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

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Homework Help Overview

The discussion revolves around calculating the expected value in a betting game involving coin tosses. The game stipulates that winning occurs with exactly one head in three tosses, yielding a reward of $27, while failing to achieve this results in a loss of $21.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the probabilities associated with different outcomes of coin tosses, specifically focusing on the number of heads obtained. Questions arise regarding the calculation of probabilities for 0, 1, 2, and 3 heads, with some participants expressing confusion about the underlying concepts.

Discussion Status

There is an ongoing exploration of the probabilities and expected value calculations. Some participants have provided guidance on understanding the probabilities of winning and losing, while others are attempting to clarify their reasoning and calculations. Multiple interpretations of the expected value are being discussed, but no consensus has been reached regarding the final understanding.

Contextual Notes

Participants note the constraints of the problem, including the specific outcomes of the coin tosses and the associated probabilities. There is an emphasis on understanding the definitions and calculations involved in determining expected value, with some participants expressing frustration over their understanding of the concepts.

Karebear
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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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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?
 
Yes I do, I also understand that all possible outcomes are: (HHH,THH,HTH,HHT,TTT,HTT,THT,TTH)
 
Good. So can you rephrase your question? (Or did you already figure it out?)
 
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
 
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?
 
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.)
 
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?
 
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.
 

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