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

In summary: This problem involves finding the expected value, or the amount of money a player can expect to win or lose per game. In summary, the player has a 3/8 chance of winning $27 and a 5/8 chance of losing $21, resulting in an expected loss of $3 per game.
  • #1
Karebear
9
0

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

 
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  • #2
Karebear said:
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?
 
  • #3
Yes I do, I also understand that all possible outcomes are: (HHH,THH,HTH,HHT,TTT,HTT,THT,TTH)
 
  • #4
Good. So can you rephrase your question? (Or did you already figure it out?)
 
  • #5
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
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
Karebear said:
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.)
 
  • #8
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
The probability of not getting exactly 1 head is 7/8... should I have used the 1-P(loose) formula?
 
  • #10
Karebear said:
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.
 
  • #11
Karebear said:
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.
 
  • #12
ok so I will try again

P(win) = 1/8
P(loose) = 7/8
?
ugh, confused again... sorry is this at all right?
 
  • #13
Karebear said:
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.
 
  • #14
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
Karebear said:
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.
 
  • #16
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
Karebear said:
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.
 

1. What is expected value?

Expected value is a way to calculate the average outcome of a random event by multiplying the possible outcomes by their respective probabilities and adding them together. It represents the long-term average result of repeating the event many times.

2. How is expected value calculated?

To calculate expected value, you multiply each possible outcome by its probability and then add all of the results together. For example, if a coin flip has a 50% chance of landing on heads (with a value of $2) and a 50% chance of landing on tails (with a value of $0), the expected value would be ($2 x 0.5) + ($0 x 0.5) = $1.

3. What is the significance of expected value in statistics?

Expected value is a useful tool for understanding the likelihood of certain outcomes and making decisions based on those probabilities. It is also an important concept in probability theory and can be applied to many real-world scenarios, such as gambling, insurance, and business decisions.

4. Can expected value be negative?

Yes, expected value can be negative if the possible outcomes have negative values or if the probabilities are skewed in a way that results in a negative overall value. This is often the case in risky or uncertain situations.

5. How is expected value used in decision making?

Expected value can be used in decision making to determine the most favorable option based on the potential outcomes and their probabilities. For example, a business may use expected value to determine which investment or project will likely result in the highest return. However, it should be noted that expected value is not the only factor to consider in decision making and should be used in conjunction with other information and considerations.

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