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08-13-2007
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Thinking
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The Six Dice Puzzle
This is reminiscent of similar dice riddles, but the answer to this one appears to defy logic....
Last edited by Simon; 08-13-2007 at 02:55 PM..
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Re: The Six Dice Puzzle
The um... 'absolute' probability doesn't increase though. All that happens is that you calculate the probability of there being a 6 in the next cup given that the first is a dud.
Let's not forget the original question:
Quote:
Originally Posted by Simon
If one of the dice landed as 6, under which cup is it most likely to be found?
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Re: The Six Dice Puzzle
I agree with ron, as successive cups are labelled with letters the overall chance of labelling a six has increased, but each individual cup still would have an equal chance of having the six..
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Are the conditions stated correctly?
I suspect this puzzle is being slightly misstated or misquoted.
Quote:
Originally Posted by Simon
If one of the dice landed as 6, under which cup is it most likely to be found?
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If there are zero or one sixes, regardless of the order in which you reveal them, the probability of a six being under each cup is the same: 1/6 is there is exactly one 6, or 3125/46656 if there is either one or zero sixes. This can be shown empirically, by actually rolling the dice and checking the cups in order (or, more practically, having a computer program simulate doing so), or by arguing from systematic counting principles.
The question is more interesting if the condition “you are informed that no more than one of them landed a 6 - but it may be that none of them did” (which seems contradicted by the final question, which is qualified “if one of the dice landed as 6”) is removed. If the dice are rolled without anyone looking at them, and placed under the cups, the probability of finding a six (and immediately stopping looking) is greatest under the first cup revealed, least under the last, the precise probabilities being 7776/46656, 6480/46656, 5400/46656, 4500/46656, 3750/46656, and 3125/46656, respectively.
There are also several interesting variations of this kind of question, where you are asked how much you should bet on the next die revealed being a six after a given number have been revealed as not sixes.
Of course, I may be misunderstanding something, but it seems a pretty clearly stated problem, and actual counting rarely leads one astray. |
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Last edited by CraigD; 08-14-2007 at 04:56 AM..
Reason: Fixed number typo
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Re: The Six Dice Puzzle
Quote:
Originally Posted by sanctus
I don't agree, just imagine in the first cup there is no 6 but you know there is one hence the probability increases. The easiest way to see this is by supposing that you do the experiment and the first 5 cups have something different from 6, hence the last cup has a probability of 1 having a six.
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Sandro, the question states that " no more than one of them has a six", and that it is possible that none has a 6.
Hence my looking for the probability that none shows a 6, and the probability of (about) 1/9 instead of 1/6 for showing a 6.
Still, I am not completely happy with my own answer, but I can't find anything better right now. |
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Re: The Six Dice Puzzle
Ron and JQ are correct. If one of them is a six, then each one has an equal probability of being the six. If you go through the exercise of using *not* logic, then you can initially determine that the probability increases from A to F like so...
A = 83.333% not 6, 16.666% is 6
B = 80% not 6, 20% is 6
C = 75% not 6, 25% is 6
D = 66.666% not 6, 33.333% is 6
E = 50% not 6, 50% is 6
F = 0% not 6, 100% is 6
of course those odds are only forward looking, and don't account for the chances that have already passed. Properly calculated you get this...
A = 16.666% is 6
B = 83.333% x 20% = 16.666% is 6 (83.333% are the remaining odds after the first choice)
C = 83.333% x 80% x 25% = 16.666% is 6 (83.333% remaining from first choice, 80% of that remaining from second choice)
D = 83.333% x 80% x 75% x 33.333% = 16.666% is 6
E = 83.333% x 80% x 75% x 66.666% x 50% = 16.666% is 6
F = 83.333% x 80% x 75% x 66.666% x 50% x 100% = 16.666% is 6
They all have an equal chance of being 6. In my logic class this was described as "The Gambler's Fallacy".
Bill
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Last edited by TheBigDog; 08-14-2007 at 11:31 AM..
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Re: The Six Dice Puzzle
I say the answer is C.
In multiple choice questions, I've always heard "C" is most often the correct answer. 
Actually, I agree with BigDog. With each letter placement there is a 1/6 chance of finding the six. To assume your odds increase with each successive letter is to assume that each previous letter placement was not six.
And we all know what happens when we assume....  |
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Re: The Six Dice Puzzle
That's my point! At the start you have 1/6 of chance of having the 6 in any cup, but when you start lifting them then every time you don't have the six the probabilities change.
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Re: The Six Dice Puzzle
Suppose you had, say six dollars, and I took away one dollar every time you picked a cup.
then, even though you've picked more cups up, and the probability of finding the 6 in the next cup increases every time, you lose money while picking the cups up.
When you reach the last cup, you know that you're gonna find the 6, but that won't quite be useful, because you've got no cash left.
Take it in this way. Te amount of cash you've got left depends on the initial selection only, and not on when you pick the cups up.
Maybe this won't be all that totally related, but it rephrases the original question:
What is the amount of money you are most likely to recieve?
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Rephrased as an expected value question
Quote:
Originally Posted by sanctus
That's my point! At the start you have 1/6 of chance of having the 6 in any cup, but when you start lifting them then every time you don't have the six the probabilities change.
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Yes, but as TBD explained, the probability of getting to the state where the probability of the remaining cups has changed exactly counteracts their change in probability, so in the end, if you repeat the game many times (with the assurance that exactly 1 six was rolled), you’ll find that each cup is exactly as likely (probability 1/6) to contain the six.
Quote:
Originally Posted by ronthepon
Suppose you had, say six dollars, and I took away one dollar every time you picked a cup.
…
Maybe this won't be all that totally related, but it rephrases the original question:
What is the amount of money you are most likely to recieve?
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You’ve rephrased the question as an expected value question.
So, for your example, and assuming the condition that the dice are randomly rolled, but you’re only allowed to play if exactly one six is rolled,  . You will, on average, have $2.50 after playing the game.
As I noted in post #13, the game is much more interesting if you don’t impose any conditions other than the dice are randomly rolled. Can anyone calculate the expected value of Ron’s game, with these conditions? |
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Re: Rephrased as an expected value question
Ok I get it, prob. theory was never my strength but I was sure this time I got it right  |
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