Answers (I\'m pretty sure at least) - Don\'t peek!
Answers (?)
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In the first problem everyone has an equal chance of pulling the ace of spades (odds will be 51:1 for everyone because the cards are not replaced). I'm pretty sure it doesn't matter where you stand, although the argument to go first seems logical.
The second problem requires more math than I can handle, but here is my general logic. Odds for the first player are 51:1, the second player will have the combined odds of 51:1 and 50:1 but he will only get to pick 51 times in 52 (because the first player will have won the other time).
I'm sure there is a way to figure out your max EV by comparing your combined odds of pulling the ace of spades with the chances that someone in front of you will win first. I'm almost positive this problem will have a specific answer for a position in line, but we might need bruceZ to give us the answer. [img]/images/graemlins/blush.gif[/img]
J
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