Min/max dice problem

Question

Suppose Anne throws two fair dice and the faces show $4$ and $3$. It is your turn and Anne states that if both dice show a number greater than $4$ and $3$, you will win the game. What is the probability of you winning the game? (i.e. if your outcome is $4$ and $5$ you will win, as the biggest gets paired with the biggest and smallest with the smallest).

By listing the favorable possibilities, we obtain:

$$\begin{aligned} &(4,5)&\qquad (4,6) &\qquad (5,6) &\qquad (5,5)&\\ &(5,4)&\qquad (6,4) &\qquad (6,5) &\qquad (6,6)& \end{aligned}$$

so the proability must be $8/36$ or $2/9$. However, I am interested in knowing why the following way does not work. Let $X_1\sim\text{Unif}\,\{1,6\}$ and $X_2\sim\text{Unif}\,\{1,6\}$. I need

$$\begin{aligned} P(\max\{X_1,X_2\}>4,\min\{X_1,X_2\}>3)&=P(\max\{X_1,X_2\}>4)P(\min\{X_1,X_2\}>3)\\ &=\left[1-P(\max\{X_1,X_2\}\leq4)\right]P(X_1>3,X_2>3)\\ &=\left[1-P(X_1\leq4,X_2\leq4)\right]P(X_1>3)P(X_2>3)\\ &=\left[1-P(X_1\leq4)P(X_2\leq4)\right]\left[1-P(X_1\leq3)\right]\left[1-P(X_2\leq3)\right]\\ &=\left(1-\frac{4}{6}\cdot\frac{4}{6}\right)\left(1-\frac36\right)\left(1-\frac36\right)\\ &=\left[1-\left(\frac{4}{6}\right)^2\right]\left(1-\frac36\right)^2=\frac5{36}. \end{aligned} $$

What is wrong with this approach?

Another approach I considered is, $$\begin{aligned} P(\max\{X_1,X_2\}>4,\min\{X_1,X_2\}>3)&=P(\{X_1>4,X_2>3\}\cup\{X_1>3\,X_2>4\})\\ &=P(\{X_1>4,X_2>3\})+P(\{X_1>3\,X_2>4\})\\ &=P(X_1>4)P(X_2>3)+P(X_1>3)P(X_2>4)\\ &=\frac26\cdot\frac36+\frac36\cdot\frac26=\frac13. \end{aligned}$$

Why are the above different? Is it not true that $\max\{X_1,X_2\}>4,\min\{X_1,X_2\}>3 \implies \{X_1>4,X_2>3\}\cup\{X_1>3\,X_2>4\}$?

Answer 1

The problem is the first line of your approach. $P(A,B)=P(A)P(B)$ is only true if the events $A$ and $B$ are independent. Here the events $\max\{X_1,X_2\}>4$ and $\min\{X_1,X_2\}>3$ are not independent, since if we know the maximum is small that reduces the range of values the minimum can take.

Similarly in the second approach you use $P(C\cup D)=P(C)+P(D)$. This only works if the events $C$ and $D$ are disjoint, which they are not in this case.

Answer 2

Your second approach is right, but formula you have used only works for two disjoint events. In general, for two events A and B, we have $$P(A ∪ B) = P(A) + P(B) - P(A ∩ B)$$

Answer 3

You can use:

$$P(A\cap B) = P(A|B)P(B)$$

Wikipedia : Conditional_probability

So, $P(\max|\min)=\frac89$, and $P(\min)=\frac14$, or, alternatively, $P(\min|\max)=\frac{8}{20}$, and $P(\max)=\frac{20}{36}$.