Answer to Question #318046 in Statistics and Probability for Kheyt

Question

1.You play a game with two six-sided dice. If you roll a sum of 4 or 7, you win P 400. If you roll a sum of 10,you win 200. However, you lose 300 for anything else. If you continue to play the game, how much do you expect to win or lose in the game?

Answer 1

Since the probability of the coin landing on heads or tails are both 1/2, we can substitute it to the equation with

So the expected value of a coin flip is 

Let us set up first the values of the results. We can assign the following variables for easier understanding of a problem:

400 for winning 400 (rolling a sum of 4 or 7)

200 for winning 200 (rolling a sum of 10)

-300 for losing 300 (rolling a sum of something else)

Now we need to know the probabilities of getting the events.

If 2 dice are thrown, then it has 36 possible results because the first dice has 6 results, and the second also has 6 results. We can tabulate each result and add them to find out how many satisfy our probabilities.

If 2 dice are thrown, the results can be assigned an ordered pair (x,y) with x the result of the first die, and y the result of the second die. The tabulated results can be found on the attached image.

We can add the results since we our concern is the sum of adding the two dice. The tabulated results can again be found on the attached image. The highlighted numbers are the "winning numbers". There are 36 possible results.

For the first scenario of rolling a sum of 4 or 7, we add the probabilities of both 4 and 7. This is because of the sum rule of the fundamental principle of counting. There are 3 ways of rolling a sum of 4, and then 6 ways for rolling a sum of 4. That means:

P(rolling a sum of 4 or 7) = P(rolling a sum of 4) + P(rolling a sum of 7)

$P(rolling\ a\ sum \ of \ 4 \ or \ 7) = \frac{3}{36} + \frac{6}{36} =\frac{1}{4}$

For the second scenario, there are 3 times the sum comes up to 10. Since there are 36 results, the probability of getting a sum of 10 is.

$P (rolling\ a\ sum\ of \ 10) = \frac{3}{36} =\frac{1}{12}$

For the last scenario, we can either count the non-winning numbers or subtract the winning numbers from the total number of results. Let us do the latter. There are 36 possible results of rolling 2 dice. There are 9 winning results from rolling a sum of 4 or 7, and 3 winning results from rolling a sum of 10. Therefore, the non-winning numbers are =36−9−3

=36−12

=24

There are 24 ways of not winning. The probability of that happening is

$P(not \ rolling \ a \ sum \ of \ 4, 7\ or \ 10) = \frac{24}{36} =\frac{2}{3}$

We now have the probabilities of all scenarios. We can multiply each probability with its assigned variable to compute the expected value of the situation.

E[X] = (value if the sum is 4 or 7)*(probability that the sum is 4 or 7) +(value if the sum is 10)*(probability that the sum is 10) +

(value if the sum is not 4,7, nor 10)*(probability that the sum is not 4,7, nor 10)

$E[X] = 400*(\frac{1}{4}) + 200*(\frac{1}{12})+ (-300)*(\frac{2}{3})= 100 + \frac{200}{12} - 200\\\\E[X] = -100 +\frac{50}{3} \\\\E[X] = \frac{-250}{3}\\\\E[X] = -83\frac{1}{3}$

That means, if I play the game, I would lose 83.33

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