Two Games of Chance

Two friends spend a wet afternoon with a pair of dice.

1. In the first game each rolls a single die. What is the chance that one named player rolls strictly higher than the other?

2. In the second they roll both dice and add the scores. Which total should they bet on?

3. In the third they put the dice away and instead draw a rectangle, choosing each of its two side lengths at random somewhere between $0$ and $1$ inch. What is the chance that the rectangle’s diagonal is shorter than an inch?

Solution

Part 1. There are $36$ equally likely pairs of rolls. The two players tie in $6$ of them, a chance of one in six. By symmetry the remaining $30$ split evenly between "the first rolls higher" and "the second rolls higher", so each has probability $15/36 = 5/12$.

Part 2. The total ranges from $2$ to $12$. Counting the ways to make each, the total $7$ arises most often, in $6$ of the $36$ cases (again one in six), more than any other total; $6$ and $8$ come next, with five ways each. So bet on $7$.

Part 3. Picture the two side lengths as a single point chosen uniformly in the unit square, its coordinates the two lengths. The diagonal $\sqrt{x^2 + y^2}$ is shorter than $1$ exactly when the point lies inside the quarter circle of radius $1$ drawn from the corner. The chance is the quarter circle’s area divided by the square’s area, $$\frac{\pi/4}{1} = \frac{\pi}{4} \approx 0.79,$$ a little under four times in five.