# Expected value Problem

• May 20th 2007, 10:03 PM
solace
Expected value Problem
For a game played using the dice in Problem #13, you make a 2$bet on the outcome of a roll of the dice. If the roll is 5 or lower you win 3$. If the roll is 11 or 12, you win 10$. What is the expected value of this game?(note the winning amounts are net amounts.) Here is the dice probability distribution for problem 13 http://img72.imageshack.us/img72/3042/dicerollrs4.png Here is what i did http://img527.imageshack.us/img527/9...edvalueet9.png Is this correct? • May 20th 2007, 10:47 PM CaptainBlack Quote: Originally Posted by solace For a game played using the dice in Problem #13, you make a 2$ bet on the outcome of a roll of the dice. If the roll is 5 or lower you win 3$. If the roll is 11 or 12, you win 10$. What is the expected value of this game?(note the winning amounts are net amounts.)

Here is the dice probability distribution for problem 13

http://img72.imageshack.us/img72/3042/dicerollrs4.png

Here is what i did

http://img527.imageshack.us/img527/9...edvalueet9.png

Is this correct?

I can't read this last table, but the probability of rolling a 5 or less is 0.3225
and the prob of rolling a 11 or 12 is 0.0175, and the probability of any other
result is 1-0.3225-0.0175 = 0.66.

Now the value of the game is:

0.3225*5 + 0.0175*10 - 0.66*2 = +0.4675

RonL
• May 20th 2007, 11:18 PM
solace
Quote:

Originally Posted by CaptainBlack
I can't read this last table, but the probability of rolling a 5 or less is 0.3225
and the prob of rolling a 11 or 12 is 0.0175, and the probability of any other
result is 1-0.3225-0.0175 = 0.66.

Now the value of the game is:

0.3225*5 + 0.0175*10 - 0.66*2 = +0.4675

RonL

Why .3225*5? Should it be *3 since thats how much money you get if you roll a 5 or lower?

Also why .3325?

.09+.09+.1425+.15= .4725 right?

THen 1-.4725+.0175 = .5100

so: .4725*3+.0175*10-.51*2= .5725

Also, doesnt the net value mean that you lose the 2$you bet and any winnings are -2? So the 3$ win is actually 1$net profit, and the 10$ win is actually 8$net profit? So should i times the probabilities by 1 and 8 instead? • May 21st 2007, 05:19 AM CaptainBlack Quote: Originally Posted by solace Why .3225*5? Should it be *3 since thats how much money you get if you roll a 5 or lower? If that is what you mean net amount. It is an odd way of expressing it but if you have it defined that way OK. Quote: Also why .3325? .09+.09+.1425+.15= .4725 right? THen 1-.4725+.0175 = .5100 so: .4725*3+.0175*10-.51*2= .5725 yes, slip of the calculator Quote: Also, doesnt the net value mean that you lose the 2$ you bet and any winnings are -2? So the 3$win is actually 1$ net profit, and the 10$win is actually 8$ net profit?

So should i times the probabilities by 1 and 8 instead?
If that is what you mean by net winnings (I would interpret net winnings
to be how much better off you are after compared to before the bet), then:

0.4725*1 + 0.0175*8 - 0.51*2 = +0.4675

My experience is that when you bet your returned stake is not counted
part of your winnings. Just think of the fun the tax man would have if it did.

RonL
• May 25th 2007, 12:55 PM
Soroban
Hello, solace!

Quote:

You make a $2 bet on the outcome of a roll of the dice. If the roll is 5 or lower you win$3.
If the roll is 11 or 12, you win $10. What is the expected value of this game? Rolling a pair of dice, there are 36 possible outcomes. The totals and their respective probabilities are: . .$\displaystyle \begin{array}{cc}2 & \frac{1}{36} \\ 3 & \frac{2}{36} \\ 4 & \frac{3}{36} \\ 5 & \frac{4}{36} \\ 6 & \frac{5}{36} \\ 7 & \frac{6}{36}\end{array}$. .$\displaystyle \begin{array}{cc}8 & \frac{5}{36} \\ 9 & \frac{4}{36} \\ 10 & \frac{3}{36} \\ 11 & \frac{2}{36} \\ 12 & \frac{1}{36}
\end{array}\displaystyle P(\text{5 or lower: win \$3}) \;=\; \frac{1}{36} + \frac{2}{36} + \frac{3}{36} + \frac{4}{36} \;=\;\frac{10}{36}$

$\displaystyle P(\text{11 or 12: win \$10}) \;=\;\frac{2}{36} + \frac{1}{36}\;=\;\frac{3}{36}\displaystyle P(\text{lose \$2}) \;=\;\frac{5}{36} + \frac{6}{36} + \frac{5}{36} + \frac{4}{36} + \frac{3}{36} \;=\;\frac{23}{36}$

Hence: .$\displaystyle E \;=\;\frac{10}{36}(3) + \frac{3}{36}(10) + \frac{23}{36}(-2) \;=\;\frac{14}{36} \;=\;0.30555...$

Therefore, you can expect to win an average of about 30½¢ per game.

• May 25th 2007, 01:04 PM
CaptainBlack
Quote:

Originally Posted by Soroban
Hello, solace!

Rolling a pair of dice, there are 36 possible outcomes.

The totals and their respective probabilities are:

. . $\displaystyle \begin{array}{cc}2 & \frac{1}{36} \\ 3 & \frac{2}{36} \\ 4 & \frac{3}{36} \\ 5 & \frac{4}{36} \\ 6 & \frac{5}{36} \\ 7 & \frac{6}{36}\end{array}$
. .$\displaystyle \begin{array}{cc}8 & \frac{5}{36} \\ 9 & \frac{4}{36} \\ 10 & \frac{3}{36} \\ 11 & \frac{2}{36} \\ 12 & \frac{1}{36}$$\displaystyle \end{array} \displaystyle P(\text{5 or lower: win \3}) \;=\;$$\displaystyle \frac{1}{36} + \frac{2}{36} + \frac{3}{36} + \frac{4}{36} \;=\;\frac{10}{36}$

$\displaystyle P(\text{11 or 12: win \$10}) \;=\;\frac{2}{36} + \frac{1}{36}\;=\;\frac{3}{36}\displaystyle P(\text{lose \$2}) \;=\;\frac{5}{36} + \frac{6}{36} + \frac{5}{36} + \frac{4}{36} + \frac{3}{36} \;=\;\frac{23}{36}$

Hence: .$\displaystyle E \;=\;\frac{10}{36}(3) + \frac{3}{36}(10) + \frac{23}{36}(-2) \;=\;\frac{14}{36} \;=\;0.30555...$

Therefore, you can expect to win an average of about 30½¢ per game.

They are not fair dice.

RonL