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I am working out a problem which states: "Let X1, X2, and X3 be independent and identically distributed random variables with the uniform distribution on [0, 1]. What is the probability P{X1 is the largest}?" I checked the answer and it is equal to 1/3. Now,clearly, the values of all three random variables will lie in the range [0,1].

I then let X1=a, X2=b and X3=c, and consider the following NINE cases: a,b and c all equal; b and c are equal and a is greater than them; b and c are equal and a is less than them; a and b are equal and they are greater than c; a and b are equal and they are less than c; a and c are equal and they are less than b; a and c are equal and they are greater than b; a is greater than b which is greater than c; c is greater than b which is greater than a.

Then I work out the probability as 2/9, since there are only two outcomes out of the above nine where a is largest(the ones in bold). How can the answer then be equal to 1/9? Please help

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    You cannot simply count the number of distinctt events because they have different probability masses. In particular, all but three of the above have zero probability mass. – 2017-01-18
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    I buy a ticket in a million dollar lottery. There are only two outcomes: I win and I don't. Therefore do I have a 1/2 probability for winning? – 2017-01-18

4 Answers 4

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Because the variables are independent and identically distributed random variables over $[0, 1],$ the probability of any of them being equal to each other is considered 0 (they might be off by $0.0000001,$ or $0.1,$ or anything else but you don't know; all cases have an infinitesimally low probability). So the only cases are $a>b>c,$ $a>c>b,$ $b>a>c,$ $b>c>a,$ $c>a>b,$ and $c>b>a.$ So the probability is indeed $\frac{2}{6}=\frac{1}{3}.$

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    That makes sense. I was missing out on some cases also. Thanks! – 2017-01-18
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Only exchangeability and the fact that $P(X_i=X_j)=0$ for $i\neq j$ is needed. Indeed, in that case $$(X_{p_1},X_{p_2},X_{p_3})$$ has the same distribution for any permutation $p$ of $\{1,2,3\}$. There are $3!=6$ such permutations, and $2$ of those have $X_1$ in the same fixed (say, last) position.

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By symmetry we have \begin{align*} P(\max(X_i) = X_1) = P(\max(X_i) = X_2) = P(\max(X_i) = X_3) \end{align*} Since these three sets disjointly cover the state space, we must have \begin{align*} 1 = P(\Omega) &= P(\{\max(X_i) = X_1\} \cup \{\max(X_i) = X_2\} \cup \{\max(X_i) = X_3\}) \\ &= P(\max(X_i) = X_1) + P(\max(X_i) = X_2) + P(\max(X_i) = X_3) \\ &= 3 P(\max(X_i) = X_1) \end{align*} Dividing both sides by 3, we have $P(\max(X_i) = X_1) = \frac{1}{3}$.

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A simulation of a million triples $(X_1, X_2, X_3):$ On any one iteration $X_1$ has one chance in three of being the maximum. (With a million iterations, one can expect at least three-place accuracy.)

m = 10^6;  x1=runif(m);  x2=runif(m);  x3=runif(m)
w = pmax(x1, x2, x3);  mean(x1==w)
## 0.333267

Also (+1) each for @DanielXiang (logically closest to this), @MartinBladt, and @pie314271.