0
$\begingroup$

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

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

enter image description here

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

$\endgroup$
4
$\begingroup$

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}

whereupon

\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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.