Answered

Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

Suppose you are playing a game with two dice at a casino. If you roll double ones, you win $1, if you roll double twos, you win $2. This pattern continues, up to winning $6 for double sixes. If you roll anything besides doubles, you lose $1. What is the expected value of the game? Can you expect to win money by playing this game over and over again? *Hint: Find the total number of outcomes of this game first. Only six outcomes win money. The other outcomes lose money. You’ll find the expected value with these outcomes. Good luck

Sagot :

We have here a case in which we need to find the expected value, and this value is obtained by having the probability of the event times the amount of money (in this case).

We also need to remember that the expected value is a value that we are about to have after making a statistic experiment for a large number of trials.

We can see that the possible outcomes for two dice are 36, and we can see that the probability for double ones, double twos, double threes, and so for is:

[tex]P(doubles)=\frac{6}{36}=\frac{1}{6}[/tex]

Now, the probability of not having doubles is:

[tex]P(\text{ not-doubles\rparen}=\frac{30}{36}=\frac{5}{6}[/tex]

However, the probability for each double is:

[tex]P(\text{ each-double\rparen}=\frac{1}{36}[/tex]

And since we have six possible doubles, then the sum of the probabilities is 6/36 = 1/6.

Now, the expected value here can be found as follows:

[tex]E(X)=\frac{1}{36}(1)+\frac{1}{36}(2)+\frac{1}{36}(3)+\frac{1}{36}(4)+\frac{1}{36}(5)+\frac{1}{36}(6)+\frac{30}{36}(-1)[/tex]

We can use -$1 since we lose money besides doubles. Then, the expected value is as follows:

[tex]\begin{gathered} \frac{1}{36}(1+2+3+4+5+6)+\frac{30}{36}(-1) \\ \\ \frac{1}{36}(21)-\frac{30}{36} \\ \\ \frac{21}{36}-\frac{30}{36}=-\frac{9}{36}=-0.25 \end{gathered}[/tex]

Then we can expect that we will lose about a quarter of a dollar playing this game.

Therefore, in summary, the expected value of the game is -0.25, or to lose 25 cents if we play this game over and over again, and we will expect to lose money.

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.