle with an ambigous solotion that should have been 2/3, but the wording supports 1/2 as a more "correct" answer.
[SFW] Riddles / Puzzles / Brain Teasers - Page 25
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Zaphod Beeblebrox
Denmark697 Posts
I think this is a poorly worded riddle with an ambigous solotion that should have been 2/3, but the wording supports 1/2 as a more "correct" answer. | ||
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frogrubdown
1266 Posts
On May 11 2012 04:43 Heh_ wrote: 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. On May 11 2012 05:09 Zaphod Beeblebrox wrote: Agreed - Two ways to read this puzzle, one gives 1/2 the other 2/3. I would say 1/2 because the possibility of you picking a white ball first is not even mentioned. BUT this is a riddle, and it is too simple math to say 1/2, while the 2/3 case actually requires thinking and considering more than: "well there are only two options". I think this is a poorly worded riddle with an ambigous solotion that should have been 2/3, but the wording supports 1/2 as a more "correct" answer. No offense, but did either of you read the original problem? Yes, there are two ways the problem could have been asked. It could have been asked such that it was possible that your friend did not get the orange ball using a random procedure but instead guaranteed it, and in that case (if you suitably fill it out) the answer will be 1/2. The problem is that this is unambiguously not what was asked. The question unambiguously stated that the first orange ball was drawn at random. This explicitly rules out the scenario that you need to make 1/2 correct. There may have been an ambiguity in your head, but there was none in the question. | ||
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kittensrcute
United States617 Posts
OO WW OW 2/3 chance you get the Orange Orange pocket. So there's a 2/3 chance you get the Orange. Edit: Sorry, I wasn't "low" enough the first time through. Derp, downs moment. | ||
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pandabearguy
United States252 Posts
On May 11 2012 05:21 kittensrcute wrote: How are you guys even arguing? OO OW WW If the first ball is orange, the next one is either W or O. If you picked OO, you have a 100% chance of getting O. If you picked OW, you have a 100% chance of getting W. You're not switching pockets, so how could you possibly think that you could have a 2/3 chance of getting O. "how are you guys even arguing?" :: proceeds to answer question wrong:: | ||
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pandabearguy
United States252 Posts
i get how this can be confusing but that's how it works | ||
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Lounge
537 Posts
Instead of orange and white balls let's replace them with money. Gray and brown coins. (For you non-Americans all coins are gray colored except for pennies, which are brown.) One pocket has two gray coins. A quarter and a dime. One pocket has 1 gray coin and a brown coin. A nickel and a penny One pocket has 2 brown coins. 2 pennies. You pick a coin out of one of the pockets and it's gray. The reason I'm not stating the value of the coin is because you have to treat each orange ball as a separate entity, likewise each gray coin as a separate entity. Here are your possibilities: You drew a quarter from the quarter/dime pair. You drew a dime from the dime/quarter You drew a nickel from the nickel/penny pair. What are the odds that the other coin is gray? 2/3rds of these possibilities will be a gray coin. These three possibilities are eliminated: You drew a penny from the nickel/penny pair. You drew a penny from the penny/penny pair. You drew the OTHER penny from the penny/penny pair. | ||
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Slithe
United States985 Posts
BTW the answer is 2/3. | ||
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zaikantos
Netherlands43 Posts
![[image loading]](http://img600.imageshack.us/img600/1813/riddle.png) | ||
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Slithe
United States985 Posts
On May 11 2012 07:07 zaikantos wrote: Perhaps a short drawing can help explain why I think it is a 50% chance. Your assumption that "because a and b are in the same pocket we can say you picked a or c" is your mistake. You say there are 2 possibilities, but there are actually 3 possibilities. You picked a, so the other ball is b. You picked b, so the other ball is a. You picked c, so the other ball is d. 2/3 | ||
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CptZouglou
France146 Posts
You are a russian bomber, and you want to send a bomb on a nuclear submarine. The submarine is located on a line and has an integer position (can be negative). It moves at a constant integer speed (can be negative too) each second. The good part is that you have an unlimited amount of bombs, and you can send one each second at a any position. The problem is that you don't know where is the submarine at time 0, and what its speed is. Is there a way to ensure that you can send hit the submarine in a finite amount of time ? If yes, what is your strategy ? hint: + Show Spoiler + Yes! Consider the problem as finding two parameters initial position (p) and speed (s). After t seconds, the submarine is at position p + t*s. Try and find a way to explore all possible initial positions and speeds ! | ||
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zaikantos
Netherlands43 Posts
On May 11 2012 07:13 Slithe wrote: Your assumption that "because a and b are in the same pocket we can say you picked a or c" is your mistake. You say there are 2 possibilities, but there are actually 3 possibilities. You picked a, so the other ball is b. You picked b, so the other ball is a. You picked c, so the other ball is d. 2/3 That's what I thought untill i looked closer at the wording of the riddle. 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? You do not pick a random orange ball, which would lead to your 3 possibilites and a 2/3rd chance, but you reach into a random pocket. If we label the two pockets with at least one orange ball X and Y, ![[image loading]](http://img823.imageshack.us/img823/6989/pocketxory.png) You then "reach into one random pocket and pull out an orange ping pong ball". You pick one of the two pockets, X or Y. Let's say X has two orange balls, and Y has one orange ball and one white ball. ![[image loading]](http://img819.imageshack.us/img819/212/pocketxaaandyab.png) Two options: You picked pocket X and the second ball is also orange. You picked pocket Y and the second ball is white. 50% or ½. If the wording came down to "you pick a random orange ball" instead of "you pick a random pocket and take a orange ball out", it would indeed be 2/3. | ||
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CyDe
United States1010 Posts
 A bridge has a weight limit of 10,000 KG. A truck weighing exactly 10,000 KG drives onto the bridge, and stops at the center, where a bird weighing 30g lands on it. Why doesn't the bridge collapse? Hint: + Show Spoiler + It has nothing to do with the manner that the bridge was constructed (support placements, etc). Answer: + Show Spoiler + The truck would have burned more than 30g of gasoline by the time it reached the center of the bridge. | ||
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AndyGB4
Canada156 Posts
I think it's quite simple, tree pockets O1+O2, O3+W1, W2+W3. You drew an O for sure, so either you drew: - O1 which means your next ball is O2 (Success!) - O2 which means your next ball is O1 (Success!) - O3 which means your next ball is W1 (failure) it's 2/3, there's really no other way... if the both pockets had same number of Oranges in them, then it would 50% cuz they have the same probabilities of being picked, but in this case there's more chances of u getting a ball from the OO pocket than the OW pocket. | ||
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zaikantos
Netherlands43 Posts
On May 11 2012 23:30 AndyGB4 wrote: wow, Last time i commented it seemed like we were done with the ping pong balls.. originally I thought 1/2, but I realized shortly after I was wrong. I think it's quite simple, tree pockets O1+O2, O3+W1, W2+W3. You drew an O for sure, so either you drew: - O1 which means your next ball is O2 (Success!) - O2 which means your next ball is O1 (Success!) - O3 which means your next ball is W1 (failure) it's 2/3, there's really no other way... if the both pockets had same number of Oranges in them, then it would 50% cuz they have the same probabilities of being picked, but in this case there's more chances of u getting a ball from the OO pocket than the OW pocket. Please read my last post: + Show Spoiler + On May 11 2012 21:22 zaikantos wrote: That's what I thought untill i looked closer at the wording of the riddle. You do not pick a random orange ball, which would lead to your 3 possibilites and a 2/3rd chance, but you reach into a random pocket. If we label the two pockets with at least one orange ball X and Y, You then "reach into one random pocket and pull out an orange ping pong ball". You pick one of the two pockets, X or Y. Let's say X has two orange balls, and Y has one orange ball and one white ball. Two options: You picked pocket X and the second ball is also orange. You picked pocket Y and the second ball is white. 50% or ½. If the wording came down to "you pick a random orange ball" instead of "you pick a random pocket and take a orange ball out", it would indeed be 2/3. | ||
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AndyGB4
Canada156 Posts
On May 11 2012 23:32 zaikantos wrote: Please read my last post: + Show Spoiler + On May 11 2012 21:22 zaikantos wrote: That's what I thought untill i looked closer at the wording of the riddle. You do not pick a random orange ball, which would lead to your 3 possibilites and a 2/3rd chance, but you reach into a random pocket. If we label the two pockets with at least one orange ball X and Y, You then "reach into one random pocket and pull out an orange ping pong ball". You pick one of the two pockets, X or Y. Let's say X has two orange balls, and Y has one orange ball and one white ball. Two options: You picked pocket X and the second ball is also orange. You picked pocket Y and the second ball is white. 50% or ½. If the wording came down to "you pick a random orange ball" instead of "you pick a random pocket and take a orange ball out", it would indeed be 2/3. I read your last post before posting mine, and I disagree with your reasoning. You can't just say its 50% because they say "you reach into a random pocket"... The fact that you grabbed an Orange ball AFTER picking a random pocket affects the probability of which pocket you picked at random. Here are some perfectly good examples to prove my point: If the two pockets had 99 balls in each: P1 = 99 Orange balls P2 = 1 Orange ball, 98 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 99% chance you drew your ball from P1, There's a 1% chance you drew your ball from P2, therefore a 99%(99/100) chance the next ball will also be Orange. and a 1%(1/100) chance the next ball will be White. If the two pockets had 9 balls in each: P1 = 9 Orange balls P2 = 1 Orange ball, 8 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 90% chance you drew your ball from P1, There's a 10% chance you drew your ball from P2, therefore a 90%(9/10) chance the next ball will also be Orange. and a 10%(1/10) chance the next ball will be White. If the two pockets had 4 balls in each: P1 = 4 Orange balls P2 = 1 Orange ball, 3 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 80% chance you drew your ball from P1, There's a 20% chance you drew your ball from P2, therefore a 80%(4/5) chance the next ball will also be Orange. and a 20%(1/5) chance the next ball will be White. If the two pockets had 3 balls in each: P1 = 3 Orange balls P2 = 1 Orange ball, 2 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 75% chance you drew your ball from P1, There's a 25% chance you drew your ball from P2, therefore a 75%(3/4) chance the next ball will also be Orange. and a 25%(1/4) chance the next ball will be White. *** And this is the ACTUAL riddle: *** If the two pockets had 2 balls in each: P1 = 2 Orange balls P2 = 1 Orange ball, 1 White ball And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 66% chance you drew your ball from P1, There's a 33% chance you drew your ball from P2, therefore a 66%(2/3) chance the next ball will also be Orange. and a 33%(1/3) chance the next ball will be White. Under your logic from earlier, all these examples would still be 1/2 to you, correct? Don't you see how silly that logic looks on a larger scale though? Honestly, for anyone who still believes its 1/2, please try it out yourself, recreate this riddle with real objects. Take bottle caps, or pencils, or whatever, and split them up like in the problem, and record each draw. You will see that about 66% of the time, you will draw another Orange ball. The more times you do it, the clearer the probability will be. PLEASEEEEEE try it out | ||
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zaikantos
Netherlands43 Posts
On May 12 2012 00:02 AndyGB4 wrote: I read your last post before posting mine, and I disagree with your reasoning. You can't just say its 50% because they say "you reach into a random pocket"... The fact that you grabbed an Orange ball AFTER picking a random pocket affects the probability of which pocket you picked at random. Here are some perfectly good examples to prove my point: If the two pockets had 99 balls in each: P1 = 99 Orange balls P2 = 1 Orange ball, 98 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 99% chance you drew your ball from P1, There's a 1% chance you drew your ball from P2, therefore a 99%(99/100) chance the next ball will also be Orange. and a 1%(1/100) chance the next ball will be White. If the two pockets had 9 balls in each: P1 = 9 Orange balls P2 = 1 Orange ball, 8 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 90% chance you drew your ball from P1, There's a 10% chance you drew your ball from P2, therefore a 90%(9/10) chance the next ball will also be Orange. and a 10%(1/10) chance the next ball will be White. If the two pockets had 4 balls in each: P1 = 4 Orange balls P2 = 1 Orange ball, 3 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 80% chance you drew your ball from P1, There's a 20% chance you drew your ball from P2, therefore a 80%(4/5) chance the next ball will also be Orange. and a 20%(1/5) chance the next ball will be White. If the two pockets had 3 balls in each: P1 = 3 Orange balls P2 = 1 Orange ball, 2 White balls And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 75% chance you drew your ball from P1, There's a 25% chance you drew your ball from P2, therefore a 75%(3/4) chance the next ball will also be Orange. and a 25%(1/4) chance the next ball will be White. *** And this is the ACTUAL riddle: *** If the two pockets had 2 balls in each: P1 = 2 Orange balls P2 = 1 Orange ball, 1 White ball And you drew an Orange Ball from a RANDOM pocket, what are the chances your next ball is Orange? There's a 66% chance you drew your ball from P1, There's a 33% chance you drew your ball from P2, therefore a 66%(2/3) chance the next ball will also be Orange. and a 33%(1/3) chance the next ball will be White. Under your logic from earlier, all these examples would still be 1/2 to you, correct? Don't you see how silly that logic looks on a larger scale though? Honestly, for anyone who still believes its 1/2, please try it out yourself, recreate this riddle with real objects. Take bottle caps, or pencils, or whatever, and split them up like in the problem, and record each draw. You will see that about 66% of the time, you will draw another Orange ball. The more times you do it, the clearer the probability will be. PLEASEEEEEE try it out You didn't understand what I meant. You do not draw a random orange ball, you open a random pocket and take a orange ball out. 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? If the two pockets had 2 balls in each: P1 = 2 Orange balls P2 = 1 Orange ball, 1 White ball If you randomly open one of the two pockets and take an orange ball out, what are the chances your next ball is Orange? There's a 50% chance you drew your ball from P1, There's a 50% chance you drew your ball from P2, therefore a 50%(1/2) chance the next ball will also be Orange. and a 50%(1/2) chance the next ball will be White. | ||
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AndyGB4
Canada156 Posts
You didn't understand what I meant. You do not draw a random orange ball, you open a random pocket and take a orange ball out. If the two pockets had 2 balls in each: P1 = 2 Orange balls P2 = 1 Orange ball, 1 White ball If you randomly open one of the two pockets and take an orange ball out, what are the chances your next ball is Orange? There's a 50% chance you drew your ball from P1, There's a 50% chance you drew your ball from P2, therefore a 50%(1/2) chance the next ball will also be Orange. and a 50%(1/2) chance the next ball will be White. I understand what you mean, but the fact that you pulled out an orange ball gives you some info on what Pocket you may have picked. here's another way to look at it: You know how we both agree that the WW pocket is out of the question? WHY? because we drew an orange ball. So you NEED to take into account the orange ball that was picked. So basically this is what it looks like.. Since we KNOW that we drew an orange ball from our Random Pocket, and there's 3 orange balls in all: P1 = 2 orange balls = (2/3 orange balls) = 66% P2 = 1 orange ball = (1/3 orange balls) = 33% P3 = 0 orange balls = (0/3 orange balls) = 0% P3 is 0% because it has no orange balls. thats why we rule it out. The fact that P1 has more orange balls in it, gives it better odds of being that pocket chosen than being P2. Ok let's try something else. If we start by ONLY opening a pocket at random (dont choose a ball yet). (since you said we're just picking a pocket at random) This means there's a 1/3 chance it can be P1, P2 and P3 respectively, correct? Ok I think we all agree on that. Now once inside that random pocket chosen, you pulled out an Orange Ball. There are 3 possible orange balls that you may now be holding. You may have picked P1 and got O1, You may have picked P1 and got O2, You may have picked P2 and got O3. There are 2 events here, (1) You picked a pocket. (2) You drew an orange ball. Both events must be taken into consideration. Do you see what I mean? you can't just ignore the fact that you pulled an orange ball after you picked a pocket. If you want, here are ALL the possibilities BEFORE picking a ball. (With possible outcomes from that pocket after picking a ball) You picked P1 - and picked ball O1. - and picked ball O2. You picked P2 - and picked ball O3. - and picked ball W1. You picked P3 - and picked ball W2. - and picked ball W3. Now obviously any chances of picking a white ball are out of the question because we KNOW we got an orange ball, right? So all that remains is this: You picked P1 - and picked ball O1. - and picked ball O2. You picked P2 - and picked ball O3. Picking an Orange ball is the info we have to figure out which pocket we're in. You can't ignore the possibilites listed. The only way the pockets are all equal is if we DIDNT pick a ball yet, but we did, and it was Orange, so that tells us that theres 2 chances that ball in our hands came from P1, and 1 chance it came from P2. Oh I just thought of another example I really like (Sorry I'm writing so much! my mind just gets going lol): List the balls as A,B,C,D,E,F and they're grouped, as usual, like so: A+B, C+D, E+F. If you picked a random pocket and i tell you the ball in your hands is either A, B or C, (which is like our Orange/White problem above) Do you still think that there's a 50/50 chance of being in P1 and P2? I'm actually really curious to hear your answer to this last one. | ||
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Gnosis
Scotland912 Posts
On April 18 2012 22:47 yeaR wrote: If Pinochio said my nose will grow right now, what would happen ? + Show Spoiler + Nothing will happen regardless of how this poorly punctuated sentence is understood. The difference comes in the form of the distinction between a falsehood and a lie. A falsehood may be a belief one sincerely believes is true, such as an individual of the middle ages stating the falsehood: "the sun revolves around the earth". Pinochio is ignorant in much the same way, as he has no knowledge of whether or not my own or his nose will grow "right now". The statement is one of a falsehood proclaimed in ignorance, not a lie spoken in the face of what is known to be the truth. In short, Pinochio isn't lying. | ||
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zaikantos
Netherlands43 Posts
On May 12 2012 00:59 AndyGB4 wrote: Ok let's try something else. If we start by ONLY opening a pocket at random (dont choose a ball yet). (since you said we're just picking a pocket at random) This means there's a 1/3 chance it can be P1, P2 and P3 respectively, correct? Ok I think we all agree on that. Now once inside that random pocket chosen, you pulled out an Orange Ball. There are 3 possible orange balls that you may now be holding. You may have picked P1 and got O1, You may have picked P1 and got O2, You may have picked P2 and got O3. There are 2 events here, (1) You picked a pocket. (2) You drew an orange ball. Both events must be taken into consideration. Do you see what I mean? you can't just ignore the fact that you pulled an orange ball after you picked a pocket. If you want, here are ALL the possibilities BEFORE picking a ball. (With possible outcomes from that pocket after picking a ball) You picked P1 - and picked ball O1. - and picked ball O2. You picked P2 - and picked ball O3. - and picked ball W1. You picked P3 - and picked ball W2. - and picked ball W3. Now obviously any chances of picking a white ball are out of the question because we KNOW we got an orange ball, right? So all that remains is this: You picked P1 - and picked ball O1. - and picked ball O2. You picked P2 - and picked ball O3. Picking an Orange ball is the info we have to figure out which pocket we're in. You can't ignore the possibilites listed. The only way the pockets are all equal is if we DIDNT pick a ball yet, but we did, and it was Orange, so that tells us that theres 2 chances that ball in our hands came from P1, and 1 chance it came from P2. Oh I just thought of another example I really like (Sorry I'm writing so much! my mind just gets going lol): List the balls as A,B,C,D,E,F and they're grouped, as usual, like so: A+B, C+D, E+F. If you picked a random pocket and i tell you the ball in your hands is either A, B or C, (which is like our Orange/White problem above) Do you still think that there's a 50/50 chance of being in P1 and P2? I'm actually really curious to hear your answer to this last one. I left the first part of your post out, as that is just the same as before. My view: We can simplify the question to "are there two orange balls in the pocket you picked". You first pick one of two pockets without needing knowledge of there even being balls in there. This will give you a 50% chance of picking either pocket. P1 gives you the correct answer, both orange P2 gives you the wrong answer, one orange one white. 50/50. Your explanation of why there is a 2/3 chance of picking P1 comes down to once again, picking a random orange ball and not picking a random pocket and THEN taking orange out of that pocket. List the balls as A,B,C,D,E,F and they're grouped, as usual, like so: A+B, C+D, E+F. If you picked a random pocket and i tell you the ball in your hands is either A, B or C, (which is like our Orange/White problem above) Do you still think that there's a 50/50 chance of being in P1 and P2? I'm actually really curious to hear your answer to this last one. I don't think that this is the same as our situation. In your example, we start with P1:AB P2:CD P3:EF and then rule out P3:EF leaving us with P1:AB and P2:CD. Then we pick a A, B or C at random, this means that instead of picking a random pocket like in our riddle, you pick a random orange ball. In our riddle, we start with P1:OO P2:OW P3: WW and then also rule out P3 leaving us with P1:OO P2:OW. Then we randomly pick one of these two pockets, and see if both balls are orange. Like I said before, picking a random orange ball means a 66% chance of the other being orange as well, but picking a random pocket that has at least one orange ball means a 50% chance of the other being orange, and I read the riddle as: + Show Spoiler [Original riddle] + 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 version: A friend packed four ping pong balls for you, 3 orange, 1 white. He's placed them in two pockets of your sports bag. One pocket has two orange balls, the other has one white one orange. You pick a random pocket and take out an orange ping pong ball. What's the probability of the pocket having two orange balls? | ||
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AndyGB4
Canada156 Posts
But I dont understand how you can use the info of having an orange ball to rule out the WW pocket, but you will not use it to calculate the probability of which pocket was picked... Before you post another post, can you PLEASE try it in person. take ping pong balls and try it. If you dont have ping pong balls, take pieces of paper, it doesnt really matter. And post your results here when you try. The results won't lie. you will pick from the OO pocket more often than the OW pocket. This isn't a word play riddle. This is a logical riddle, a mathematical riddle if you will. And for the record, the ABCDEF example can still be exactly like our riddle now, you just changed the wording of it. let me merge the 2 examples and you'll see they're still the same: In our riddle, we start with P1:AB P2:CD P3: EF and then also rule out P3 leaving us with P1:AB P2:CD. Then we randomly pick one of these two pockets, and see if both balls AB or CD. Because think about it, each ball is it's own object, just like each letter in this case. And when you draw an orange ball, we don't know which ball it is, correct? all we know is that it's orange. So like in the case above, all we know is that its etiher A, B or C. (just like its either O1, O2, O3 but we don't know which) You just twisted the words around. Because in the end, were just trying to find out if its pocket AB or pocket CD. Both problems are the same (like you said before, -. But that's not that important i guess. I disagree with this following part of your answer. In our riddle, we start with P1:OO P2:OW P3: WW and then also rule out P3 leaving us with P1:OO P2:OW. Then we randomly pick one of these two pockets, and see if both balls are orange. How could you rule out P3 if you haven't picked a ball yet? The only way your 50% would be correct was if the question was if we did not pick a ball at all. It would have to be "There are 2 pockets (OO,OW) Pick one of the pockets. Whats the probability of getting OO pocket." That would be 50%. But the fact that you draw an Orange ball after you choose a Pocket in our real example, means there are 2 chances it was from OO and 1 chance it was from OW. You just can't ignore the orange ball. We are NOT just picking a pocket. here's a diagram I drew up showing our problem, let me know if you agree with it: ![[image loading]](http://i48.tinypic.com/jqmb7r.jpg) And here's my solution, going through it step by step, starting by picking a random pocket, like you advise. ![[image loading]](http://i48.tinypic.com/24brib7.jpg) You simply can't ignore the fact that AFTER picking a random pocket, you PULL a ball from it, and that ball happens to be orange. Theres a 2/3 chance you're in Pocket 1 and a 1/3 chance you're in Pocket 2, because Pocket 1 has more orange balls that Pocket 2. Answer me this as well, If one pocket has 100 orange balls, and the other pocket has 1 orange ball and 99 white balls. And then you pick a random pocket, and then draw an Orange ball, is it still a 1/2 chance that your next ball is Orange? EDIT: And I found the flaw in your logic from an image you posted earlier on. (I also made my post a bit cleaner with quotes) ![[image loading]](http://i48.tinypic.com/2hzn85h.png) | ||
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![[image loading]](http://i.imgur.com/DhiBS.png)
