[pdurrant]

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:

jbjb's analysis is right as far as it goes (although perhaps overly mathematical).

And he is right - following the strategy, if you draw more coins than the number of heads originally showing, you win. If you draw fewer, you lose.

So if (say) you could only draw 20 coins, it is the wrong strategy to take.

I don't know off-hand what the correct strategy would be. It sounds like a good follow-on puzzle.