This question is really hard for me to get my head around.

A fair six-sided die is rolled and let $X$ denote the score. The die is rolled again repeatedly until the number on the die is greater than or equal to $X$. let $Y$ denote this number.

We have to find $P(Y=5|X=2)$ and to find the the value of $P(Y=5)$ using the theorem of total probability.

So am I looking for the probability that the first roll greater than 2 is a roll of 5?

Any help is appreciated.

  • $\begingroup$ the probability that the first roll greater than or equal to $2$ is is a roll of $5$ $\endgroup$ – Henry Feb 14 '18 at 20:37

That is essentially what the question asks.

After the die is rolled for the first time and the number seen is $X$, the value of $Y$ essentially acts like when rolling a fair die with $6 - X + 1$ faces. (For instance, if your first roll was a 6, you effectively roll a 1-sided die with one face being six, and you will always get 6; if the first roll was 5, you roll a two-sided die--or flip a coin if you prefer--with one side labeled "5" and the other labeled "6".) Use this to compute that conditional probability.

To use the law of total probability, get all of the conditional probabilities $P(Y = 5 | X = x)$ using the trick above. You already know that $P(X = x) = 1/6$ for $x \in \{1, \ldots, 6\}$. So you want to compute the sum:

$$\sum_{x = 1}^6 P(Y = 5 | X = x) P(X = x) = \frac{1}{6} \sum_{x = 1}^6 P(Y = 5 | X = x)$$

I've already computed the answer: it is $\frac{29}{120}$. You can show the work.

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