Solving the Six Dice Puzzle: A Logic Defying Riddle

[Simon 6]

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 as 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?

 

[dodo]

Under none. You just removed all the cups. Please.

 

[Simon 6]

Sorry. It's not that kind of riddle!

I'll rephrase.

"Having labelled them, you will remove the cups accordingly, from A to F.
Under which cup is the dice that landed a 6 most likely to be?"

 

[mathman]

This sounds like a trick question, but my guess is all are equally likely.

 

[Simon 6]

This sounds like a trick question, but my guess is all are equally likely.

I think it's a question with a counter-intuitive answer.

If I'm correct, they are not all equally likely and one of the cups has a higher probability of revealing the 6 than each of the others.

But there may be a flaw in my logic.

Also I amended the wording shortly after I first posted it. The clue may lie in not knowing whether any cup holds a six.

 

[CRGreathouse]

There are $5^6$ ways for the dice to land without any 6s. There are $5^5$ ways for each die to land on a 6. Thus the chance that any die lands on a 6 is 1/11 and the chance that no die lands on a 6 is 5/11.

 

[HallsofIvy]

But I don't see how that has anything to do with Simon 6's assertion that a "6" is more likely to be under a specific one of the cups.

 

[Rogerio]

...one of the cups has a higher probability of revealing the 6 than each of the others.

Instead just saying they are equally likely, I prefer you answer me:
Why should one cup be different from the others?

 

[CRGreathouse]

But I don't see how that has anything to do with Simon 6's assertion that a "6" is more likely to be under a specific one of the cups.

If it's right, he's wrong. I tried to make it as transparent as possible.

 

[dodo]

I presume it will have something to do with your outcomes changing as you remove each cup. In the extreme, "if I removed all but one cup and still no 6, then the last cup has 100% blah blah".

For I can't see the a priori probability of having a six somewhere to be altered by the cup removal order.

 

[CRGreathouse]

For I can't see the a priori probability of having a six somewhere to be altered by the cup removal order.

I'll admit that I actually though the probabilities would differ, until I actually tried to model it. It seems to me that it should be symmetric.

 

[robert Ihnot]

What you have to do, I think, is find the sample space. Now we can consider the binomial expansion where S=6,andN = not a six: $$N^6+6(N^5)S+15(N^4)(S^2)+20...$$ but of all the variations, we are only interested in the first two above: n^6+6(N^5)S, this is in fact only 7 varations, but the probability of (5/6)^6+(5/6)^5 is about 74% of all cases.

So all we have to do is consider the case of all N's = NNNNNN, about 33%, and then consider 6 more cases as the S moves from the first position to the last, about 40% total, 6.67% for each position. Thus if we had 100 trials under consideration, we would exclude all but 74 of them, and then find 6.67/74 = 9% of those would have a 6 in a given position. or to be exact we could use CRGreathouse's figures:

$$\frac{(5/6)^5*1/6}{(5/6)^6+(5/6)^5} =1/11$$

 

[Simon 6]

Having made further calcuations, I draw the following conclusion

The question as stated - where you're told in advance that no more than one dice landed 6 - does yield the intuitive answer: all have equal probability. I was mistaken to think otherwise.

However: if you're not limited to rolling one 6, a very counterintuitive answer applies.
Suppose you're informed: "Two die landed 6".
Everything else in the puzzle stays the same.
You label the cups A-F in the order you will remove them.
The question is then asked: "Under which cup are you most likely to discover a 6?"

The answer is always "A".

In other words: before you start removing cups, whichever cup you choose to lift first is most likely to contain one of the 6s. If cup "A" is found not to have a 6, then the highest probability moves to cup "B". This continues alphabetically until the first 6 is discovered. After that, the remaining cups have equal probability.

That may seem unbeleivable.

Do others agree?

 

[CRGreathouse]

In other words: before you start removing cups, whichever cup you choose to lift first is most likely to contain one of the 6s. If cup "A" is found not to have a 6, then the highest probability moves to cup "B". This continues alphabetically until the first 6 is discovered. After that, the remaining cups have equal probability.

The probabilities should be
A: 1/3
B: 4/15
C: 1/5
D: 2/15
E: 1/15
F: 0

 

[CRGreathouse]

However: if you're not limited to rolling one 6, a very counterintuitive answer applies.

I'm not sure it's counterintuitive. If you're told nothing, A is still the most likely and F the least likely. The only cases when they should be equal is when you know there are less than 2 sixes (3 cases).

0-6 sixes:
A: 1/6
B: 5/36
C: 25/216
D: 125/1296
E: 625/7776
F: 3125/46656

 

[dodo]

If cup "A" is found not to have a 6, then the highest probability moves to cup "B".

This sounds to me more like conditional probability, something on the order of:
P(A) = 1/3
P(B|~A) = 2/5
P(C|~(A.B)) = 1/2
P(D|~(A,B,C)) = 2/3
P(E|~(A,B,C,D)) = 1
[Edit: corrected this last one:]
P(F|~(A,B,C,D,E)) = 0/0 (undefined)

 

[chronon]

I don't think it's counterintuitive at all. If I have two cups and put a marble under each, then I'm most likely (i.e certain) to first find a marble under the first cup I look under. Now the probability that there is a marble underneath is the same for both cups (i.e. 1), just as the probability that there is a six underneath is the same for all cups in the question posed.

 

[Simon 6]

If I have two cups and put a marble under each, then I'm most likely (i.e certain) to first find a marble under the first cup I look under.

Naturally! Certainties dispel counter-intuition.

Translate this to six cups hiding two marbles.

You're about to lift the cups one at a time from A to F.

I ask: "Under which cup are you most likely to find the first marble?". Most people won't assume it's the first cup you lift.

However, suppose the initial question is: "Under which cup are you most likely to find the second marble?".

Again, I believe the answer will surprise most people.

 

[CRGreathouse]

However, suppose the initial question is: "Under which cup are you most likely to find the second marble?".

I'd expect the last cup, naturally.

 

[b3hr057]

There are six dice, one under each cup. The number of sixes is less than or equal to one.
The odds of finding a six under cup number A are therefore (less than or equal to)one in six.
Either it is a 1:6 chance or there is a 0% chance depending on if there is even one six out of all the dice!

It then follows that as the cups are eliminated, the odds of a six being revealed increase.

Cup B = 1:5, C = 1:4, &c. so that if you eliminate the first five cups, and there is indeed a six, the last cup has a 100% probability of containing that six.


This is similar to another riddle/math problem I call "The Game Show" You choose one of three doors, two of which have nothing of value behind them, one containing the "grand prize". The game show host shows you one of the two doors tht you did NOT pick contains nothing. So now it is either the door you picked originally or you could change your guess to the other door. Do you stick to your original guess or switch?

The best bet is to switch because your first guess you had a 1:3 chance of getting it right. Now you have a 1:2 chance so it would be foolish not to change your bet!
Think this doesn't have logic to it? Okay, take it on a grander scale. Let's say you pick a number between 1 and 1,000,000,000. Let's say you pick 42. Now the administrator of the "lottery" tells you that they have eliminated every number except 42, 576,341,100, and 923,128,001. The grand prize is $500,000 and you can now either stick to your original guess or switch to either two of the remaining numbers. What do you do?

Same thing, just more extreme odds. You went from a 1:1,000,000,000 chance to a 1:3 chance.

 

[ThirstyDog]

I am going on that it depends if you keep looking under cups once you have found a six. The logic defying component might just be dependent on this.