# What is the probability that $X<Y$?

If two dice are rolled repeatedly, and $X$ is the number of tosses until $3$ is rolled, and $Y$ is the number of tosses until a $5$ is rolled, what is the probability$(X < Y)$? Also, if $Z$ is the number of rolls until a $10$ is rolled, what the probability$(X < Y < Z)$?

Is $P(X < Y) = \sum_{n=2}^{\infty}\sum_{k=1}^{\infty}((32/36)^{(n-1)}(4/36)-(34/36)^{(k-1)}(2/36)]$?

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Is it assumed that $X$, $Y$ and $Z$ are independent ? – Sasha Sep 23 '11 at 16:50
Yes X Y and Z are independent as the dice are balanced – lord12 Sep 23 '11 at 16:50
No, $X$, $Y$ and $Z$ are not independent. For example, if $X = 1$ then $Y \ne 1$ and $Z \ne 1$. – Robert Israel Sep 23 '11 at 18:40
It's not clear whether "a 3 is rolled" is referring to a 3 coming up on at least one of the two dice, or the sum of the two being 3. But since "a 10 is rolled" can only refer to the sum, I guess we can assume the 3 and 5 are also referring to the sum. – Robert Israel Sep 23 '11 at 18:43
@Robert: Those were my thoughts too. But also notice that the first line says "until 3 is rolled" and not "until a 3 is rolled". So I suppose this indeed refers to the sum being 3, rather than one of them being 3. – TMM Sep 23 '11 at 18:47

(a) $P(X<Y)$

Suppose $q = P(3 \text{ before } 5)$. When will we see a $3$ before we see a $5$? That is if we first see a $3$ on the first roll (we are done), or if we don't see a $3$ or a $5$ on the first roll and then see a $3$ before a $5$ after that. In other words:

$$q = P(3) \cdot 1 + P(5) \cdot 0 + P(\text{neither } 3 \text{ nor } 5) \cdot q$$

Since we know that $P(3) = \frac{2}{36}$, $P(5) = \frac{4}{36}$ we get $P(\text{neither } 3 \text{ nor } 5) = 1 - \frac{2}{36} - \frac{4}{36} = \frac{30}{36}$, so

$$q = \frac{2}{36} + \frac{30}{36} q \quad \Longleftrightarrow \quad q = \frac{1}{3}$$

Note that we can also rewrite the first equation as

$$q = P(3) \cdot 1 + P(5) \cdot 0 + P(\text{neither } 3 \text{ nor } 5) \cdot q \quad \Longleftrightarrow \quad q = \frac{P(3)}{P(3) + P(5)} = \frac{P(3)}{P(3 \text{ or } 5)} = P(3\ |\ 3 \text{ or } 5)$$

This is in accordance with a comment made by Dilip: We can simply ignore all throws which are neither a $3$ or a $5$, as they are irrelevant. Then the probability of seeing a $3$ first is simply the conditional probability of throwing a $3$, given that it's either a $3$ or a $5$.

(b) $P(X<Y<Z)$

We can use the same approach as above. First, let us ignore all throws which are not $3$, $5$ or $10$. We now need that the first event is a $3$, which happens with probability

$$P(X<Y,X<Z) = \frac{P(X)}{P(X \text{ or } Y \text{ or } Z)} = \frac{P(X)}{P(X) + P(Y) + P(Z)} = \frac{2}{2+4+3} = \frac{2}{9}$$

Now if this happens, we only need that in all successive events of $3,5,10$ we first see some number of $3$s, and then see a $5$ (and not a $10$ yet). But that means that after this first event, we can again ignore all events $X$, and look at the first throw resulting in either a $5$ or a $10$. Then

$$P(Y<Z | X<Y,X<Z) = \frac{P(Y)}{P(Y \text{ or } Z)} = \frac{P(Y)}{P(Y)+P(Z)} = \frac{4}{4+3} = \frac{4}{7}$$

So the combined probability is then

$$P(X<Y<Z) = P(X<Y,X<Z) \cdot P(Y<Z | X<Y, X<Z) = \frac{2}{9} \cdot \frac{4}{7} = \frac{8}{63}$$

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Here's a simpler way than one involving infinite series (such as that posted by Sasha). Roll a die until you get either a 3 or a 5. What's the probability that it's a 3? It's just the conditional probability of getting a 3, given that you've got either a 3 or a 5: $$\Pr(W=3\mid W=3\text{ or }W=5) = \frac{\Pr(W=3)}{\Pr(W=3\text{ or }W=5)} = \frac{2/36}{6/36}= \frac13.$$

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Following on from @Thijs analysis, which is actually a tool called "first step analysis", we can get the probability that $P(X<Y<Z)$. So we have to wait until one of these events happens. $P(X)=\frac{2}{36}$ $P(Y)=\frac{4}{36}$ $P(Z)=\frac{3}{36}$. So we have probability of $\frac{27}{36}$ of achieving "nothing" each trial, and effectively going back to the first step.

So at effective trial one, if $X$ occurs we have success, if $Y$ or $Z$ occurs, we have failure.

$$P(X<Y<Z)=\frac{2}{36}P(X<Y<Z|X<Y,Z)+\frac{3}{36}P(X<Y<Z|Z<X,Y)$$ $$+\frac{4}{36}P(X<Y<Z|Y<X,Z)+\frac{27}{36}P(X<Y<Z)$$

The middle two probabilities are zero, and the first is simply $P(Y<Z)$, and applying first step analysis again we have

$$P(Y<Z)=\frac{4}{7}$$

Plugging back into the previous equation and solving we get

$$P(X<Y<Z)=\frac{8}{36(7)}+\frac{27}{36}P(X<Y<Z)=\frac{8}{63}$$

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Nice, your follow-up is more in line with my previous analysis than my own follow-up! :) – TMM Sep 23 '11 at 18:37