# Conditional probability question?

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A game is played with one standard die. On a turn, if a player rolls a 1, he gets two more rolls. If he rolls a 2, he gets one more roll. If he rolls anything else, his turn ends. His score is the sum of all the rolls on his turn (up to 3 rolls). If a player scores 5 points on a turn, what is the probability that it took 2 rolls?

A game is played with one standard die. On a turn, if a player rolls a 1, he gets two more rolls. If he rolls a 2, he gets one more roll. If he rolls anything else, his turn ends. His score is the sum of all the rolls on his turn (up to 3 rolls). If a player scores 5 points on a turn, what is the probability that it took 2 rolls?

##### 2 Answers

We use our good friend Bayes' rule to help us deduce:

#### Explanation:

**Warning: Long answer ahead!**

One form of Bayes' rule states the following:

This allows us to write a conditional probability of "B given A" in terms of "A given B", which may be easier to calculate. In this question,

B = "die rolled twice", and

A = "score is 5".

Let's write this into the equation:

The numerator on the RHS is easy to deduce; however, the denominator, in its current state, is not. But we can make it easier. We recall that, if

We actually have that here; the

Kind of a "the whole is equal to the sum of its parts" thing.

From hereon, let's use the shorthand

Now, we do the calculations!

Finally, we place these values back into the equation for

For completeness, *had* to take either 1, 2, or 3 rolls.

## Bonus:

There's a great Numberphile video on YouTube discussing Bayes' rule here:

Alternate explanation to the long version I previously posted.

We still get

#### Explanation:

What are the different ways to score 5 points, and what are their probabilities? Well, we could roll:

- a 5 right away;
- a 2, then a 3;
- a 1, then a 1, then a 3;
- a 1, then a 2, then a 2; or
- a 1, then a 3, then a 1.

So the probability of scoring 5 points is simply the sum of the probabilities above. The final answer we seek is **the fraction of this sum belonging to the two-roll method**—in other words, the relative probability of "scoring 5 in two rolls" to "scoring 5 in any way".

Using conditional probability, we get:

which is the same answer as before.

This method is easier in this case; however, the Bayesian method is more general and has much broader applicability.