- Here is a possible strategy for the seven prisonners, if each of them

is able to see the 6 other coins :

(I write H for Head and T for Head, for example 4H2T means "4 Heads and 2

Tails")

Each prisonner seeing 3H3T, 5H1T or 1H5T should pass

Each prisonner seeing 4H2T or 6H should guess T

Each prisonner seeing 2H4T or 6T should guess H

Using this strategy, the prisonners will win if the seven coins are 4H3T,

3H4T, 6H1T or 1H6T, and lost if they are 5H2T, 2H5T, 7H or 7T.

Their probability to win is 21/32, something like 65%. This is quite high,

but is it the best possible strategy ???

Myriam - In search of an optimal strategy, 65% looks good, but... Just driving home the dimensions of the problem...If each person can see the other six coins, (and not know which is whose), then what each person sees can be mapped to one of A = {0, 1, 2, 3, 4, 5, 6} - by assigning 1 to heads and 0 to tails. (Of course, if you knew whose coin was each, this set will include 2^6 choices, not just 7). Any strategy must be then a function from A to B = {H, T, P}, where P is pass. Here, the optimal strategy for each prisoner appears to be the same (- this wouldn't be the case if they knew which coin was whose). For any strategy S: A -> B, one can evaluate the associated probability of success p(S), and we need to find S which is maximises p(S).There are 3^7 = 2187 choices for S. To evaluate p(S), we need to work out how each S performs in 2^7 = 128 cases. I guess one could write a program and find the best S, but trying out by hand seems unwieldy. We may be able to cut things down two by using the symmetry between H & T, etc.There is the another possible dimension - if you include mixed/ random strategies - say a strategy could be if observed total is odd, the prisoner guesses H with probability p, T with probability q, and passes every other time. This of course complicates things, but seems legitimate strategy to me.I think the problem should be solved by analysis rather than enumeration, and the proof should be able to prove the optimum is valid even over the mixed strategies... Where did this come up, BTW?-----Original Message-----
**From:**Myriam Fradon [mailto:myriam.fradon@...]**Sent:**Thursday, February 01, 2001 1:43 PM**To:**probability@yahoogroups.com**Cc:**fradon@...-lille1.fr**Subject:**[probability] Re:Laws of Probability of a Coin Toss`Here is a possible strategy for the seven prisonners, if each of them`

is able to see the 6 other coins :

(I write H for Head and T for Head, for example 4H2T means "4 Heads and 2

Tails")

Each prisonner seeing 3H3T, 5H1T or 1H5T should pass

Each prisonner seeing 4H2T or 6H should guess T

Each prisonner seeing 2H4T or 6T should guess H

Using this strategy, the prisonners will win if the seven coins are 4H3T,

3H4T, 6H1T or 1H6T, and lost if they are 5H2T, 2H5T, 7H or 7T.

Their probability to win is 21/32, something like 65%. This is quite high,

but is it the best possible strategy ???

Myriam