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A,B,C tosses a coin in turns. The first one to throw a head wins game. What are their respective chances of winning.

A,B,C tosses a coin in turns. The first one to throw a head wins game. What are their respective chances of winning.

Grade:Upto college level

1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
9 years ago
Let H be the occurrence of Head, P(H) = p = 1/2
and T be the occurence of Tail, P(T) = q = 1/2
A starts the game,

A wins in the following cases :
----------H in the 1st toss
Or------T in 1st toss & T in 2nd toss & T 3rd toss & H in 4th toss
Or------T in 1st toss & T in 2nd toss & T 3rd toss & ... H in 7th toss
Or------T in 1st toss & T in 2nd toss & T 3rd toss & ... H in 10th toss
.
.
ad infinitum

So, P(winning of A) = p + qqqp + qqqqqqp + ...
= (1/2) + (1/2)^4 + (1/2)^7 + ...
= (1/2)/{1 - (1/2)^3} = 4/7 [ using sum of infinite terms of a GP = a/(1 - r) ]


B wins in the following cases :
----------T in the 1st toss & H in the 2nd toss
Or------T in 1st toss & T in 2nd toss & T 3rd toss & T in 4th toss & H in 5th toss
Or------T in 1st toss & T in 2nd toss & T 3rd toss & ... H in 8th toss
Or------T in 1st toss & T in 2nd toss & T 3rd toss & ... H in 11th toss
.
.
ad infinitum

So, P(winning of B) = qp + qqqqp + qqqqqqqp + ...
= (1/2)^2 + (1/2)^5 + (1/2)^8 + ...
= (1/4)/{1 - (1/2)^3} = 2/7 [ using sum of infinite terms of a GP = a/(1 - r) ]


Since winning of A, B, C are mutually exclusive and exhaustive (i.e. two of them can not win simultaneously and one of them must win)
So, P(winning of C) = 1 - [(4/7) + (2/7)] = 1/7
Thanks and Regards
Shaik Aasif
askIITians Faculty

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