Expected value of maximum of two D20 rolls

Question

I'm trying to work through a similar version of this problem. The idea is that you have a fair, 20-sided die, and $n$ turns. You are trying to decide how many times to roll it (each roll costs a turn) before taking the face value of the die in dollars for the rest of the turns.

The part I'm stuck on is a question regarding the expected value of the fair, 20 sided die after a number of turns. At first, when I thought through the problem, I saw the expected value of the maximum of two rolls as follows:

Let $x$ be the value of the die. Let $S$ be the event that $x$ is between 11 and 20 (upper $\frac{1}{2}$ of values).

P(S) = 1/2. Thus we can treat $S$ as a geometric r.v., and the expected number of rolls to get a value in $S$ is $\frac{1}{p(S)} = \frac{1}{\frac{1}{2}} = 2$. We can also say that any number in the set {11, ... , 20} is equally likely, and thus $E[x | x \in S] = 15.5$

This would mean that on average, we would expect our maximum value of the two dice rolls to be 15.5 after two rolls.

However, I was doing more work on problems regarding uniform distributions, and came across the problem of drawing multiple values $(X_1, X_2)$ from a $U(a,b)$. This solution here gives $E[Max(X_1,X_2)] = \frac{1}{3}(2b + a)$. If we were to use this on a $U(1,20)$ (outcome of a fair, 20 sided die), we would get that $E[Max(X_1,X_2)] = \frac{1}{3} \cdot (2\cdot20+1) = 13.667$.

Could someone explain why the two calculations for the expected value after two rolls are produce different values?

It seems to me that in both cases, two values are being drawn independently from $U(1,20)$, however, with one of them I expect the maximum value to be 15.5 after two rolls, and the other 13.667.

Answer 1

I take that your question is to compute the expected larger outcome from two throws of a 20 sided die.

Let's make things simpler and use a die with 4 sides instead. For 2 rolls the first method yields $3.5$, the second yields $3$. Now we count all cases directly.

The equally possible outcomes are

$(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)$

their corresponding largest values are

$1, 2, 3, 4, 2, 2, 3, 4, 3, 3, 3, 4, 4, 4, 4, 4$

with average $\frac{25}{8}$, which indicates that both answers above are wrong.

You may check the above distribution to see why the first one is wrong. The second one is wrong because we are working with discrete variables instead of continuous variables in the other post.

The correct method is to sum through all the pairs carefully, let's say we have $X_1,X_2$ drawn uniform and independently from integers on $[a,b]$, the first sum is for $X_1\neq X_2$, the $2$ factor takes care about that each one could be larger. The second sum is for cases that $X_1=X_2$.

$E(Max(X_1,X_2))=\frac{1}{(b-a+1)^2}[2\sum_{i=a+1}^{b}\sum_{j=a}^{i-1}i+\sum_{i=a}^{b}i]$

$=\frac{1}{(b-a+1)^2}[2\sum_{i=a+1}^{b}(i-a)i+\frac{(b-a+1)(a+b)}{2}]$

$=\frac{1}{(b-a+1)^2}[2\sum_{i=1}^{b-a}(i^2+ai)+\frac{(b-a+1)(a+b)}{2}]$

$=\frac{1}{(b-a+1)^2}[2(\frac{(b-a)(b-a+1)(2b-2a+1)}{6}+a\frac{(b-a)(b-a+1))}{2})+\frac{(b-a+1)(a+b)}{2}]$

$=\frac{1}{(b-a+1)}[2(\frac{(b-a)(2b-2a+1)}{6}+a\frac{(b-a)}{2})+\frac{(a+b)}{2}]$

substituting $a=1,b=20$ gives the answer $\frac{553}{40}=13.825$.