How Do You Calculate Expected Value in a Coin Toss Betting Game?

[Karebear]

Homework Statement

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

Homework Equations

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

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

 

[Doc Al]

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

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?

 

[Karebear]

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

 

[Doc Al]

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

 

[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

 

[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?

 

[Doc Al]

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

Exactly.

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

 

[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?

 

[Karebear]

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

 

[Doc Al]

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

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.

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?

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.

 

[Doc Al]

The probability of not getting exactly 1 head is 7/8...

Nope.

should I have used the 1-P(loose) formula?

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.

 

[Karebear]

ok so I will try again

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

 

[Doc Al]

ok so I will try again

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

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

 

[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

 

[Doc Al]

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

Exactly! Keep going.

 

[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?

 

[Doc Al]

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?

Perfect!

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