Expected value of a die

Question

In a game a fair die is rolled three times. If a 6 first appears on the first, second or third roll you receive £5, £2 or £1 respectively. If no 6 appears you receive nothing. If the game is fair, how much should you pay to play it?

And my working out:here

But apparently the answer is £1.23 and my answer is £1.33

What am I doing wrong? Could you please suggest?

Thanks, Yasme

Answer 1

The question is ambiguous on this point, but it seems to be making the assumption that the game ends when you get a 6 on any roll. This means that you can't assume a 1/6 probability of getting a 6 on the second or third roll, because there might not be one. So:

• The probability of getting a 6 on the first roll is $\frac{1}{6}$.
• The probability of not getting a 6 on the first roll but getting one on the second roll is $\frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$.
• The probability of not getting a 6 on the first or second roll but getting one on the third roll is $\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{25}{216}$.

For an expected payout of $\frac{1}{6}(£5) + \frac{5}{36}(£2) + \frac{25}{216}(£1) = £\frac{265}{216} \approx £1.226852$.

A “fair” game means that your bet is equal to the expected payout. But since you can't bet a fraction of a penny, the amount is rounded up to £1.23.