Question 712456: A and B alternatively toss a coin. The first one to turn up a head wins. If no more than five tosses are allowed for a single game, find the probability that the person who tosses first will win the game. What are the odds against A's loosing if he goes first?
I think A has a greater chance to win, if he goes first, because A will have 5 turns, while B will have 4 turns only. So, A will have a head start by 1 turn. So, it's more likely that A will win.
p(A get first Head) = (1/2)^5/{(1/2)^5+(1/2)^4}=1/3
Probability that A will win (since A tosses first) is 1/3. However, correct answer is 21/32. And for part b, answer is 21:11. Thank you for your help.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A and B alternatively toss a coin. The first one to turn up a head wins. If no more than five tosses are allowed for a single game, find the probability that the person who tosses first will win the game. What are the odds against A's loosing if he goes first?
I think A has a greater chance to win, if he goes first, because A will have 5 turns, while B will have 4 turns only. So, A will have a head start by 1 turn. So, it's more likely that A will win.
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If A tosses 1st his winning patterns are:
h (after that game is over) ; probability of this is 1/2
tth (after that game is over) ; probability of this is 1/8
tttth (after that game is over) ; probability of this is 1/32
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p(A gets first Head) = p(h or tth or tttth) = (1/2) + 1/8 + 1/32
= [16 + 4 + 1)/32 = 21/32
P(B gets first head when A tosses 1st) = P(th or ttth) = 1/4 + 1/16 = 5/16
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Odd's against A's winning if A tosses 1st are (31/32)/(1/32) = 31:1
Odd's against B's winning if A tosses 1st ae (11/16)/(5/16) = 11:5
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Cheers,
Stan H.
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