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More colorful foreheads


Go to solution Solved by k-man,

Question

Here is yet another puzzle based on

Suppose that there is a game as follows

* There is a host with 10 stamps, 5 red and 5 blue. There are 4 players- A, B, C, and D.
* In the beginning, the host affixes two stamps to each of the 4 players' head. The choice of stamps for each player is completely random (i.e. the host puts all stamps into an opaque bag and then draws them one by one). The remaining 2 stamps go into the host's pocket. Each player can see the stamps on the remaining 3 players, but can not see his own stamps nor the two in the host's pocket.
* Starting from A to D (and then looping back to A and so on), the host asks if each player definitively knows his color (RR, BB, or RB). If the player does not know, the host goes on to the next player. First player to know his color wins. No guessing is allowed.

Suppose that the host likes you, so he secretly offers you a side bet before the game. You have to pay the host 1
dollar before the game starts, and then you can choose whether to be A, B, C, or D. The payout by position if you win is as follows

A: 3.5 dollars
B: 2.5 dollars
C: 6 dollars
D: 7 dollars

Which position should you choose for the greatest expected winnings?

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Interesting puzzle, Bushino. Can you clarify something for me?

Based on the method of selecting the stamps described in the OP, the probabilities of different combinations are not equal. For example, BB BB BR RR (RR in the pocket) is not as likely as BB BB RR RR (BR in the pocket). That's because after drawing 4 blue stamps out of the bag the chance of drawing another blue among the next 2 stamps is lower that drawing 2 red stamps.

Does the OP intend to account for these unequal probabilities or should it be assumed that all combinations are equally probable?

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Interesting puzzle, Bushino. Can you clarify something for me?

Based on the method of selecting the stamps described in the OP, the probabilities of different combinations are not equal. For example, BB BB BR RR (RR in the pocket) is not as likely as BB BB RR RR (BR in the pocket). That's because after drawing 4 blue stamps out of the bag the chance of drawing another blue among the next 2 stamps is lower that drawing 2 red stamps.

Does the OP intend to account for these unequal probabilities or should it be assumed that all combinations are equally probable?

Definitely, k-man. Clarification are as follows

Let the patterns on the 4 players' head be the 8-dimensional binary vector V (e.g., (B,B,R,R,B,B,R,R) ), the drawing method in the OP is mean to assure that all instances of V are equally likely. As for the example you gave, the chance of drawing BB BB BR RR (RR in the pocket) is the same as drawing BB BB RR RR (BR in the pocket) since if we ignore the first (BB BB) on both, then

P[ BB BB BR RR (RR in the pocket) ] = (1/6) + (5/6)(1/5) = 2/6

P[ BB BB RR RR (BR in the pocket) ] = (5/6)(4/5)(3/4)*(2/3) = 2/6

Note that even though V is equally likely, the aggregated distribution of Blue = BB, Red = RR, and Mixed = BR or RB are not equally likely. For instance, the distribution (Mixed, Mixed, Mixed, Mixed) is a lot more likely than (Blue, Blue, Red, Red).

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Interesting puzzle, Bushino. Can you clarify something for me?

Based on the method of selecting the stamps described in the OP, the probabilities of different combinations are not equal. For example, BB BB BR RR (RR in the pocket) is not as likely as BB BB RR RR (BR in the pocket). That's because after drawing 4 blue stamps out of the bag the chance of drawing another blue among the next 2 stamps is lower that drawing 2 red stamps.

Does the OP intend to account for these unequal probabilities or should it be assumed that all combinations are equally probable?

Definitely, k-man. Clarification are as follows

Let the patterns on the 4 players' head be the 8-dimensional binary vector V (e.g., (B,B,R,R,B,B,R,R) ), the drawing method in the OP is mean to assure that all instances of V are equally likely. As for the example you gave, the chance of drawing BB BB BR RR (RR in the pocket) is the same as drawing BB BB RR RR (BR in the pocket) since if we ignore the first (BB BB) on both, then

P[ BB BB BR RR (RR in the pocket) ] = (1/6) + (5/6)(1/5) = 2/6

P[ BB BB RR RR (BR in the pocket) ] = (5/6)(4/5)(3/4)*(2/3) = 2/6

Note that even though V is equally likely, the aggregated distribution of Blue = BB, Red = RR, and Mixed = BR or RB are not equally likely. For instance, the distribution (Mixed, Mixed, Mixed, Mixed) is a lot more likely than (Blue, Blue, Red, Red).

I picked a bad example. I didn't actually calculate the probabilities in this case. but I noticed that the probabilities of different distributions are not the same. So, I take it from your response that this difference in probabililties should be accounted for.

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...choose position B. Unless I messed up in the calculations somewhere, the odds of winning are roughly

A: 30.16%

B: 49.21%

C: 12.70%

D: 7.94%

Multiplying the payouts by the odds, we get the best expected payout of $1.23 (per game) for position B.

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...choose position B. Unless I messed up in the calculations somewhere, the odds of winning are roughly

A: 30.16%

B: 49.21%

C: 12.70%

D: 7.94%

Multiplying the payouts by the odds, we get the best expected payout of $1.23 (per game) for position B.

Some comments

Since we are choosing 5 stamps out of 10, the total number of unique unaggregated stamp distributions is 10C5 = 252. We would then expect the probability of winning for each participant to be a fraction with 252 in the denominator.

Looking at the winning percentages above, I don't think they can be converted into such a fractional form.

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...choose position B. Unless I messed up in the calculations somewhere, the odds of winning are roughly

A: 30.16%

B: 49.21%

C: 12.70%

D: 7.94%

Multiplying the payouts by the odds, we get the best expected payout of $1.23 (per game) for position B.

Some comments

Since we are choosing 5 stamps out of 10, the total number of unique unaggregated stamp distributions is 10C5 = 252. We would then expect the probability of winning for each participant to be a fraction with 252 in the denominator.

Looking at the winning percentages above, I don't think they can be converted into such a fractional form.

Here are the fractions:

A: 19/63

B: 31/63

C: 8/63

D: 5/63

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Here are the fractions:

A: 19/63

B: 31/63

C: 8/63

D: 5/63

Sorry about that, my bad. Since D has the lowest probability in the above table,

Could you list out the distributions that would lead to D winning?

Sure.

Pocket stamps are in parenthesis. in all below combinations D wins in the first round. All have equal probability of 1/126.

A  B  C  D
BB BB BR RR (RR)
BB BB RR RR (BR)
BB BR BB RR (RR)
BB RR BB RR (BR)
RR RR BR BB (BB)
RR RR BB BB (BR)
RR BR RR BB (BB)
RR BB RR BB (BR)
BR BB BB RR (RR)
BR RR RR BB (BB)

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Sure.

Pocket stamps are in parenthesis. in all below combinations D wins in the first round. All have equal probability of 1/126.

A  B  C  DBB BB BR RR (RR)BB BB RR RR (BR)BB BR BB RR (RR)BB RR BB RR (BR)RR RR BR BB (BB)RR RR BB BB (BR)RR BR RR BB (BB)RR BB RR BB (BR)BR BB BB RR (RR)BR RR RR BB (BB)

Comments

A B C D

BB BB BR RR (RR) *

BB BB RR RR (BR) *

BB BR BB RR (RR)

BB RR BB RR (BR) *

RR RR BR BB (BB) *

RR RR BB BB (BR) *

RR BR RR BB (BB)

RR BB RR BB (BR) *

BR BB BB RR (RR)

BR RR RR BB (BB)

You may want to check the distribution with the the star next to them. For instance, for the second distribution

BB BB RR RR (BR)

On his turn, B will reason that his hat is either BB or BR. If A saw [bB *BR* RR RR (BB)], he would have called his hat at turn 1. But A didn't, so B knows his hat.

Let's look at the first distribution, [ BB BB BR RR (RR) ]. On his turn, C reasons that his hat is either RR or BR. If it is RR, then B would have won on turn 2 due to the reasoning for distribution 2. There on his turn C would know his color.

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Sure.

Pocket stamps are in parenthesis. in all below combinations D wins in the first round. All have equal probability of 1/126.

A  B  C  DBB BB BR RR (RR)BB BB RR RR (BR)BB BR BB RR (RR)BB RR BB RR (BR)RR RR BR BB (BB)RR RR BB BB (BR)RR BR RR BB (BB)RR BB RR BB (BR)BR BB BB RR (RR)BR RR RR BB (BB)

Comments

A B C D

BB BB BR RR (RR) *

BB BB RR RR (BR) *

BB BR BB RR (RR)

BB RR BB RR (BR) *

RR RR BR BB (BB) *

RR RR BB BB (BR) *

RR BR RR BB (BB)

RR BB RR BB (BR) *

BR BB BB RR (RR)

BR RR RR BB (BB)

You may want to check the distribution with the the star next to them. For instance, for the second distribution

BB BB RR RR (BR)

On his turn, B will reason that his hat is either BB or BR. If A saw [bB *BR* RR RR (BB)], he would have called his hat at turn 1. But A didn't, so B knows his hat.

Let's look at the first distribution, [ BB BB BR RR (RR) ]. On his turn, C reasons that his hat is either RR or BR. If it is RR, then B would have won on turn 2 due to the reasoning for distribution 2. There on his turn C would know his color.

You're right. I messed up. Here are the corrected probabilities...

A: 19/63

B: 32/63

C: 10/63

D: 2/63

Based on these odds, B is still the pest position with the expected payout of almost $1.27.

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You're right. I messed up. Here are the corrected probabilities...

A: 19/63

B: 32/63

C: 10/63

D: 2/63

Based on these odds, B is still the pest position with the expected payout of almost $1.27.

Comments

A B C D

BB BR BB RR (RR)

RR BR RR BB (BB)

BR BB BB RR (RR)

BR RR RR BB (BB)

So it seems that the sequences that lead to D winning (probability 2/63) are above. But there are a few more sequences that would lead to D winning. For instance, consider the following distributions

BR RR BB BR BR.

There are more missing distributions for D, so you might want to look for the winning cases for A, B, C and find them.

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You're right. I messed up. Here are the corrected probabilities...

A: 19/63

B: 32/63

C: 10/63

D: 2/63

Based on these odds, B is still the pest position with the expected payout of almost $1.27.

Comments

A B C D

BB BR BB RR (RR)

RR BR RR BB (BB)

BR BB BB RR (RR)

BR RR RR BB (BB)

So it seems that the sequences that lead to D winning (probability 2/63) are above. But there are a few more sequences that would lead to D winning. For instance, consider the following distributions

BR RR BB BR BR.

There are more missing distributions for D, so you might want to look for the winning cases for A, B, C and find them.

Doh! :duh:

I went through everything again and found a couple of mistakes, so now the odds and expected payouts are as follows...

A: 5/21 $0.83

B: 4/9 $1.11

C: 10/63 $ 0.95

D: 10/63 $ 1.11

According to this result positions B and D are equal in the expected payout, but this makes me think that I'm still missing something.

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You're right. I messed up. Here are the corrected probabilities...

A: 19/63

B: 32/63

C: 10/63

D: 2/63

Based on these odds, B is still the pest position with the expected payout of almost $1.27.

Comments

A B C D

BB BR BB RR (RR)

RR BR RR BB (BB)

BR BB BB RR (RR)

BR RR RR BB (BB)

So it seems that the sequences that lead to D winning (probability 2/63) are above. But there are a few more sequences that would lead to D winning. For instance, consider the following distributions

BR RR BB BR BR.

There are more missing distributions for D, so you might want to look for the winning cases for A, B, C and find them.

Doh! :duh:

I went through everything again and found a couple of mistakes, so now the odds and expected payouts are as follows...

A: 5/21 $0.83

B: 4/9 $1.11

C: 10/63 $ 0.95

D: 10/63 $ 1.11

According to this result positions B and D are equal in the expected payout, but this makes me think that I'm still missing something.

Comments

You're very close. I would recommend checking the cases under B again.

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I think I finally got it!

D is the best position with approx $1.11 expected payout. The odds and payouts for each position are (rounded):

A: 19/63 $1.06

B: 8/21 $0.95

C: 10/63 $0.95

D: 10/63 $1.11

Great job, k-man. You got it.

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