The Six Dice Puzzle

[ronthepon]

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?

[CraigD]

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.

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.

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?

You’ve rephrased the question as an expected value question.

 

[math]\mbox{Expected value} = \sum{ \mbox{Value_{event}}} \cdot \mbox{Probability_{event}}[/math]

 

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, [math]\mbox{Expected value} = \frac16 \cdot 5 + \frac16 \cdot 4 + \frac16 \cdot 3 + \frac16 \cdot 2 + \frac16 \cdot 1 + \frac16 \cdot 0 = 2.5[/math]. 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?

[sanctus]

Ok I get it, prob. theory was never my strength but I was sure this time I got it right :weather_storm:

[Simon]

Having made further calcuations, I find I agree with Eric.

 

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.

 

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?

[TheBigDog]

Having made further calcuations, I find I agree with Eric.

 

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.

 

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.

 

Do others agree?

You have been told there is only one six. Kutan? Are you ever coming back to settle this thing?

 

Bill

[CraigD]

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".

…

Do others agree?

I agree.

 

The probabilities of discovering a six after revealing each cup are

1 3125/46656

2 2500/46656

3 1875/46656

4 1250/46656

5 625/46656

Of course, as there are exactly 2 sixes, the probability of getting to reveal the 6th and last cup is zero.

 

Since we’re given that the 6 die faces have a uniform probability (the dice are “fair”), it’s easy to calculate probabilities for any sort of selection rule with a small computer program, avoiding the need to ponder the question from a philosophical perspective :eek2:

[Simon]

The probabilities of discovering a six after revealing each cup are

1 3125/46656

2 2500/46656

3 1875/46656

4 1250/46656

5 625/46656

Of course, as there are exactly 2 sixes, the probability of getting to reveal the 6th and last cup is zero.

 

Naturally these figures only apply if the question is taken to mean "under which cup is the first 6 most likely to be found?".

 

What if you're also asked in advance (i.e before the first cup is lifted): "under which cup is the second 6 most likely to be found?"

 

Haven't worked it out yet, but I suspect cups B-F carry equal probability.

[CraigD]

The probabilities of discovering a six after revealing each cup are

1 3125/46656

2 2500/46656

3 1875/46656

4 1250/46656

5 625/46656

Of course, as there are exactly 2 sixes, the probability of getting to reveal the 6th and last cup is zero.

Naturally these figures only apply if the question is taken to mean "under which cup is the first 6 most likely to be found?".

 

What if you're also asked in advance (i.e before the first cup is lifted): "under which cup is the second 6 most likely to be found?"

Without commentary, just counting all 9375 applicable of 46656 total unique and equal-probability rolls, the probability of discovering the second 6 under each cup are:

2 625/46656

3 1250/46656

4 1875/46656

5 2500/46656

6 3125/46656

Of course, since you can’t discover 2 dice under the first cup, the probability of discovering the second 6 under the first cup revealed is zero.

 

Suppose that you’re asked to guess the cups that both the first and second six will be found under. The unwary might, based on the previous counts, answer “A and F”. Actually, any valid answer (eg: excluding impossibilities such as “C and C” or “B and A”) is equally likely, so one cannot do better at this question than a random guess.

[Simon]

Now that does confound me.

 

Let me get this right:

 

1) If asked where both sixes will probably be found - answer: any two cups.

2) If asked where the 1st six will probably be found - answer: cup A (the first removed)

3) If asked where the 2nd six will probably be found - answer: cup F (the last removed)

 

Is this your view?

 

I already agree with statement 2.

The others don't seem right to me yet.

[Simon]

Statements 1 and 3 now verfified.

 

I agree!

[Simon]

I've just realised Eric that you're exact figures are way off.

 

Where you're told that exactly two of the die are sixes and you lift the cups A-F, the probability of finding the first six under each cup is:

 

A: 1/3

B: 4/15

C: 1/5

D: 2/15

E: 1/15

F: 0

 

The probability of finding the second six under each cup is:

 

A: 0

B: 1/15

C: 2/15

D: 1/5

E: 4/15

F: 1/3

 

The principle is the same, but I think you got lost counting too many permutations.

Since you're informed in advance that there are precisely 2 sixes, it becomes immaterial what the posssible combinations are for the dice that didn't land six. Each of the six die can be treated as either "six" or "not six". There are then only 720 combinations, not 46656!

[CraigD]

I've just realised Eric that you're exact figures are way off.

You’re right – except that I’m Craig, not Eric :). I divided the counts by [math]6^6 = 46656[/math], the total number of unique roles, giving the probability that there would be exactly 2 sixes AND that a six would be under a particular cup. I should have divided by [math]3 \cdot 5^5 = 9375[/math], the number of rolls that have exactly 2 sixes. All my numbers were, therefore, overstated by [math]\frac{9375}{46656} = \frac{3125}{15552}[/math]

The principle is the same, but I think you got lost counting too many permutations.

I actually didn’t user any arithmetic more sophisticated than counting, and converting to base 6 – I just checked each of the [math]6^6[/math] possible rolls of 6 6-sided dice, and for ones with exactly 2 sixes, counted how many had the first or second 6 in each position. It’s a thoughtless, brute force approach. As usual, I wrote it in MUMPS – here’s the actual code:

k C s C=0 f N=0:1:6**6-1 s A=N,D="" x "f I=1:1:6 s D=A#6_D,A=A6" i (D,0)=3 s I=(D,0)-1,C=C+1,C(I)=(C(I))+1 ;count location of first 0
k C s C=0 f N=0:1:6**6-1 s A=N,D="" x "f I=1:1:6 s D=A#6_D,A=A6" i (D,0)=3 s I=(D,0,(D,0))-1,C=C+1,C(I)=(C(I))+1 ;count location of second 0

[Simon]

The question appears even stranger when the same principle is applied to a larger number of events.

 

I have just spun a roulette wheel exactly 36 times. A zero came up twice.

 

On which spin did the first zero occur?

On which spin did the second zero occur?

 

The most probable answers:

 

The first zero occured on the first spin.

The second zero occured on the 36th spin.

[Qfwfq]

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

This makes the probability of finding 6 under a given cup less than 1 out of 6, it is 1 out of 11. With n cups remaining, the probability for the next cup, on condition that 6 hasn't already turned up, is [imath]\frac{1}{5+n}[/imath] and increases to 1 out of 6 for the last remaining cup.

 

No Bill, it isn't the Gambler's Fallacy, it's conditional probability.