# VERY Basic Expected Value Question

1. Feb 5, 2010

### Karebear

1. The problem statement, all variables and given/known data

If you throw exactly 1 head in three tosses of a coin you win $27. If not you pay$21.

2. Relevant equations
I know, E(X) = sum[x subscript i * P(X = x subscript i)]

3. The attempt at a solution
Sorry y'all I don't really know how to use all the fancy letters and signs on here, but I'm tryin.
OK so, I get stuck at the very beginning of the problem trying to find P(X = x)
Here is what I have:

Possible Outcomes:
x = # of heads: P(X=x)
0 1/8 = 0.125
1 3/8 = 0.375
2 ?
3 ?

My problem is that I don't really understand why the P(X=x) for 0 heads is 1/8 I know I should, I just can't seem to remember and I'm getting frustrated which is making everything harder. I do understand that there are 8 possible outcomes b/c 2 sides to a coin and 3 tosses so 2^3 = 8
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 5, 2010

### Staff: Mentor

Do you further understand that each of those 8 possible outcomes is equally likely, thus the probability of each is 1/8? And that '0 heads' corresponds to just one of those outcomes, namely TTT?

3. Feb 5, 2010

### Karebear

Yes I do, I also understand that all possible outcomes are: (HHH,THH,HTH,HHT,TTT,HTT,THT,TTH)

4. Feb 5, 2010

### Staff: Mentor

Good. So can you rephrase your question? (Or did you already figure it out?)

5. Feb 5, 2010

### Karebear

Am I right in my thinking that with 1 head the probability is 3/8 because head only occurs once in (HTT,THT,TTH) so,(1/8)+(1/8)+(1/8) = 3/8

6. Feb 5, 2010

### Karebear

I figured it out, ugh... I think I just psyched myself out over it... do you mind if I work the rest of it out so you can tell me if I am on the right track?

7. Feb 5, 2010

### Staff: Mentor

Exactly.

So what's the probability of you not getting exactly 1 head in three tosses? (No need to enumerate all the possibilities.)

8. Feb 5, 2010

### Karebear

E(X) = (0.125+0.375+0.375+0.125)/4
= 1/4
= 0.25

ok, why is it called an expected value? I think I don't get it, because I don't really understand the purpose of it... what does the 0.25 actually mean?

9. Feb 5, 2010

### Karebear

The probability of not getting exactly 1 head is 7/8.... should I have used the 1-P(loose) formula?

10. Feb 5, 2010

### Staff: Mentor

I'm not sure what you are doing here, but that's not the expected value. Your first post had the correct definition of E(x)--assuming I understand your notation. Use that.
For this game, you can think of the expected value as how much you would 'expect' to win (or lose) per game--on average--if you played many, many games.

11. Feb 5, 2010

### Staff: Mentor

Nope.
Yep.

You have two situations, A and not-A. A = getting exactly 1 head in three tosses. The probability of A plus the probability of not-A must add to 1.

12. Feb 5, 2010

### Karebear

ok so I will try again

P(win) = 1/8
P(loose) = 7/8
?
ugh, confused again... sorry is this at all right?

13. Feb 5, 2010

### Staff: Mentor

No. You already figured out that the probability of winning was 3/8 in post #5. Make use of that.

14. Feb 5, 2010

### Karebear

Deep Breath and trying again... so, the probability that I win (by getting 1 head in 3 tosses) is 3/8; so the probability that I loose (by getting anything else) is 5/8... am i on the right track

15. Feb 5, 2010

### Staff: Mentor

Exactly! Keep going.

16. Feb 5, 2010

### Karebear

so...
E(X) = 27.00 *(3/8) - 21.00 * (5/8)
= 10.125 - 13.125
= - $3.00 So the player should expect to loose$3 per game?

17. Feb 5, 2010

### Staff: Mentor

Perfect!

Yes, on average the player will lose \$3 per game.