Let's assume A goes first. We can break it down:
Either:
A wins on first toss: 1/2
or A tails on first toss AND B wins on second toss: (1/2)(1/2) = 1/4
or A tails on first toss, B tails second toss, A wins on third: (1/2)(1/2)(1/2) = 1/8
or A tails, then B tails, then A tails, then B wins: 1/16
and so on.
So the probability A wins should be (since the game lasts at most 5 tosses each):
1/2 + 1/8 + 1/32 + 1/128 + 1/512 = 341/512
and the probability B wins should be:
1/4 + 1/16+ 1/64 + 1/256 + 1/1024 = 341/1024
and of course there's a 1/1024 chance that neither A nor B wins (if they both get tails, ten times).
Because there are three possible outcomes: A wins, B wins, or neither wins, one needs to be a bit careful calculating the odds against A losing. These odds should be equal to the ratio of:
(The probability A does not lose) to (the probability A does lose)
I presume the probability that A loses is equal to the probability that B wins- that is, 341/1024- so the odds against A losing should work out to
683 to 341
I wonder, however, if the question is really asking for the odds that A wins, assuming someone wins (that is assuming A and B do not both get tails every toss). Then the answer is much simpler:
682 to 341, or 2 to 1.
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