# Thread: Discrete Probability Distribution Problem

1. ## 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?

2. 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.

3. Hello, ose90!

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.

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%.)