Question

On three rolls of a single die, you will lose $17 if a 5 turns up at least once, and you will $6 otherwise. What is the expected value of the game?
a. Let x be the random variable for the amount won on a single play of this game. Complete the payoff table for this random variable x.

xi ____ ____ (Maintain the order of values as given in the problem)
pi ____ ____ (Type probabilities as simplified fractions)

Answer 1

first work out the probability of getting ZERO fives, call it P0 say. You get this by multiplying the probability of getting ZERO fives on each roll together, so it's (5/6)^3 = 125/216 ......This is by far the easiest way. Once we get that we multiply that by the expected winnings, so 6*(125/216) = $3.47

Then we multiply the probability of losing or getting a five which is just (1-125/216 = 91/216) and multiply that by the expected loss (91/216*17 = $7.16)

So you can see the expected losses far outweight the expected winnings so this is, as usual, a biased game. You should not play unless you really want the thrill!

Just buy some dice man. Then you can roll all day for free!

Answer 2

If you need more detail just ask.. probability is tricky!

Answer 3

The question you asked is somewhat different to the one I answered. I assumed you meant that you win $6 otherwise, or the question makes no sense really. Payoff is usually how much you expect to win per game so in this case it is 3.27 - 7.16.. which is going to be negative if you are the player. If you are running the game then you will expect a profit so just reverse the sign! Any good to you?

Answer 4

To fill in your table, simply put in the values of the money and the probabilities of each outcome that I multiplied together to get the expected winnings and losses. Please give me a best answer, because I need some points! Thank you very much. Daniel

Answer 5

i just need help figureing out what i should put into the blank spaces. :/

Answer 6

xi ____ ____ (Maintain the order of values as given in the problem)
pi ____ ____ (Type probabilities as simplified fractions)

Answer 7

Alrighty, but it would help you to gamble more successfully in the real world if you understood this stuff. The x values are the outcomes AKA the amounts you might win (I don't know if you are the player or the runner of the game so tell me that please if you know), and the pi values are the corresponding probabilities of each outcome. You need this to figure out whether the game is biased or not. You can then work out the value of the game per go, which is the sum of (the winnings times by the probability of each). If it is negative you are going to lose money in the long run. This is how casinos work out how much to charge for each game you play. They ALWAYS make sure using maths that they will make a profit. It's quite interesting maths and it will help you work out when someone is trying to cheat you in a game. Very useful!

Answer 8

Go on, please give me a medal !

Answer 9

so under pi i would put 125/216 in the first space and on the second space i would put 91/216?

Answer 10

yep and the winnings are negative if you are losing... if you're the player

Answer 11

there are two possible values of \(x\) namely \(x=-17\) or \(x=6\)

Answer 12

what about pi?

Answer 13

the probability \(x=-17\) is \(1-\left(\frac{5}{6}\right)^3\)

Answer 14

and the probability \(x=6\) is \(\left(\frac{5}{6}\right)^3\)

Answer 15

so
\[P(x=6)=\left(\frac{5}{6}\right)^3\]
\[P(x=-17)=1-\left(\frac{5}{6}\right)^3\]

Answer 16

expected value is
\[6\times \frac{5}{6}^3-17\times (1-\left(\frac{5}{6}\right)^3)\]