Dear Uncle Colin,

Player 1 rolls a fair die with 200 sides. Player 2 rolls a fair die with 300 sides. What’s the probability that player 1 rolls a higher number than player 2?

- Probably A Simple Calculation Applying Logic

Hi, PASCAL, and thanks for your message!

Let’s ignore for the moment that a D200 and D300 would be enormous, difficult to make fair and practically impossible to stop rolling – as mathematicians, we care little for such practical considerations.

This is a problem that requires a little care, but does drop out nicely in a few different ways. I’ll go through two of them here.

### Way 1: Direct counting

Suppose player 1 rolls $k$. What’s the probability that player 2 rolls a smaller number? It’s $\frac{k-1}{300}$.

So we can get the probability of player 1 winning by adding up all of those probabilities (multiplied by their probability of rolling $k$, which is always $\frac{1}{200}$.)

We have $\frac{1}{200} \sum_{k=1}^{200} \frac{k-1}{300}$. I can take the 300 outside as well, leaving a simple 200-term arithmetic series that starts at 0 and ends at 199.

I’ll write down what I’m doing: $\frac{1}{200} \times \frac{1}{300} \times (0 + 1 + 2 + \dots + 199)$

The sum of an arithmetic series with first term $a$, last term $L$ and $n$ terms is $\frac{n}{2}(a+L)$, and here that’s $\frac{200}{2}\times(199)$.

So our sum is $\frac{1}{200} \times \frac{1}{300} \times \frac{200}{2} \times 199$, which boils down nicely to $\frac{199}{600}$ – a little under a third.

### Aside: why “a little under a third” is a reasonable answer

A third of the time, player 2 rolls over 200 and wins automatically.

Draws are fairly rare, so player 1 should win about half of the remaining games, a third of the total. This is a slight overestimate, because we’ve ignored the rare draws. We can account for those carefully, though.

### Way 2: Probabilities

Let’s look at it from player 2’s point of view:

• If player 2 throws more than 200, they win automatically. That’s a one-in-three chance.
• Otherwise, whatever player 2 throws, there’s a 1/200 chance that player 1 will throw exactly the same. The chance of a draw is $\frac{2}{3} \times {1}{200} = \frac{1}{300}$.
• Of the remaining possible outcomes, player 1 and player 2 each win exactly half.

Let’s imagine they play 300 games ((I can see that 600 would work better, but I’m pretending otherwise; the 300 is a logical number to pick from the setup)). On average:

• 100 will be an automatic win for player 2
• 1 will be a tie
• The remaining 199 will be split evenly between the two players.

So on average, over 300 games, player 1 wins 99.5, which shakes out to a probability of $\frac{199}{600}$, as before.

Hope that helps!

- Uncle Colin