This math problem had me going for a bit. Looked at from a distance, it looked like one thing; and when I had the occasion to sit down and hash it out, it was quite another.

A student submitted a project that was of her game which was played with just two dice. If you roll a 2 or 12 you win; but if you roll any sum from 5 to 9 you lose; and if you roll a 3, a 4, a 10, or an 11, you survive to another round. You are limited to a maximum of three turns to roll a winning total.

It was not the normal Bernoulli trial, since this doesn’t just have the two states of success and failure; but it introduces a third state, which we will call “survival”. While you don’t “win” if you survive, you can still play again, but you can’t go past 3 turns. Three survivals in a row gets you nothing. You need to get a 2 or a 12 in three turns or less.

$latex P(2 or 12) = \frac{1}{18}$, the probability of winning on your first try. Even though there are three states, we can still discuss wait time. In this context, it can be 0, 1, or 2. With a wait time of 1, $latex P(2 or 12) = P(3, 4, 10, or 11)\times (\frac{1}{18})=(\frac{5}{18})\times(\frac{1}{18}) \approx 0.015432$.

It means that to even make it to the second turn you can only get there with a 3, 4, 10, or 11. If you got 2 or 12 on the first try, there would be no need for a second turn. Similarly, to get to the third turn, you needed to survive twice and win the third time: $latex (\frac{5}{18})^2\times(\frac{1}{18}) \approx 0.00428669$.

It made sense, but I still wasn’t sure about this. What about the probability of losing completely? Those are the numbers from 5 to 9, which has a probability of $latex \frac{2}{3}$. Why wasn’t I making use of that information?

I don’t think it was necessary in computing the probabilities I did, since the winning conditions preclude rolling any sum between 5 and 9. But it can come in handy as a check. A good indicator that I am on the right track is to see if expectations of winning and losing for 1000 trials, add up to 1000. For winning, I need to add up the probabilities for all 3 wait times: $latex \frac{1}{18} + \frac{5}{324}+\frac{25}{5832} = \frac{439}{5832}\approx 0.0752743$. For 1000 games, the expectation of winning is: 75.2743 games.

The expectation of losing is similarly calculated, based on a single-turn probability of 2/3: $latex \frac{2}{3} + \frac{2}{3}\times \frac{5}{18} + \frac{2}{3}\times \left(\frac{5}{18}\right)^2 = \frac{439}{486}\approx 0.9033$. This means you will lose, on average 903 out of every 1000 games, or an expectation of 903.3. Notice that for 1000 games, the winning and losing conditions don’t add to 1000. What we didn’t add was survival until the third turn: $latex \left(\frac{5}{18}\right)^3 = \frac{125}{5832}\approx 0.02143$. This means that you will “survive” (but not win)21.43 out of every 1000 games. With enough decimal precision, we do indeed get 1000 games when we add all these numbers up, or at the very least add the expectation using the fractions: $latex 1000 \times \left(\frac{439}{5832} + \frac{439}{486} + \frac{125}{5832}\right) = 1000 \times \left(\frac{439}{5832} + \frac{5268}{5832} + \frac{125}{5832}\right) = 1000$.