Posted 26 June 2012 - 12:30 AM

The probability of four same color (thus first chair winning):

8/8 * 3/7 * 2/6 * 1/5 = 1/35. Same answer results from using combinations.

There are 70 (ordered) permutations of the 8 stamps, each with equal chance.

There are 16 (ordered) permutations of 4 of the 8 stamps, not with equal chance.

If this doesn't occur, then first chair cannot know the answer.

Second chair now knows that first chair does not see: YYXXXX

Specifically it leaves 68 permutations with equal chance.

Then, if he sees XXYYXX he wins. As in the first case, there are two permutations that satisfy this.

So he wins this way with 1/34 chance.

I believe he learns nothing else.

Third chair now knows that second chair does not see: XXYYXX

This eliminates 2 possibilities, leaving 66 with equal chance.

If he sees XXXXYY he wins, 2/66 chance.

Also if he sees XXYY?? then he wins knowing he has XY (since the others haven't won yet).

This could happen as BBWW**** or WWBB**** where **** is permutations of BBWW. 2 * 6 = 12 ways

However XXYY?? intersects with XX??XX at XXYYXX so there are 8 ways left.

So if he didn't win by XXXXYY he wins this with 8/64.

Equivalent overall: 2/66 + 64/66 (8/64) === 10/66 = 5/33

I believe he learns nothing else.

First chair:

He's survived this far, so there are 56 possibilities left. YYXX?? is not possible.

He applies same thing as third chair, he wins if he sees ??XXYY where he must have XY. Still 12 possibilities, minus 4 intersections.

Wins with chance 8/54 = 4/27.

I can't see him learning anything else.

Second chair:

48 possibilities left, YYXX?? and ??XXYY are excluded.

If he sees XX??YY he must have XY and he wins. 8 ways => 8/48 = 1/6 chance of winning like this.

If he doesn't win like that, he knows the XX, YY, XY pairs aren't there (no one sees them). XXYYXY and XXYYXX and XXYYYY aren't there, it must mean XX YY isn't there. What about XX XY? (XX XY YY, XX XY XX can't be). One mixed pair doesn't exist, this means there are two or three mixed pairs.

Thus if he sees XX??MM or MM??XX where MM = mixed pair, then he knows he has XY.

XXMMMM has 8 forms: 2x for swapping X with Y, 4x for two MM pairs as XY or YX. No more multipliers because the rest must be YY.

MMMMXX has 8 ways also.

He wins with 16/40 chance now = 2/5

Overall: 8/48 + 40/48 (16/40) = 24/48 = 1/2 chance.

Third chair: 24 left.

So I think there are 2 or 3 mixed pairs. Second chair would have won if one of 1st or 3rd pair was not mixed, therefore they must both be mixed.

I believe possibilities left are MMMMMM or MMXXMM. This would mewan third must have XY no matter what.

Can check this by seeing how many possibilities MMMMMM and MMXXMM form:

MMXXMM: 2x for XY or YX on third pair, 2x for XY/YX on 1st pair, 2x for identify of X, 1x for the rest must be YY. 8 ways.

MMMMMM: 8x for three pairs XY/YX. Remaining is also a pair, 2x for that. 16 ways.

Totals 24 so correct.

Patterns:

1 XXXX??, 2 ways

2 XX??XX, 2 ways

3 ??XXXX, 2 ways

3 XXYY??, 8 new ways

1 ??XXYY, 8 new ways

2 XX??YY, 8 new ways

2 XXMMMM or MMMMXX, 16 ways

3 MMXXMM or MMMMMM, 24 ways

By intuition or otherwise, the order of events is independent of the ways of winning (must be better way to explain). Anyhow, you can just add up the number of ways.

First chair: 10 ways

Second chair: 26 ways

Third chair: 34 ways

So the logical choice is third chair, with 34/70 chance of winning, assuming the other two students are not (color)blind and are equally logical.