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Fair play (Posted on 2011-12-30) Difficulty: 3 of 5
Two gamblers, A and B, divided into a few bundles the money they've recently won and agreed to toss a coin for each bundle.
A 3rd gambler, nicknamed Mathwiz remarked: "It is an even money that A will win at least five of the tosses".
How many bundles of money were there?

See The Solution Submitted by Ady TZIDON    
Rating: 3.0000 (3 votes)

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Solution Comment 3 of 3 |
Assuming that:

1. even money means that probability of winning at least five of the tosses is 1/2;
2. probability of tails and heads in each toss is the same, i.e. 1/2

then

the probability of A winning at least 5 of the tosses is the same as A losing all of them plus winning 1plus winning 2 plus winning 3 plus winning 4;

then , we have:

(1/2)**n)*[1+n+(n*(n-1)/2)+(n*(n-1)*(n-2)/6)+(n*(n-1)*(n-2)*(n-3)/24) = 1/2

The value of n that satisfies the above equation is n=9.

There were 9 bundles of money.

  Posted by zoccer on 2011-12-31 09:08:24
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