[jbjb]

Originally Posted by LazyScot

Spoiler:

My understanding is that this works because the number of coins you draw exactly matches the number of coins that were originally laid out with heads.

How would the answers be altered if the number of coins you drew where not the same as the number of heads? Could you still influence the probability of your being freed?

Spoiler:

Say Y(n) is the excess of heads that the jailer has over you after n draws. So, the jailer starts with an excess of Y(0) (which is 21 in the original puzzle). The strategy of drawing a coin and turning it over means that the excess reduces by 1 each time (either it was a head, so you've reduced the jailer's count, or it was a tail, so you've turned it over and increased your count).

i.e., (for all n >= 0):

Y(n+1) =Y(n) - 1
or:
Y(n) = Y(0)-n

Let's say that N is the number of draws you make, then the jailer's excess at the end is Y(N), and

if Y(N) > 0: you die
if Y(N) <= 0: you are freed

This means that if (as in the original puzzle) the number drawn is the same as the initial count (i.e. N = Y(0)), then the two counts end up equal and you are freed. If the number drawn is more than the initial count (i.e. N > Y(0)) then you will end up with more than the jailer and be freed. If the number drawn is less than the the initial count of heads (i.e N < Y(0)), then you die.

/JB