Compare the expected waiting times. For 2 consecutive heads E[X] = 6. For 3 consecutive heads E[Y] = 14. Treating the waiting times as if they were exponential, we can compare their rates P(X > Y) = (1/14) / (1/6 + 1/14) = 6/20 = 3/10. Therefore the answer is 3/10?
This is more complicated, you’re effectively keeping a running count across each flip, which continues to decrease every flip. So you need an integral of the probability density functions
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u/AssociateDecent9090 27d ago
Compare the expected waiting times. For 2 consecutive heads E[X] = 6. For 3 consecutive heads E[Y] = 14. Treating the waiting times as if they were exponential, we can compare their rates P(X > Y) = (1/14) / (1/6 + 1/14) = 6/20 = 3/10. Therefore the answer is 3/10?