Question

You roll a die. If an odd number comes up, you lose. If you get a 6, you win $60. If it is aneven number other than 6, you get to roll again. If you get a 6 the second time, you win$36. If not, you lose.
(a) Construct a probability model for the amount you win at this game. Explain briefly how you obtain the probabilities associated with the different amounts of winning.
(b) How much would you be willing to pay to play this game?

Answer 1

Answer:

Step-by-step explanation:

Given that if an odd number comes up, you lose.

i.e. $P(losing) = P(1,3,5) = \frac{1}{2}$

$P(win 60) = P(6) =\frac{1}{6}$

If you get 2 or 4 you roll again

Thus rolling again prob = $\frac{2}{6} =\frac{1}{3}$

After rolling II time, if get 6, win otherwise lose.

Thus winning amounts A canbe

0     36   60

$P(0) = \frac{1}{2}+\frac{1}{3} (\frac{5}{6})=\frac{7}{9} \\P(36) = \frac{1}{3} (\frac{1}{6})=\frac{1}{18}\\P(60) =\frac{1}{6}$

This is a genuine pdf since prob >0 and total prob = 1.

b)Expected gain = sum of gain * prob = $\frac{7}{9} (0)+\frac{1}{18} (36)+\frac{1}{6} (60)\\=12$

Thus I would be willing to pay an amount atmost 12 dollars.