I came across the following dice-roll problem:
- You roll two dice consecutively. Rolling a 6, then a 6 wins you \$100.
- Rolling a 6, then not a 6 loses \$x
- For all other cases you, redo the dice rolling.
- When should you play the game?
If X is the value amount I receive by playing this game; that event A: (6,6); event B: (6, not 6); I calculated the expected value E[X] as follows:
E[X] = P(A)*100 – P(B)*x = $\frac{100}{36} – \frac{5x}{36}$. E[X] is positive when $5x < 100$. I should therefore play the game when $x< \$20$.
Am I missing something by not taking into account the "re-roll" information? Does that change my probabilities, or can it just be treated as the following independent re-running of the game?
Best Answer
No, you can just assume the first roll is $6$ because you reroll anything else. It becomes a single roll with $+100$ on a $6$ and $-20$ on anything else, which is fair.