# Probability and expected value of rolling two dice with a twist

Let this be a game with two phases.

In phase 1, roll two 6-sided dice and compute the diﬀerence between the rolls. Call this diﬀerence x. In phase 2, roll x dice, and add up the total of the rolls. This is the payout in dollars of the game.

In the ﬁrst phase of the game, x can take on any value 0 to 5. Compute the probability (as a fraction) Now for the second phase of the game, for each value of x, what is the average payout (expected value) of the game?

Now I believe I got the right probabilities for 0 - 5, but I'm questioning if i'm doing the second part correctly.

When $$x = 0$$ I said $$E(X) = 0$$

And for $$x = 1$$ I said that $$E(X) = \frac{10}{36} * 3.5$$

Is this correct? My reasoning was the expected value for rolling a 6 sided die, would be 3.5 and since we have a $$\frac{10}{36}$$ of getting a value of 1 in the first part we need to multiply 3.5 by it.

Is this the correct logic? Thanks!!

• How do you count the difference? It is always the absolute value or can be the difference negative as well? May 11 '19 at 16:16
• Always the absolute value May 11 '19 at 16:17
• Your reasoning is correct. Just finish the the calculation. May 11 '19 at 16:19
• The average payout is $\frac{245}{36}\approx 6.81$ May 11 '19 at 16:28

You have two random variables here: one is the random absolute difference of the two die rolls in the first step, and the second is the random payout in the second step. Let's call the first one $$X$$ and the second $$Y$$, to be consistent with the notation set forth in the question.
Then the question is asking you to calculate $$\operatorname{E}[Y \mid X = x], \quad x = 0, 1, \ldots, 5.$$ You are calculating $$\operatorname{E}[Y \mid X = x]\Pr[X = x]$$ which is fine if you want to use this information to obtain the unconditional expectation $$\operatorname{E}[Y] = \sum_{x=0}^5 \operatorname{E}[Y \mid X = x]\Pr[X = x].$$
For $$x = 0$$ and $$x = 1$$, you correctly have $$\operatorname{E}[Y \mid X = 0] = 0 \\ \operatorname{E}[Y \mid X = 1] = \frac{7}{2}$$ but what are the rest? There is a pattern. How do you justify it?
Then, how do you get $$\Pr[X = x]$$ for each $$x = 0, 1, \ldots, 5$$?
• For Pr[X=x] for each x=0,1,…,5? I got $\frac{6}{36}$$\frac{10}{36}$$\frac{8}{36}$$\frac{4}{36}$$\frac{6}{36}$$\frac{2}{36}$ Now, I believe what you are saying is that I should find the probability of each Y given a specific X? So the formula would be: $E[Y∣X=x] = \frac{p(Y \land x)}{p(x)}$ May 11 '19 at 17:27
• @theOracle The question only appears to ask for each conditional expectation, which is $\operatorname{E}[Y \mid X = x]$. So you will have six values, one for each $x$ in $\{0, 1, 2, 3, 4, 5\}$. It doesn't seem to be asking for the unconditional expectation, which is the weighted average of the conditional expectations, weighted by the probability of obtaining $x$ dice to roll from the first step. Finally, you cannot write $\operatorname{E}[Y \mid X = x] = \frac{\Pr[Y \cap X]}{\Pr[X]}$. The LHS is an expectation. The RHS is a ratio of probabilities. May 11 '19 at 17:58