5 dice are rolled, resulting in outcomes $X_1,X_2,X_3,X_4,$ and $X_5$

We need to find $Pr(min(X_1,X_2)>=min(X_3,X_4))$

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The picture below is the professor's approach to the problem, but I'm not understanding the logic in the first line. I'm pretty terrible at these dice roll questions that involve min's, so any help would be great


There's a kind of principle of indifference going on here. Suppose you have two independent random variables $X$ and $Y$ with the same distribution. There are three possibilities: $X > Y, X = Y, X < Y$; the probabilities of these three events sum up to $1$ (by total probability). By symmetry, $P(X > Y) = P(X < Y)$. Thus,

\begin{align} P(X \geq Y) & = P(X > Y) + P(X = Y) \\ & = P(X < Y) + P(X = Y) \end{align}


\begin{align} 2P(X \geq Y) & = P(X > Y) + P(X = Y) + P(X < Y) + P(X = Y) \\ & = 1 + P(X = Y) \end{align}

which is a model for the first line of your professor's reasoning.

  • $\begingroup$ makes perfect sense, thanks $\endgroup$ – Jonnathan Baquero May 2 '17 at 19:45

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