# Rolling Two Dice You have a pair of regular dice and roll them repeatedly. Which event is more likely to happen first?:
• You roll a twelve (double six; also known as ‘box cars’ in craps)?
• You roll seven, twice in a row?
• Or, are both events equally likely?
At first glance you might be tempted to say that both events are equally likely. There are 36 ways that two dice can land, and only one way to make twelve (double six), so there is a 1/36 chance of rolling box cars.
Conversely, there are six ways to achieve a score of seven: {1,6}, {2,5}, {3,4}, {4,3}, {5,2}, {6,1}, so the chances of rolling seven on one roll is 1/6. Surely this make the chances of rolling two sevens in a row as (1/6)2, which is also 1/36 (the same chance as rolling the twelve)? … … Not so fast!
If you roll twelve, you win with a single independent roll. However, to win by rolling a seven you need to have previously already rolled a seven. (After the first seven is rolled, you could end up rolling a twelve and ending the game at that point). Remember, we’re looking for the chance that one event occurs before the other.

## Probabilities

Lets enumerated the possibilities. First, we’ll look at the basic probabilities. On any single roll, there is a:
• 1/36 chance of rolling {a count of 12}
• 1/6 chance of rolling {a count of 7}
• 29/36 chance of rolling something else {not a seven, and not a twelve}

## Multiple Rolls

The key to solving this is to realize that, other than the winning states, most of the rolls are unimportant, and rolling one of these other numbers simply sets resets things to try again. We can build up a formula that is self-referential.
Let’s look at the probabilities of a twelve winning. One of four things could happen:
1. You roll a twelve. There is a 1/36 chance of this. Game over.
2. You roll a seven, and then roll a twelve (1/6 × 1/36). Game over.
3. You roll a seven, then roll something that is neither a seven nor a twelve (1/6 × 29/36). Neither wins, so we start again.
4. You roll something that is neither a seven nor a twelve (29/36); again we start again at the initial state.
If P12 is the probability of twelve winning, then:
P12 = (1/36) + (1/6 × 1/36) + (1/6 × 29/36) P12 + (29/36)P12
P12 = (1/36) + (1/216) + (203/216)P12
216P12 = 7 + 203P12
P12 = 7/13
Now let’s look at the probabilities of consecutive sevens winning. One of three things can happen:
1. You roll a seven, followed by another seven. (1/6 × 1/6). Game over.
2. You roll a seven, then roll something that is neither a seven nor a twelve (1/6 × 29/36). Neither wins, so we start again.
3. You roll something that is neither a seven nor a twelve (29/36); again we start again at the initial state.
P7 = (1/36) + (1/6 × 29/36)P7 + (29/36)P7
P7 = (1/36) + (203/216)P7
216P7 = 6 + 203P7
P7 = 6/13