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PS- probability-Coin

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Joined: 05 Jun 2008
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02 Dec 2008, 04:40
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A and B alternately toss a coin. The first one to turn up a head wins. if no more than five tosses each are allowed for a single game.

1- Find the probability that the person who tosses first will win the game?

2- What are the odds against A's losing if she goes first?
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Joined: 27 Apr 2008
Posts: 110

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03 Dec 2008, 00:16
1. 341/512

prob(a to win)= 1/2+1/8+1/32+1/128+1/512=(2^8+2^6+2^4+2^2+1)/(2^9)=341/512

2. 341/1024

prob(b to win)= 1/4+1/16+1/64+1/256+1/1024=(2^8+2^6+2^4+2^2+1)/(2^9)=341/1024

the probabilities do not sum to one because there is the chance fo a draw
Senior Manager
Joined: 05 Jun 2008
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03 Dec 2008, 04:41
vivektripathi wrote:
A and B alternately toss a coin. The first one to turn up a head wins. if no more than five tosses each are allowed for a single game.

1- Find the probability that the person who tosses first will win the game?

2- What are the odds against A's losing if she goes first?

Let’s say A tosses first.

what is the probability that he wins

H + TTH + TTTTH + TTTTTTH + TTTTTTTTH

i.e. either the first toss is head,
or the first time A tosses the coin he gets a tail and B also gets a tail , n in the second throw A gets a head.....

This continues for a max till 5 throws, because the game is for 5 throws only.
So, 1. 1/2 + (1/2)^3 + (1/2)^5 + (1/2)^7 + (1/2)^9

2. (1/2)^2 + (1/2)^4 + (1/2)^6 + (1/2)^8 + (1/2)^10
Re: PS- probability-Coin   [#permalink] 03 Dec 2008, 04:41
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