# What are the probabilities of winning for each player?

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Robin and Williams are playing a game. An unbiased coin is tossed repeatedly. Robin wins as soon as the sequence of tosses HHT appears. Williams wins as soon as the sequence of tosses HTH appears. The game ends when one of them wins. What are the probabilities of winning for each player?

posted Dec 17, 2014

Ans. (Robin) HHT - 2/3 (Williams) HTH - 1/3

Let the probability of Robin winning be p. The probability of Williams winning is (1-p). If the first toss is tails, it is as good as the game has not started, hence the probability of Robin winning is p after the first tail.

p = 1/2*p + ….

Let the first toss be heads. If the second toss is heads, then Robin definitely wins. Since HH has occurred, and at some point, tails will occur, so HHT will occur. Hence Robin wins with probability 1 for HH.

p = 1/2*p + 1/2 (1/2*1 + ….)

Let the second toss be tails. If the third toss is heads, Robin loses as HTH occurs. If the third toss is tails (HTT) - since two tails have occurred in a row, now it is as good as the game has started from the beginning, so the chances of Robin winning are back to p.

p = 1/2*p + 1/2 (1/2*1 + 1/2*(1/2*0 + 1/2*p))

solving this equation gives us p=2/3

Ans: 0.5 (50%)

Probability that HHT will appear = 1/8
Probability that HTH will appear = 1/8
Probability that both will not appear = 6/8

Robin can win the game in first trial or in second trial or in third........
So probability =
1/8 + (6/8 ×1/8) +(6/8×6/8×1/8)+ (6/8×6/8×6/8×1/8)......
= sum of this infinite GP whose common ratio is 6/8 and first term is 1/8.
Sum= first term/(1- common ratio)
= (1/8)/(1 - 6/8 )
=1/2
= 0.5

Same for Williams...

Why we should toss again say we tossed three times and got some H/T combination and none wins. Now we should toss only once more and last two (some H/T combination) plus this toss will decide the result.

Sorry for my understanding of the question and your answer, though not checked the working....
How second toss will decide the result.
Take this case... In first toss TTT , second toss HHH , again third time TTT.....

Sorry but I didn't understood what were you said exactly.
Both of us has not understood each other :)

One time only one toss either T or H so first result would be after third toss after their after every toss can generate the result along with previous two toss.
Ohh! You are right sir.
Means he can win in first three tosses or in first four tosses or in first five tosses and so on.....
So I think probability will be-
(1/2×1/2×1/2)+(1/2×1/2×1/2×1/2)+(1/2×1/2×1/2×1/2×1/2).......
Sum= (1/8)/(1- 1/2)
=1/4
= 0.25
I think 0.25 be the answer.
Am I right sir?
I dont know the answer, it been 20 years I left probability :)
Probably answer again so Shashi Kant get the intimation and hope he know the answer.

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