Question

Three friends go out for a coffee. To choose the one that will pay the coffee...

Three friends go out for a coffee. To choose the one that will pay the coffee they play the following game: each one flips a coin, and whoever gets a different result will have to pay, if everyone gets the same result they will continue to throw until someone loses. The coin is not balanced, the face probability is 0.60.
Calculate the probability that less than 5 games have to be played before one of them has to pay.

Math   Statistics-And-Probability   
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Answer #1

Given that

p = 0.60

q = 1-0.60 = 0.40

flips a coin

Probability that one of them will pay

(HHT) : (HTT) : (HTH) : (THH) ;(TTH); (THT)

(HHH) (TTT)

probability of HHT or HTT :

=3(p^2 *q) + 3(q^2 * p)

= 3(0.6*0.6*0.4) +3(0.6*0.4*0.4)

=3*0.144 + 3*0.096

=0.432 + 0.288

=0.72

Probability of HHT or HTT (successess) = 0.72

probability (failure) = 1-0.72 = 0.28

Probability that they get success in 1st game = 0.72

Probability that they get success in 2nd game = 0.28 * 0.72 = 0.202

Probability that they get success in 3rd game = 0.28*0.28 * 0.72 = 0.056

Probability that they get success in 4th game = 0.28 * 0.28 * 0.28 * 0.72 = 0.01568

probability that less than 5 games have to be played before one of them has to pay. = 0.72 + 0.202 + 0.056 + 0.01568

probability that less than 5 games have to be played before one of them has to pay. = 0.9937

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