EPTOn July 22 2007 14:18 MyLostTemple wrote:
i suck at math : [
i suck at math : [
That's why you have a brother.
can someone solve this math riddle for me - Page 4
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nemY
United States3119 Posts
That's why you have a brother. | ||
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thoraxe
United States1449 Posts
On July 22 2007 14:00 Gatsu wrote: It s not 27$ + 2$ = 29$ but 27$ (what they pay) - 2$ (what was stolen) = 25$ (the real price of the room) this is about right. You don't add the 2 cuz they were stolen i.e. subtract | ||
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niteReloaded
Croatia5281 Posts
You have 3 boxes, one of them is the right box. Still, you have only 2 options: - Pick the right box - 33% chance. - pick the wrong box - 66% chance. Lets now analyze the odds for the switching strategy. If you pick the right box, the host will remove one of the false boxes, you will switch and pick the remaining wrong box. As mentioned, this will happen whenever you pick the right box at the start. However, if you initially pick the wrong box (66% chance), 2 boxes will remain, with one of them being the right box. Out of these 2 boxes, the host will remove the wrong one, and when you switch, u score the right box. This scenario occurs whenever your first choice is the wrong box, which happens in 66% of the cases. | ||
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Sr18
Netherlands1141 Posts
On July 23 2007 11:56 Chill wrote: Counter example: Monty Hall with 10 doors. You can switch or not switch. Don't switch = 10% chance to win. By your explination above, switch should = 90%. Switching would be 90%, if the host eliminates all but one of the doors you didn´t pick. | ||
Chill
Calgary25980 Posts
On July 23 2007 16:52 Sr18 wrote: Switching would be 90%, if the host eliminates all but one of the doors you didn´t pick. Right. | ||
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CharlieMurphy
United States22895 Posts
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mikeymoo
Canada7170 Posts
On July 23 2007 17:17 CharlieMurphy wrote: If your odds start at 33.3333% to fail on 3 doors, they obviously get better on 2 doors (50%) but if you add them both together its 83.333% chance to fail. ...wrong. It's the ORDER that the things happen in. You pick a door. Let's say it's the wrong door. There are two left, one of which is the correct one, the other, he eliminates. So switching after picking the wrong door is 100% to become the correct one. So if you have a 2/3 of picking the wrong door initially, you have 2/3 of a chance to win after switching. | ||
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oneofthem
Cayman Islands24199 Posts
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JeeJee
Canada5652 Posts
you only lose by switching if you picked the right box to begin with. what are the odds of picking the right box to begin with? that's usually enough =p | ||
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[r]h_probe
United States188 Posts
On July 23 2007 17:17 CharlieMurphy wrote: If your odds start at 33.3333% to fail on 3 doors, they obviously get better on 2 doors (50%) but if you add them both together its 83.333% chance to fail. LOL | ||
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Frits
11782 Posts
On July 23 2007 11:56 Chill wrote: Counter example: Monty Hall with 10 doors. You can switch or not switch. Don't switch = 10% chance to win. By your explination above, switch should = 90%. Nope it still counts, what you forgot is when adding doors, obviously you have to divide the chance by that added amount. Choice 1 will be 10% no matter what they do like I said right? And there's still sure to be a price so: 100 - 10 = 90 % chance of ONE of the 8 doors left being the right one. 90/8 = 11,25% of getting it right when switching. On July 23 2007 17:17 CharlieMurphy wrote: If your odds start at 33.3333% to fail on 3 doors, they obviously get better on 2 doors (50%) but if you add them both together its 83.333% chance to fail. What are you even calculating here? The question is should he switch or not. I don't understand what you did here at all. :p | ||
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CharlieMurphy
United States22895 Posts
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Frits
11782 Posts
On July 23 2007 20:36 CharlieMurphy wrote: Maybe I'm not sure how the show works but what I meant is: The first choice you make between 3 doors is 66% chance to win and a 33% chance to lose out of 3 doors. If you get a winner the first time and try to get the other winner which is now a 50/50 shot you've just made your odds worse because essentially its like you picked 2 doors at once, Which is going to be a fail at 83% of the time, correct? or is it 66%? I still have no idea what you mean. Here's how the show works: -3 doors, 1 prize (so 33% of getting it right if picking a random). -You pick 1 door. -The host removes 1 of the remaining 2 doors, and you know he removed the one which isn't the one with a prize. Now the question is this: Will he have more chance of winning if he switches or not. answer: Yes he should, 66% of the other one being right, 33% of getting it right if he keeps the original pick | ||
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[r]h_probe
United States188 Posts
On July 23 2007 20:17 Frits wrote: Nope it still counts, what you forgot is when adding doors, obviously you have to divide the chance by that added amount. Choice 1 will be 10% no matter what they do like I said right? And there's still sure to be a price so: 100 - 10 = 90 % chance of ONE of the 8 doors left being the right one. 90/8 = 11,25% of getting it right when switching. Wrong! Look at it this way: 10 doors You choose one door ----> 1/10 chance of getting it right no matter what! Out of the 9 other doors the host reveals 8 So essentially you are either choosing 1 of two groups: The group of one or the group of 9 (of which 8 have been revealed) Therefore it is a 90% prob of getting it right if you switch! There is no "dividing the chance by that added mount"! To further illustrate my point, lets make the problem even bigger: say you to choose a number between 1 in a million. There is 1 random correct number. Then someone (who obviously knows what the number is) says it is either your number or it is a different one they tell you. You would have to be absolutely insane to stick with your number because it is a 1/1000000 shot no matter what where as the other number is 999999/1000000 | ||
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oneofthem
Cayman Islands24199 Posts
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Cambium
United States16368 Posts
On July 22 2007 13:53 zizou21 wrote: ok i need someone to break this down and explain it to me because i am retarded -3 guys go to a hotel which is 30 bucks a night, so they each pay 10 -the clerk forgets that its a special night or some bullshit and its only $25 a nyte -he gives the bellboy $5 to give it to the 3 dudes which r prob renting a room for hot gay sex but thats irrelevant -the bellboy rips them off and pockets 2$ and gives 1$ to each one OK so they each payed 10$ at first but got 1$ back each ($9 x 3 = $27). the bellboy pocketed 2$ 27 + 2 = 29! o noez wheres the missing dollar? anyway i kno theres some kind of faulty calculation in there, but i just dont remember where and since im an idiot i cant figure it out. dont try to explain this to me by solving it backwards. I KNOW THERES NO MISSING DOLLAR, i just want to understand where and how the riddle tricks the reader i know tl.net is here for my problems I don't get why this makes sense... Ok, they paid 9$ each, so they are out 27 dollars. The bellboy takes 2$. 27-2=25, which is how much the room costs... | ||
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Cambium
United States16368 Posts
On July 23 2007 20:43 Frits wrote: I still have no idea what you mean. Here's how the show works: -3 doors, 1 prize (so 33% of getting it right if picking a random). -You pick 1 door. -The host removes 1 of the remaining 2 doors, and you know he removed the one which isn't the one with a prize. Now the question is this: Will he have more chance of winning if he switches or not. answer: Yes he should, 66% of the other one being right, 33% of getting it right if he keeps the original pick Nice how this turned into a Monty Hall problem. I didn't really understand this problem until I read "Curious Incident of a Dog at a Nighttime" (or something like that). IIRC, they used pictures to illustrate the problem, and it really cleared things up. Like someone already mentioned above, if you switch, you'll win 66% of the time. For instance, the boxes are: L L W You choose 1st box, host removes 2nd box, you switch, you win. You choose 2nd box, host removes 1st box, you switch, you win. You choose 3rd box, host removes either 1st or 2nd box, you switch, you lose. So you win 2/3 of the time, and you lose 1/3 of the time. Whereas if you don't switch L L W You choose 1st box, host removes 2nd box, you lose. You choose 2nd box, host removes 1st box, you lose. You choose 3rd box, host removes 1st or 2nd box, you win. You lost 2/3 of the time, and you win 1/3 of the time. By switching, you double your chance of winning. | ||
Hot_Bid
Braavos36375 Posts
you pick door #1 host eliminates doors #2-99 would you switch to door #100 or would you stick with #1? | ||
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geometryb
United States1249 Posts
On July 23 2007 12:45 evanthebouncy! wrote: Or you can think this way: You have 100000000000000000 boxes You pick one, knowing you have 1/100000000000000000 chances Host eliminate 999999999999998 boxes, 2 remains You better switch. if you want more experimental results then i guess. | ||
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Cambium
United States16368 Posts
On July 23 2007 20:56 [r]h_probe wrote: Wrong! Look at it this way: 10 doors You choose one door ----> 1/10 chance of getting it right no matter what! Out of the 9 other doors the host reveals 8 So essentially you are either choosing 1 of two groups: The group of one or the group of 9 (of which 8 have been revealed) Therefore it is a 90% prob of getting it right if you switch! There is no "dividing the chance by that added mount"! To further illustrate my point, lets make the problem even bigger: say you to choose a number between 1 in a million. There is 1 random correct number. Then someone (who obviously knows what the number is) says it is either your number or it is a different one they tell you. You would have to be absolutely insane to stick with your number because it is a 1/1000000 shot no matter what where as the other number is 999999/1000000 I don't understand the problem, do you guys reveal one more door or reveal eight more (say we have ten doors). If you only reveal one more, it doesn't affect the total outcome whether you switch or you don't switch. Say you have ten doors and one goat. GGGGGGGGGC If you don't switch, you have 10% chance of winning. If you switch: If you pick door 1-9, you switch, each case generates 9 more subcases, one that you win, and eight that you lose. If you pick door 10, you switch, you have 9 more subcases, all of wich you lose. So... 9/90 = 10% Either way, switch or no switch, you have 10% chance of winning. nvm, it still improves the probability slightly. It's still 1/10 if you don't switch. If you switch, since he kills a door, You get a 1/8 probability of getting it right. So... 9/80 which is 11.25% vs the original 10% Same goes for 100 obviously 1% vs ( 99 / 980 = ) ~1.0102 % So for 10 doors, what if we remove 2? If we don't switch, still 10% If we switch, now you have a 1/7 chance if you switch if you choose the wrong door first, which is 9/10. But we lose (0%) if we have the right one at first (so 1/10) What's the probability? 9/70 = 0.129 What about three doors? Still 10% if we don't switch, but 9/60 = 15% if we do. What about removing eight doors? Still 10% if we don't switch, If we do, after removing eight doors, we have only two doors left! If our first door has a goat (9/10), then we win (100%) If our first door has a car (1/10), then we lose (0%) So, this gives us 90% chance to win ( 9/10 * 1 + 1/10 * 0 ). In other words, if we don't switch, we are not taking advantage of the "removing a door" rule, by removing a door, we naturally increase our likelihood of winning. | ||
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