# Discrete Probability Distribution Problem

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• Jun 29th 2009, 01:28 AM
ose90
Discrete Probability Distribution Problem
Hi everyone, I hope you can help me with the following problem, especially the part 2 of the problem.

Three dice are thrown in a game. If 1 or 6 turns up, you will be paid 1p. If neither 1 nor 6 turn up, you pay 5p. How much gain or loss is expected in 9 games?

Let say you are given opportunity to change the rule of the game if 1 or 6 appears. To make the game worthwhile, what is the minimum amount in everyday currency that you would suggest for both the payments?

Thanks in advance.
• Jun 29th 2009, 02:44 PM
Marth
Outcome | Probability | Payoff
1 or 6 | 1/3 | 1
2-5 | 2/3 | -5

I apologize for the appearance of my table. The '|' lines are supposed to represent column separators Unfortunately, I cannot get the spacing correctly.

To find the expected return, multiply the probabilities with the payoffs and add.

(1/3)*1 + (2/3)*-5 = 1/3 -10/3 = -9/3 = -3

That would be for 1 game. For 9 games, multiply it by 9: -3*9 = -27

The changing the game question has many possible answers. Choose the expected return you want, then change the payoff from either or both of the outcomes to match that return.
• Jun 29th 2009, 05:02 PM
Soroban
Hello, ose90!

Quote:

Three dice are thrown in a game.
If 1 or 6 turns up, you will be paid $1. If neither 1 nor 6 turn up, you pay$5.
How much gain or loss is expected in 9 games?

You will lose $5 if no 1 or 6 appear on any of the three dice. The probability is: . $\tfrac{4}{6}\cdot\tfrac{4}{6}\cdot\tfrac{4}{6} \:=\:\tfrac{8}{27}$ Otherwise, you win$1.
The probability is: . $1 - \tfrac{8}{27} \:=\:\tfrac{19}{27}$

We have: . $\begin{array}{|c|c|c|} \hline
\text{Outcome} & \text{Prob.} & \text{Payoff} \\ \hline \hline \\[-4mm]
\text{1 or 6} & \frac{19}{27} & +1 \\ \\[-4mm] \text{Other} & \frac{8}{27} & \text{-}5 \\ \hline \end{array}$

$EV \;=\;\left(\frac{19}{27}\right)(+1) + \left(\frac{8}{27}\right)(\text{-}5) \;=\;-\frac{7}{9}$

In 9 games, your expectation is: . $(9)\left(\text{-}\frac{7}{9}\right) \;=\;-7$

You can expect to lose \$7.

Quote:

Say you are given opportunity to change the rule of the game if 1 or 6 appears.
To make the game worthwhile, what is the minimum amount for both the payments?

A sloppy question! . . . There is no one right answer.
Besides, what does "worthwhile" mean?

Let $W$ = amount we win, $L$ = amount we lose.

Then: . $EV \:=\:\frac{19}{27}W + \frac{8}{27}(-L) \;=\;\frac{19W - 8L}{27}$

If the game is to be "fair", the EV should be zero:

. . $\frac{19W - 8L}{27} \:=\:0 \quad\Rightarrow\quad W \:=\:\frac{8}{19}L$

The amount won should be about 42.1% of the amount lost.
. . (In the original game, it is only 20%.)