MHB How Do You Calculate Expected Value in a Dice Game?

[AnnBean]

I really need help on how to solve this (Sweating):

A dice game involves rolling 2 dice. If you roll a 2, 3 , 4, 10, 11 or a 12, you win \$5. If you roll a 5, 6, 7, 8, or 9, you lose \$5. Find the expected value you win (or lose) per game.

 

[MarkFL]

Re: Probability

Hello, and welcome to MHB, AnnBean! (Wave)

I would begin by constructing a table as follows:

[table="width: 800, class: grid, align: left"]
[tr]
[td]Sum $S$[/td]
[td]Probability of $S$: $P(S)$[/td]
[td]Net Gain/Loss (in dollars) $G$[/td]
[td]Product $G\cdot P(S)$[/td]
[/tr]
[tr]
[td]2[/td]
[td]$\dfrac{1}{36}$[/td]
[td]5[/td]
[td]$\dfrac{5}{36}$[/td]
[/tr]
[tr]
[td]3[/td]
[td]$\dfrac{1}{18}$[/td]
[td]5[/td]
[td]$\dfrac{5}{18}$[/td]
[/tr]
[/table]

Can you complete the table?

 

[MarkFL]

Re: Probability

Just to follow up, here is the completed table:

[table="width: 800, class: grid, align: left"]
[tr]
[td]Sum $S$[/td]
[td]Probability of $S$: $P(S)$[/td]
[td]Net Gain/Loss (in dollars) $G$[/td]
[td]Product $G\cdot P(S)$[/td]
[/tr]
[tr]
[td]2[/td]
[td]$\dfrac{1}{36}$[/td]
[td]5[/td]
[td]$\dfrac{5}{36}$[/td]
[/tr]
[tr]
[td]3[/td]
[td]$\dfrac{1}{18}$[/td]
[td]5[/td]
[td]$\dfrac{5}{18}$[/td]
[/tr]
[tr]
[td]4[/td]
[td]$\dfrac{1}{12}$[/td]
[td]5[/td]
[td]$\dfrac{5}{12}$[/td]
[/tr]
[tr]
[td]5[/td]
[td]$\dfrac{1}{9}$[/td]
[td]-5[/td]
[td]$-\dfrac{5}{9}$[/td]
[/tr]
[tr]
[td]6[/td]
[td]$\dfrac{5}{36}$[/td]
[td]-5[/td]
[td]$-\dfrac{25}{36}$[/td]
[/tr]
[tr]
[td]7[/td]
[td]$\dfrac{1}{6}$[/td]
[td]-5[/td]
[td]$-\dfrac{5}{6}$[/td]
[/tr]
[tr]
[td]8[/td]
[td]$\dfrac{5}{36}$[/td]
[td]-5[/td]
[td]$-\dfrac{25}{36}$[/td]
[/tr]
[tr]
[td]9[/td]
[td]$\dfrac{1}{9}$[/td]
[td]-5[/td]
[td]$-\dfrac{5}{9}$[/td]
[/tr]
[tr]
[td]10[/td]
[td]$\dfrac{1}{12}$[/td]
[td]5[/td]
[td]$\dfrac{5}{12}$[/td]
[/tr]
[tr]
[td]11[/td]
[td]$\dfrac{1}{18}$[/td]
[td]5[/td]
[td]$\dfrac{5}{18}$[/td]
[/tr]
[tr]
[td]12[/td]
[td]$\dfrac{1}{36}$[/td]
[td]5[/td]
[td]$\dfrac{5}{36}$[/td]
[/tr]

[/table]

Summing up the $G\cdot P(S)$, column, we find the expected value is:

$$\text{EP}=-\frac{5}{3}$$