Question

You play a game where you roll two six-sided dice and sum the pips showing. It costs you $2 to play the game. If your dice sum to 7 or 11, you win $20. If your dice sum to 2 or 12, you win $2. If your dice sum to any other number, you lose. What is your expected winnings for this game?

Answer 1

Answer:

$2.56

Step-by-step explanation:

P(7 or 11) = 8/36 = 2/9

P(winning 20-2 = 18) = 2/9

P(2 or 12) = 2/36 = 1/18

P(winning 2-2 = 0) = 1/18

P(others) = 26/36 = 13/18

P(winning 0-2 = -2) = 13/18

Expected win

= 18(2/9) + 0(1/18) - 2(13/18)

= $23/9 or $2.56

Answer 2

First of all, we need the probabilities of each sum showing. You can see in the attached pictures which couple of dice produce which sum, and thus the probability of each sum showing.

So, we know that we we win 20$ with probability 8/36. In fact, we get 7 with probability 6/36 and 11 with probability 2/36.

Also, we win 2$ with probability 2/36, because both 2 and 12 appear with probability 1/36.

In the remaining 26 cases, we win nothing.

Finally, we have to consider that we lose 2$ with probability 1, because we're sure to pay 2$ to play.

Now we can compute the expected winning, summing all the "gain times probability" pieces:

$\mathbb{E}=-2\cdot 1 + 20\cdot \dfrac{8}{36} + 2\cdot\dfrac{2}{36}=\dfrac{23}{9}=2.\bar{5}$

So, you expect to win about 2.5$