Q. 265.0( 4 Votes )

# A, B and C in order toss a coin. The one to throw a head wins. What are their respective chances of winning assuming that the game may continue indefinitely?

Answer :

Given that A, B and C toss a coin until one of them gets a head to win the game.

Let us find the probability of getting the head.

⇒ P(A_{H}) = P(A getting a head on tossing a coin)

⇒

⇒ P(A_{N}) = P(A not getting head on tossing a coin)

⇒

⇒ P(B_{H}) = P(B getting a head on tossing a coin)

⇒

⇒ P(B_{N}) = P(B not getting head on tossing a coin)

⇒

⇒ P(C_{H}) = P(C getting a head on tossing a coin)

⇒

⇒ P(C_{N}) = P(C not getting head on tossing a coin)

⇒

It is told that A starts the game.

A tosses in1^{st},4^{th},7^{th},…… tosses.

This can be shown as follows:

⇒ P(W_{A}) = P(A wins the game)

⇒ P(W_{A}) = P(A_{H}) + P(A_{N}B_{N}C_{N}A_{H}) + P(A_{N}B_{N}C_{N}A_{N}B_{N}C_{N}A_{H}) + …………………

Since tossing a coin by each person is an independent event, the probabilities multiply each other.

⇒ P(W_{A}) = (P(A_{H})) + (P(A_{N})P(B_{N})P(C_{N})P(A_{H})) + ………………

⇒

⇒

The terms in the bracket resembles the infinite geometric series sequence:

We know that the sum of a Infinite geometric series with first term ‘a’ and common ratio ‘r’ is

⇒

⇒

⇒

⇒

B tosses in2^{nd},5^{th},8^{th},…… tosses.

This can be shown as follows:

⇒ P(W_{B}) = P(B wins the game)

⇒ P(W_{B}) = P(A_{N}B_{H}) + P(A_{N}B_{N}C_{N}A_{N}B_{H}) + P(A_{N}B_{N}C_{N}A_{N}B_{N}C_{N}A_{N}B_{H}) + …………………

Since tossing a coin by each person is an independent event, the probabilities multiply each other.

⇒ P(W_{B}) = (P(A_{N})P(B_{H})) + (P(A_{N})P(B_{N})P(C_{N})P(A_{N})P(B_{H})) + ………………

⇒

⇒

The terms in the bracket resembles the infinite geometric series sequence:

We know that the sum of a Infinite geometric series with first term ‘a’ and common ratio ‘r’ is

⇒

⇒

⇒

⇒

⇒ P(W_{C}) = P(winning of C)

⇒ P(W_{C}) = 1-P(W_{A})-P(W_{B})

⇒

⇒

∴ The chances of winning of A,B and C are .

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