# Expected Values of Random Variables

#### masterofchupi

##### New Member
Hi all,

As you can imagine, I am working on this problem.

Background
In the carnival game Under-or-Over-Seven, a pair of fair dice is rolled once, and the resulting sum determintes whether the player wins or loses his or her bet. For example, the player can bet $1 that the sum will be Under 7 and loses$1 if the sum is 7 or over. Similarly, the player can bet $1 that the sum will be over 7, and loses$1 if the sum is 7 or under. A third method of play is to bet $1 that the sum will be exactly 7. In this case, the player would win$4, but loses his $1 if he does not get exactly 7. Question Show that the expected long run profit or loss to the player is the same, no matter which method of play is used. My Work Under 7 Expected Value of "Lose" Frequency$1 21/36
$0 (you don't lose money if you win) 15/36 Exactly 7 Expected Value of "Lose" Frequency$1 30/36
$0 6/36 Over 7 Expected Value of "Lose" Frequency$1 21/36
$0 15/36 So I'm asked to show that the expected long-run win or loss is the same, regardless of which method is the same. I have to compare and contrast the three different games. Showing the win: What I did for the "wins" is set up how much I would win by the probability of that happening. So (2)(15/36) vs (5)(6/(36) vs (2)(15/36). They all equaled 0.833, so I know the probability of winning in the long run is the same. Now I would like to check my work to see if the probability of losing in the long run is the same. I tried to set up my probability of losing the same way, but my probability of losses isn't matching. I set up that I would lose$1 if I lost in each of the games:

E(XOver 7) = (-1 * 21/36) + (0 * 15/36) =-0.583
E(XExactly 7) = (-1 * 30/36) + (0 * 6/36) = -0.833
E(XUnder 7) = (-1 * 21/36) + (0 * 15/36) = -0.583

To me, this shows that the long-run loss is different. The Over/Under games are the same, but the loss for exactly 7 is smaller.

So I tried showing how much potential I would lose:
E(XOver 7) = (-2 * 21/36) + (0 * 15/36) = -1.167
E(XExactly 7) = (-5 * 30/36) + (0 * 6/36) = -4.167
E(XUnder 7) = (-2 * 21/36) + (0 * 15/36) = -1.167

Again, this shows that the long-term loss is different. But I think this is closer. Does anyone see what I am missing here?

Sorry for the length of the question, but I do want to let people know exactly how I am thinking about the problem.

Thank you,
Dani

#### fxp

##### New Member
I think the misunderstanding comes from the way the problem is phrased:

Betting $1 and Losing$1 should be interpreted as
Payoffs: (+1) and (-1) and not (0) and (-1)

Then you will see that:

The expected value of the first and second gamble is:
P(winning) * Outcome(winning) + P(losing) * Outcome(losing)

15/36 * 1 + 21/36 * (-1) = -0,1666666667

And similarly for the third gamble:

6/36 * 4 + 30/36 * (-1) = -0,1666666667

I hope that makes it clearer!