Question

You and a friend play a spin-the-wheel game. You have a wheel with 20 numbers (1-20) which can be spun, and with each spin, one of the numbers is randomly selected. Your friend offers you a bet: If the numbers on the next three spins contain at least two odd numbers, then they will pay you €15, otherwise you will pay them €15. To the nearest euro how, many euro would you expect to win/lose if you gambled 92 times on this bet?

Answer 1

Solution

$P(odd \space number) =0.5$

The 3 trials follow a binomial distribution.

X~Bin(3,0.5)

$$P(X \ge 2)=P(X=2)+P(X=3)$$

$$= \binom{3}{2} {0.5}^2 * {0.5}^{1} + \binom{3}{3} {0.5}^3 *{0.5}^0$$ $$=0.375+0.125=0.5$$

Thus, there's a 0.5 chance of winning €15 and 0.5 chance of losing €15 on one bet.

$$E(X) = \sum {xP(x)}$$ $$=15(0.5)-15(0.5) =0$$

Therefore, in 92 bets, the expected win/loss is:

$$=92 *0 =0$$

You should expect to have gained €0 in the end