# Guessing the outcome of a pair of dies using the outcome of another pair of dies

I have two dies with four faces, in wich are written the numbers ${1,2,3,4}$. The total outcome, $X$, is the sum of the outcomes of each die, $X=X_1+X_2$. If a friend wants to guess the outcome of my dies, $X$, using the outcome of his own dies (which have also four faces), how can I calculate the probability that his outcome $Y$ has the same value that the outcome I got, $P(Y=x|X=x)$?

I see that $X$ and $Y$ have the same probability distribution, but I don’t see how to get this conditional probability…

• Maybe I’m misunderstanding, but $X$ and $Y$ seem independent, so the conditional probability would be the same as the probability. Rolling a pair of dice doesn’t help anyone improve their guess as to what someone else rolled with their pair of dice. Unless each of you have identical dice which are not fair (but with unknown probability distributions). If so, maybe this is more interesting. – Steve Kass May 16 '18 at 18:16

The most reasonable hypothesis in absence of more information is that you and your friend's results are independent, which results in $$P(Y=x|X=x)=P(Y=x).$$

Basic approach. Presumably, you know how to obtain the probability distribution

$$f_X(x) = P(X = x)$$

And you have correctly recognized that your friend's dice will produce the same distribution

$$f_Y(y) = P(Y = y) = f_X(y)$$

So then you simply have to add up the probabilities that they are equal, for every value that they might take on (from $2$ through $8$). Assuming independence, this is simply the square that your set of dice produce each value. That is, the desired probability is

$$p = \sum_{x = 2}^8 [f_X(x)]^2$$

HINT

Find that probability distribution (e.g. $P(X=2)=\frac{1}{16}$, $P(X=3)=\frac{2}{16}$, etc.), and then calculate

$$\sum_{x=2}^8 P(X=Y=x)$$

where by $P(X=Y=x)$ I of course mean $P(X=x \cap Y=x)$, which works out to:

$$P(X=x \cap Y=x)= P(X=x)\cdot P(Y=x)=P(X=x)\cdot P(X=x) = P(X=x)^2$$

So, you need to calculate:

$$\sum_{x=2}^8 P(X=x)^2$$

Intuitively, this makes sense: what is the probability of getting the same outcome? It is the probability of both of you getting $2$, or both of you getting $3$, or both getting $4$ ...