Conditional probability - dice problem

Question

Two dice are rolled. Let X be the lower of the two scores and Y be the larger of the scores. What is the conditional probability P(Y=6|X=4)?

Answer 1

I am taking the assumption that $Y > X$ always based off of wording in OP's statement: 'Let X be the lower of the two scores and Y be the larger of the scores.'

We are given the information that $Y > X$ and now we want $P(Y = 6 \mid X=4)$.

This takes us to a subset of the sample space: $\{ 5,6\} \subset \{1,2,3,4,5,6\}$. Either value is just as likely; namely, it reduces to $Y$ is uniform over $\{5,6\}$ given this new information thus $\displaystyle P(Y=6 \mid X=4) = \frac{1}{2}$.

More generally,

If we are given a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ if we have a $B \in \mathcal{F}$ such that $\mathbb{P}(B) > 0$.

Then we may define a new probability space: $(\Omega, \mathcal{F}, \mathbb{Q})$ where $\forall A \in \mathcal{F}, \mathbb{Q}(A) = \mathbb{P}(A \mid B)$.

Since we know $Y > X$ and $\mathbb{P}(X=4) = \frac{1}{6} > 0 $ we can do the above transformation to a new space.

Additional note: Sometimes trying to solve by $\displaystyle P(A\mid B) = \frac{P(A,B)}{P(B)} $ doesn't work so well when there's a dependence. So it is better to consider what does $P(A \mid B)$ mean given some information that relates $A$ and $B$.

How this relates to the above is let $A : = Y = 6$ and $B : = X = 4$ then we should be using $P(A,B) = P(A\mid B)P(B)$ form of the above since we have a dependency here; namely, that $Y > X$.

Answer 2

The other answers and comments are incorrect as they falsely assume that $Y$ must be strictly greater than $X$ in every occasion, however that is invalid. It is perfectly acceptable for $Y$ to equal $X$ in the event that both dice show the same value.

To clarify and formalize, I am working with the assumption that $A$ and $B$ are independent random variables, each of which uniformly distributed over $\{1,2,3,4,5,6\}$ and $X=\min(A,B)$ and $Y=\max(A,B)$... that is $A$ is the result of the "first die" and $B$ the result of the "second die", that $X$ is colloquially the result of the "smaller die" and $Y$ the result of the "larger die" (yes, even in the event that $A=B$)

Let us imagine that the dice are both differently colored or that they are rolled in sequence, or similar... e.g. one blue and one red die. There are $36$ equally likely different outcomes.

$\Pr(Y=6\mid X=4) = \dfrac{\Pr(Y=6\cap X=4)}{\Pr(X=4)}$

The probability that the larger die is a $6$ and the smaller die is a $4$, well that corresponds to the two outcomes where the red die is $6$ and the blue die is $4$ or to the outcome that the red die is $4$ and the blue die is $6$. We have then that $\Pr(Y=6\cap X=4)=\dfrac{2}{36}$

The probability that the smaller die is a $4$ is going to be the probability that both dice are greater than or equal to $4$ while not being the case that both dice are simultaneously greater than or equal to $5$ (as then $4$ would not have been the minimum). There are $3\times 3$ outcomes corresponding to both dice being greater than or equal to $4$ and $2\times 2$ outcomes corresponding to both dice being greater than or equal to $5$. As such, there are $3\times 3-2\times 2=9-4=5$ outcomes corresponding to the minimum of the dice rolls being a $4$ and so $\Pr(X=4)=\dfrac{5}{36}$

This used the general result that $\Pr(X= k) = \Pr(X\geq k)-\Pr(X>k)$

Combining the results, the probability is then:

$$\Pr(Y=6\mid X=4)=\dfrac{\frac{2}{36}}{\frac{5}{36}}=\dfrac{2}{5}=0.4$$

Answer 3

There is serious confusion about a basic condition of the problem. The OP is explicitly interested in this case: "Let X be the lower of the two scores and Y be the larger of the scores." This a fortiori excludes cases where neither of the values is higher, nor lower, e.g., cases of the form $(n,n)$.

What could be clearer?

Imagine the OP wrote: "Two dice are rolled. Let X be the odd score and Y be the even score."

You simply cannot come back and say: "What about the case $(4,4)$?" or "What about the case $(3,3)$?" "What about the case $(4,6)$?"

Those cases are explicitly excluded from consideration. Hence any calculation of the probability that involves those excluded cases would of course be in error.

Imagine the OP wrote: "Two dice are rolled. Let $X$ be a roll above 3.5 and $Y$ be a roll below 3.5. What is the average difference between the rolls?"

You cannot come back and say "But what about $(5,5)$?" "What about $(2,2)$?" Those cases are explicitly excluded.

Nobody is saying that you will never roll a $(4,4)$ or $(4,6)$, etc., just that this is not part of the question. Just as a question about the rolls being relatively prime, or perfect numbers, or...

Improperly including cases such as $(4,4)$ leads to such confusion. Suppose you roll a red die and green die and get $(4,4)$ and hear: "Tell me the color of the single die that is larger than the other." A perfectly reasonable request. (It would be a deliberate mis-representation to change this clear question to a different one, such as "Tell the color of the largest." It would be to deliberately exclude the key word "than.")

Back to the explicit case asked by the OP: We are told explicitly that one of the rolls is "larger" and the other "smaller." This is of course not the same as "maximum of the set." It is of course not correct to say that $4$ is "larger of" the other element of the set $(4,4)$ and that $4$ is "smaller of"... (um... which?)... just that $4$ is the maximum value. The author was clear in using "larger of" and any attempt at an answer that deliberately tries to negate those clear words is confused and in error.

Big difference.