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A and B alternately toss a coin. The first one to turn up a

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A and B alternately toss a coin. The first one to turn up a [#permalink] New post 11 Apr 2005, 04:32
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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Re: Probability A and B Toss a Coin? [#permalink] New post 11 Apr 2005, 09:12
1. The person who starts the toss wins if either of these sequences occur:

H, T T H, or T T T T H

The probability of these sequences occuring is 1/2, 1/8 and 1/32 respectively.

Therefore, the total probability of the one who starts the tosses to win is 21/32.

2. The A's probability of winning if she starts first = 21/32. The probability of a draw is 1/32 (the sequence T T T T T). Therefore, A's probability of losing is 10/32 = 5/16 is she starts first.

Hope that helps.
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 [#permalink] New post 11 Apr 2005, 09:38
kapslock, take a look at my reasoning.
What do you think? It seems to me that you forgot the second player or limited to 3 tosses each instead of 5 ones.
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 [#permalink] New post 11 Apr 2005, 11:05
thearch wrote:
kapslock, take a look at my reasoning.
What do you think? It seems to me that you forgot the second player or limited to 3 tosses each instead of 5 ones.


Holy Cow !!!

You're right, it was 5 tosses each. So dumb of me to overlook that. I really really need to apply myself, else I am going to screw it all up.

I don't really see your reasoning theArch. Were they only the 2nd and 3rd lines in this post of yours (that I am quoting) that you referred to as reasoning, or some separate post? (I can't see it, as of now).

Would post my corrected solution in a minute.

Sorry for the goof up.
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 [#permalink] New post 11 Apr 2005, 11:13
I didn't write my reasoning out, just drew it (in the sense that you should add the bolded probabilities)
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Re: Probability A and B Toss a Coin? [#permalink] New post 11 Apr 2005, 11:14
(1). The person who starts wins in the following sequences (with the probability of occurring given in brackets)

H (1/2)
TTH (1/8)
TTTTH (1/32)
TTTTTTH (1/128)
TTTTTTTTH (1/512)

Therefore total probability of winning
= (256 + 64 + 16 + 4 + 1)/512
= 341/512

(2) The chance of a draw is in the sequence
TTTTTTTTTT (1/1024)

Since the starter's chance of a win + starter's chance of a draw + starter's chance of a defeat = 1,
341/512 + 1/1024 + starter's chance of a defeat = 1
or 683/1024 + starter's chance of a defeat = 1
=> starter's chance of a defeat = 341/1024

The odds of defeat of the starter is therefore, 341 to 683.

Hope that helps.

P.S. theArch, now I see your "reasoning" (I wasn't logged in when I saw your post). Phew !! Did you use AutoCad to draw that one?? :)
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 [#permalink] New post 11 Apr 2005, 11:17
mmmm, I didn't use Autocad, only thought it was difficult to explain with my poor English
  [#permalink] 11 Apr 2005, 11:17
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