Rolls of die

All outcomes of the die are listed in black and the score added (or out) is shown in red

The average score added per round of a game if not out is $\frac{2 × 2+3 × 4+4 × 6+5 × 8+6 × 10}{30}=\frac{70}{15}$

The chance of getting out is $\frac{6}{36}=\frac{1}{6}$

Let the expected score be $Y$

Let’s assume one round is over, there would be a $\frac{1}{6}$ chance that Alex got out, and a $\frac{5}{6}$ chance he scored an average of $\frac{70}{15}$ and then went on to play the game with an expected score of $Y$ again.

So we have the equation: $Y=\frac{1}{6} × 0 + \frac{5}{6} × ( \frac{70}{15} + Y )$

Solving we get: $Y=\frac{70}{3}$

Here’s a program where a computer plays the game with itself and prints the average score after every 100,000 games

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from random import *
score=0
game=0
while True:
    game+=1
    while True:
        die1=randint(1,6)
        die2=randint(1,6)
        if die1<die2:
            score+=die2
        elif die1>die2:
            score+=die1
        elif die1==die2:
            break
    if game%100000==0:
        print(score/game)
#It converges very slowly to the value of 23.33… 
#first two decimals are stable (staying at 23.33 and third decimal varying) 
#only after ten million games