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In MethodA, the cards are all distinct. In MethodB, the cards can either be all distinct, or Francine can get a repeated card.
<ul type="square">[*]the distinct case, she loses 1/(X+1) of the time. (Where X = number of opponents.)[/list]
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I agree with this.
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<ul type="square">[*]In the repeated case, she loses 1/2X of the time. 1/X to tie with the lowest, 1/2 to lose the coinflip.[/list]
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This doesn't make sense to me...how do you get P(tying for low) = 1/X ?
In the limiting case, this doesn't make any sense. If X=51, then the entire deck will be dealt out (save for 1 card) so either 3 or 4 people will get a deuce.
Method A:
She has a 1/52 chance of losing
Details:
1/13th of the time she has a deuce, and if they flip coins in a fair way she will lose 1/4th of the time. so 1/13 x 1/4 = 1/52.
Method B:
Every time, there will be either 3 or 4 players who draw a deuce, and they will break the tie by coinflip. Now she draws a card from the complete, shuffled deck. 1/13th of the time she will draw a deuce. Then she flips a coin with the loser from the original 51. So now she has a 1/26th chance of being the loser. The reason for this is because she effectively only had to lose 1 coin flip in this method instead of losing two coinflips in method A, when she was included. At least this is true if I read your initial post correctly (translation: Method B picks a loser and then Francine plays off against him/her.)
Even if Francine got to play against all the losers the 1/13th of the time she drew a two, the probability of losing would be 1/52, as in method A, not 1/(2*51) = 1/102
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MethodB is a weighted average of 1/(X+1) and 1/2X, so she loses less often in MethodB.
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Please explain your reasoning for assigning P(tying for low) = 1/2X
and
please explain why method B is the weighted average of 1/X+1 and 1/2x
Thanks.