The Six Dice Puzzle

[Simon]

This is reminiscent of similar dice riddles, but the answer to this one appears to defy logic. Unless I'm mistaken, this is not as obvious as it looks.

 

There are six unloaded die.

Each has been rolled out of your sight and hidden behind a cup.

You are informed that no more than one of them landed a 6 - but it may be that none of them did.

 

You are going to remove the cups one at a time.

 

Before you do, you label each cup with letters A to F. This designates the order in which you will remove them. You have complete discretion which cup is assigned each letter, but once labeled they must be removed in alphabetical sequence.

 

Having labelled them, it is time to remove the cups accordingly from A to F.

 

If one of the dice landed as 6, under which cup is it most likely to be found?

[freeztar]

None of them, the dice are behind the cup. :shrug:

[Simon]

It not that kind of a riddle! :shrug:

[Kutan]

None of them..? The way it's worded is a little weird to me but I figure if you don't even know what numbers were rolled, they all have an equal chance of having a 6?

[Simon]

None of them..? The way it's worded is a little weird to me but I figure if you don't even know what numbers were rolled, they all have an equal chance of having a 6?

 

That is the intuitive answer. But is it correct?

[Jay-qu]

I would say so..

[sanctus]

I don't think it is correct, every time you lift a cup and it is not a 6 you have a higher probability of having a six...but to do the probability calculus you need to what is the probability of having a six in the first place (which is less than 1/6 since you know that 1 or zero dice are six (while to have the probability of 1/6 you would need to know that 1 dice is 6), hence there is a probability given in advance...).

For example if we have the probability of 75% to have one six then:

cupA: p(6|A)=prob(A)prob(throw 6)prob(external75%)=1/6*1/6*3/4

cupB: p(6|B)=1/5*1/6*3/4, since now only five ways to choose the order of cups

...

cupF: p(6|F)=1*1/6*3/4

-->hence it seems to me that in F it should be the most probable one. Now, if I understood wrong and you meant that there is the probability of 1/6 any dice gives 6, than they same argument holds without the term 3/4=75%...

 

Just note, I don't know if this is correct, but I think it is the way to think.

[eric l]

My guess :

The chances of finding a six are the same for each of the dice.

The chances that you have NO six at the first dice are 5/6

The chances that you have NO six by the time you reach the 6th dice are (5/6)^6 or 15625/466656

So, the chances that you do have a 6 by that time are 1 - this or 31031/46656 or about 2 out of 3. This would mean 2 of of 18 or 1/9 per dice.

[ronthepon]

If one of the dice landed as 6, under which cup is it most likely to be found?

Assuming that one of the dice has indeed landed as a six, we are free to label the cups in whatever order we want.

 

So the probability of the 6 cup getting anything from A to F will probably be equal.

[sanctus]

Assuming that one of the dice has indeed landed as a six, we are free to label the cups in whatever order we want.

 

So the probability of the 6 cup getting anything from A to F will probably be equal.

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.

[ronthepon]

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:

If one of the dice landed as 6, under which cup is it most likely to be found?

[Jay-qu]

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..

[CraigD]

I suspect this puzzle is being slightly misstated or misquoted.

If one of the dice landed as 6, under which cup is it most likely to be found?

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.

[eric l]

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.

 

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.

[TheBigDog]

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

[REASON]

I say the answer is C.

 

In multiple choice questions, I've always heard "C" is most often the correct answer. B)

 

 

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....B)

[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.