What is the Probability that Two Independent Coin Tossers Stop on the Same Toss?

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SUMMARY

The probability that two independent coin tossers, A and B, stop on the same toss can be calculated using the negative binomial distribution. Specifically, the solution involves summing the probabilities from n = 1 to ∞ for both A and B stopping on the nth toss. This approach leverages the independence of the tosses and the defined probabilities of heads (p) and tails (q = 1 - p). The discussion clarifies the method for determining this probability effectively.

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Problem:
Suppose that the probability that a head appears when a coin is tossed is p and the probability that a tail occurs is q=1-p. Person A tosses the coin until the first head appears and stops. Person B does likewise. The results obtained by persons A and B are assumed to be independent. What is the probability that A and B stop on exactly the same number toss?

I am not quite sure how to solve this. I'm assuming that the tosses are distributed as a negative binomial and I'm lost as to how to put everything together.Please help !

Thanks!
 
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Hi FaradayLaws! :smile:
FaradayLaws said:
What is the probability that A and B stop on exactly the same number toss?

It's the sum from n = 1 to ∞ of the probability that A and B both stop on the nth toss. :wink:
 

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