A, B, C are flipping a coin independently until they got a head (same experiment "to get a head" is repeated by these 3 people). Denote X, Y, Z stand for the number of flips they need, respectively. Find the probability P(X<Y<Z).
My solution is that they form a permutation like XYZ or XZY etc. Each combination has same probability, so the P(X<Y<Z) is equal to 1/6. But if we considered 2nd solution which integrate the multibinomial distribution $P(X=x, Y=y, Z=z) = (1/2)^{(x+y+z)}$ over the domain that $1<=x<y<z<inf$, the answer is 1/21. So what's wrong with the first solution? or is there any method without the integration? Thanks!
The first solution is wrong (because it ignores ties), but is also useful. Given that there are no ties, the probability of $X<Y<Z$ is indeed $1/6$ by symmetry. And so if the probability of there being any ties is $p$, then the correct answer should be $(1-p)/6$.
The probability of there being any ties is $$ P[X=Y]+P[X=Z]+P[Y=Z]-2P[X=Y=Z]=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}-\frac{2}{7}=\frac{5}{7}, $$ where each term is found by summing a simple geometric series. And so the correct answer is $(1-5/7)/6=(2/7)/6=1/21$.