两枚骰子取大者的期望值

让我们看一下不同最大值的概率。

为了使你的最大值达到 $1$，你的两次掷骰都必须是 $1$。在 $36$ 总结果中，有一种 $1$ 方法可以做到这一点。 $P(max = 1) = \frac1{36}$

为了使你的最大值达到 $2$，我们只能滚动 $1's$ 和 $2's$。这为我们提供了 $2$ 选项，每个卷总计为 $4$。然而，我们必须减去最高掷骰为 $1$ 的情况，从而为我们留下 $3$ 的方法，使我们的最大值为 $2$。 $P(max = 2) = \frac{4-1}{36} = \frac{3}{36}$

我们可以使用类似的过程来计算$P(max = 3) = \frac{9 - 3 - 1}{36} = \frac5{36}$

你可以继续使用此过程来计算 $P(max = n)$ 或注意图案的形成：$P(max = n) = P(max = (n-1)) + \frac2{36}$

$P(max = 4) = \frac{7}{36}$ $P(max = 5) = \frac{9}{36}$ $P(max = 6) = \frac{11}{36}$ $E(max) = \sum{P(max = n)\cdot N = \frac1{36}}\cdot1 + \frac{3}{36}\cdot2+ \frac{5}{36}\cdot3+ \frac{7}{36}\cdot4+ \frac{9}{36}\cdot5+ \frac{11}{36}\cdot6$ $E(max) = \frac{161}{36}$

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Original Explanation

Let's take a look at the probabilities of different maxes.

For your maximum value to be $1$, both of your rolls must have been $1$. There is $1$ way to do this out of the $36$ total outcomes. $P(max = 1) = \frac1{36}$

For your max to be $2$, we can only roll $1's$ and $2's$. This gives us $2$ options for each roll totalling to $4$. However, we must subtract the case where the highest roll is $1$, leaving us with $3$ ways for our max to be $2$.
$P(max = 2) = \frac{4-1}{36} = \frac{3}{36}$

We can use a similar process to calculate $P(max = 3) = \frac{9 - 3 - 1}{36} = \frac5{36}$

You can continue using this process to calculate $P(max = n)$ or notice the pattern forming: $P(max = n) = P(max = (n-1)) + \frac2{36}$

$P(max = 4) = \frac{7}{36}$ $P(max = 5) = \frac{9}{36}$ $P(max = 4) = \frac{11}{36}$ $E(max) = \sum{P(max = n)\cdot N = \frac1{36}}\cdot1 + \frac{3}{36}\cdot2+ \frac{5}{36}\cdot3+ \frac{7}{36}\cdot4+ \frac{9}{36}\cdot5+ \frac{11}{36}\cdot6$ $E(max) = \frac{161}{36}$