# Coin flip deal (stats/probability)

• bowlbase

## The Attempt at a Solution

I got the expectation value from the distribution to be exactly 2. Meaning, out of 5 flips two results can be expected to be heads. This means I have a 2/5 probability of heads (40%). Out of 20 flips this means 8 are likely to be heads and since 64>15 this is a terrible deal. So immediately I know he'll have to pay around $49 dollars to his friend. My issue is with variance. The problem hints that the variance of any variable is the E(X2)-E(X)2 and that I already have a simple formula for variance of the probability distribution. I calculated the variance of the expectation using the above equation and got 3.2. But I really don't know what to do with this or what other variance he is speaking about. Can anyone point me in the right direction? Please state the problem exactly as it was given to you, including the data you were given. bowlbase said: I calculated the variance of the expectation using the above equation and got 3.2. The expectation value of a distribution is a number and not a stochastic variable. As such its variance will be zero. Did you intend to say that you calculated the variance of the distribution? Hint: How would you express the expectation value of money lost in terms of your stochastic variable X (i.e., the number of heads)? bowlbase said: Out of 20 flips this means 8 are likely to be heads You'll get a clearer idea of concepts if you are careful with your language. It isn't "likely" that exactly 8 of the 20 flips will be heads. You mean "the expected" number of heads is 8. In many situations the probability of "the expected number" actually happening is quite small. You've assumed that because "the expected number of heads" is 8 that "the expected number of (heads squared) " must be 8 squared. That isn't true, in general. In general, (the expected value of x^2) $\ne$ (the expected value of x)^2 ( Look at the distribution you were given for 5 flips and see if the expected number of (heads squared) is 4.) My issue is with variance. The problem hints that the variance of any variable is the E(X2)-E(X)2 and that I already have a simple formula for variance of the probability distribution. I calculated the variance of the expectation using the above equation and got 3.2. But I really don't know what to do with this or what other variance he is speaking about. Can anyone point me in the right direction? You can see from that formula that if E(X^2) = (E(X))^2 then the variance would be zero. The "random" variable would need to have a constant outcome. Your were directed to consider the variance in order to help figure out the expected number of (heads^2). Have you studied a formula for the variance of the sum of independent random variables? The 20 tosses can be regarded as a sum of 4 groups of 5 tosses each. A bit more information then. I'm given the following data: (I couldn't get a latex table to work on here for some reason) X:-------- 0---------1---------2---------3---------4--------5 Pr(X): 0.07776 | 0.2592 | 0.3456 | 0.2304 | 0.0768 | 0.01024 X is the number of times heads appears in 5 rolls. From this I used the summation ## ∑_{i=0}^5 i(Pr(x)) ## and this gave me an expectation of 2. So, it is expected that 2 of the 5 flips will end in heads. This much is clear. But when it comes to the next part I simply don't understand how/why it needs to be so complicated. I expect 2/5 to be heads so I expect to pay 64-15=$49. I say this only because I don't understand and want a deeper explanation of the purpose of it. This isn't a stats course homework but rather a class that, apparently, expected that I know all of this already.
Orodruin said:
The expectation value of a distribution is a number and not a stochastic variable. As such its variance will be zero. Did you intend to say that you calculated the variance of the distribution?
Yes, this is what I meant. Given the above table I used Var=E(X2)-E(X)2 and got a distribution variance of 3.2. This might be the wrong way because the Hint says that "The variance of any random variable X is =E(X2)-E(X)2 and also we know a simple formula for the variance of a distribution." I'm not sure what formula this is suggesting but I wonder if it is from the Binomial distribution where Var(X)=Kp(1-p). Where Kp is suppose to be the Expectation p is the probability of heads. This would give variance of 1.2 for 2/5 or 4.8 for 8/20.
Stephen Tashi said:
You can see from that formula that if E(X^2) = (E(X))^2 then the variance would be zero. The "random" variable would need to have a constant outcome.

Your were directed to consider the variance in order to help figure out the expected number of (heads^2).

Have you studied a formula for the variance of the sum of independent random variables? The 20 tosses can be regarded as a sum of 4 groups of 5 tosses each.

I understand that E(X^2) = (E(X))^2 isn't true here. But I'm stuck on why this matters. If I expect 8 heads then I should expect to pay 82. My understanding of statistics fails here, clearly. Your last question seems like exactly what I did at the beginning to get the variance of the 5 tosses. If I was suppose to get the expectation squared of that distribution (4) or (16/20) I'd end up with a variance of 8. Meaning, I suppose, that he'd pay anywhere from nothing (gets $15) to$(256-15). But I can't imagine this is right. This variance seems super high to me.

Thanks for the help.

bowlbase said:
(I couldn't get a latex table to work on here for some reason)

Yes, last time I tried them, latex tables didn't work on the forum. You can try matrices.

X:-------- 0---------1---------2---------3---------4--------5
Pr(X): 0.07776 | 0.2592 | 0.3456 | 0.2304 | 0.0768 | 0.01024

X is the number of times heads appears in 5 rolls. From this I used the summation ## ∑_{i=0}^5 i(Pr(x)) ## and this gave me an expectation of 2. So, it is expected that 2 of the 5 flips will end in heads.

Let Y = X^2 and use the corresponding table:

Y:-------- 0---------1---------4---------9---------16-------25
Pr(Y): 0.07776 | 0.2592 | 0.3456 | 0.2304 | 0.0768 | 0.01024

Find the expected value of Y

Stephen Tashi said:
Yes, last time I tried them, latex tables didn't work on the forum. You can try matrices.
Let Y = X^2 and use the corresponding table:

Y:-------- 0---------1---------4---------9---------16-------25
Pr(Y): 0.07776 | 0.2592 | 0.3456 | 0.2304 | 0.0768 | 0.01024

Find the expected value of Y
For some reason I said 3.2 before, I think. But it is E(Y)=5.2. The variance was 3.2. Not sure if that was clear before.

bowlbase said:
(I couldn't get a latex table to work on here for some reason)
X:-------- 0---------1---------2---------3---------4--------5
Pr(X): 0.07776 | 0.2592 | 0.3456 | 0.2304 | 0.0768 | 0.01024

X is the number of times heads appears in 5 rolls. From this I used the summation ## ∑_{i=0}^5 i(Pr(x)) ## and this gave me an expectation of 2. So, it is expected that 2 of the 5 flips will end in heads.

This much is clear. But when it comes to the next part I simply don't understand how/why it needs to be so complicated. I expect 2/5 to be heads so I expect to pay 64-15=$49. I say this only because I don't understand and want a deeper explanation of the purpose of it. This isn't a stats course homework but rather a class that, apparently, expected that I know all of this already. Yes, this is what I meant. Given the above table I used Var=E(X2)-E(X)2 and got a distribution variance of 3.2. This might be the wrong way because the Hint says that "The variance of any random variable X is =E(X2)-E(X)2 and also we know a simple formula for the variance of a distribution." I'm not sure what formula this is suggesting but I wonder if it is from the Binomial distribution where Var(X)=Kp(1-p). Where Kp is suppose to be the Expectation p is the probability of heads. This would give variance of 1.2 for 2/5 or 4.8 for 8/20.I understand that E(X^2) = (E(X))^2 isn't true here. But I'm stuck on why this matters. If I expect 8 heads then I should expect to pay 82. My understanding of statistics fails here, clearly. Your last question seems like exactly what I did at the beginning to get the variance of the 5 tosses. If I was suppose to get the expectation squared of that distribution (4) or (16/20) I'd end up with a variance of 8. Meaning, I suppose, that he'd pay anywhere from nothing (gets$15) to \$(256-15). But I can't imagine this is right. This variance seems super high to me.

Thanks for the help.

You are misleading yourself (or being mislead) by use of the word "expectation". In probability it does not mean the same thing that it means in ordinary conversation; it is a technical term, and you should avoid ascribing too much "ordinary" meaning to it.

You ask "why this matters here". It matters because you would NOT "expect to" pay ##8^2 = 64## when you "expect to" get 8 heads; that is not how expectation works in probability. As I said already, you must guard against a too literal interpretation of the word "expect".

You can just look at your data for n = 5 to see that the expected value of ##X^2## is NOT the square of the expected value of ##X##. Your data give
$$E(X^2) = \sum_{x=0}^5 x^2 p(x) = 5.20, \;\; (EX)^2 = 2^2 = 4$$

We ALWAYS have ##E(X^2) > (EX)^2## unless ##X## is a constant, non-random quantity. Sometimes the difference is relatively minor, but in other cases in can be huge.

So E(Y) = 5.2 is the expected amount you pay for (heads)^2. It isn't equal to (expected number of heads)^2 = (2)^2 = 4.

The variance of the 5 toss distribution is E(X^2) - (E(X))^2 = E(Y) - (E(X))^2 = 5.2 - 4.

The 20 tosses T can be represented as T = X1 + X2 + X3 + X4 ,where the X's are each a result of doing 5 tosses.

Find the variance of T from the fact you know the variance of each of the X's.

-------------------------

Suppose we have a random variable X and another random variable Y that is a function of X, Y = F(X). Then, in general, the expected value of Y is not equal to F evaluated at the expected value of X. An important exception to this case is when F is a linear function of X like F(X) = 3X + 5. In the linear case E( F(X)) = F(E(X)).

In your problem, the cost Y is not a linear function of the number of heads. So, unless you can show otherwise, you should assume the expected value of the cost will require some effort to compute. As to why things should be so complicated, that's a philosophical question. It's just the way they are.

Computing with mathematical expected values is more complicated than thinking in terms of "the typical case". If you think of 8 heads as the "the typical number of heads", you will think of 64 as "the typical cost". But this kind of thinking doesn't work if you want to compute the "expected cost" in the sense of the average cost.

Stephen Tashi said:
So E(Y) = 5.2 is the expected amount you pay for (heads)^2. It isn't equal to (expected number of heads)^2 = (2)^2 = 4.

The variance of the 5 toss distribution is E(X^2) - (E(X))^2 = E(Y) - (E(X))^2 = 5.2 - 4.

The 20 tosses T can be represented as T = X1 + X2 + X3 + X4 ,where the X's are each a result of doing 5 tosses.

Find the variance of T from the fact you know the variance of each of the X's.

I honestly don't know what to do with this. I get that I know the variance of each Xn (1.2). But I don't know the total number of heads for each. I the expected value is 2, I think, but this just brings me in a giant circle. Somehow I need to get E(T) without know any X values... I feel like this is exactly the same table as before just done 4 times. The variance would be exactly the same. Or is T the sum of variances (4.8)?

I'm not getting this at all.

I'm not getting this at all.

Have you studied sums of random variables? ( It seems like you have missed an important lecture.)

T = X1 + X2 + X3 + X4 is an equation involving random variables, not ordinary variables. Random variables don't have specific values. They do have specific expectations and variances. The expectation of a sum is the sum of the expectations E(T) = E(X1) + E(X2) + E(X3) + E(X4) Each the Xi is random number of heads in 5 tosses of the coin. So each Xi has expectaion 2.

The variance of a sum of independent random variables is the sum of the variances.

Var(T) = Var(X1) + Var(X2) + Var(X3) + Var(X4)

bowlbase said:
I honestly don't know what to do with this. I get that I know the variance of each Xn (1.2). But I don't know the total number of heads for each. I the expected value is 2, I think, but this just brings me in a giant circle. Somehow I need to get E(T) without know any X values... I feel like this is exactly the same table as before just done 4 times. The variance would be exactly the same. Or is T the sum of variances (4.8)?

I'm not getting this at all.
The expected value of X2 means its average value (what you will pay out on average here). That means it's the sum ##\Sigma x^2 P[X=x]##, usually written E(X2). What equation relates that to the variance?

Stephen Tashi said:
Have you studied sums of random variables? ( It seems like you have missed an important lecture.)

T = X1 + X2 + X3 + X4 is an equation involving random variables, not ordinary variables. Random variables don't have specific values. They do have specific expectations and variances. The expectation of a sum is the sum of the expectations E(T) = E(X1) + E(X2) + E(X3) + E(X4) Each the Xi is random number of heads in 5 tosses of the coin. So each Xi has expectaion 2.

The variance of a sum of independent random variables is the sum of the variances.

Var(T) = Var(X1) + Var(X2) + Var(X3) + Var(X4)

I promise I haven't missed a lecture (there was only 1 on this), we just didn't discuss anything in great detail as, apparently, it was expected that I have some background in this.

It makes sense to me that the V(T) is the sum of the variances for each Xn and, likewise, it makes sense that E(T) is the sum of E(Xn). So, I should be able to say, I think, that ##V(T)=4.8=E(T^2)-E(T)^2## Where ##E(T)^2=64##. Meaning that ##E(T^2)=59.2## And this is the expected payout. Have I got it now?

bowlbase said:
I promise I haven't missed a lecture (there was only 1 on this), we just didn't discuss anything in great detail as, apparently, it was expected that I have some background in this.
That explains the situation. The problem looks like the kind that would be given to students after they'd studied random variables, sums of random variables, and expectations and variances of sums. (There are also important results concerning the expectation of products of random variables.)

It makes sense to me that the V(T) is the sum of the variances for each Xn and, likewise, it makes sense that E(T) is the sum of E(Xn). So, I should be able to say, I think, that ##V(T)=4.8=E(T^2)-E(T)^2## Where ##E(T)^2=64##.
Yes

Meaning that ##E(T^2)=59.2## And this is the expected payout.

From the above equation E(T^2) = E(T)^2 + 4.8 instead of -4.8.

bowlbase said:
I promise I haven't missed a lecture (there was only 1 on this), we just didn't discuss anything in great detail as, apparently, it was expected that I have some background in this.

It makes sense to me that the V(T) is the sum of the variances for each Xn and, likewise, it makes sense that E(T) is the sum of E(Xn). So, I should be able to say, I think, that ##V(T)=4.8=E(T^2)-E(T)^2## Where ##E(T)^2=64##. Meaning that ##E(T^2)=59.2## And this is the expected payout. Have I got it now?

Not quite: ##4.8+64 \neq 59.2##.

Oh geez, I don't know why I did that. I got it, now. I'm going to think on this a bit and come back if I have any other questions. This is going to take some amount of googling to figure out.

Thanks for the help, everyone!

## 1. What is the probability of getting heads on a coin flip?

The probability of getting heads on a coin flip is 50%, or 1 in 2. This is because there are only two possible outcomes - heads or tails - and they are equally likely.

## 2. What is the expected outcome of 10 coin flips?

The expected outcome of 10 coin flips is 5 heads and 5 tails. This is based on the law of large numbers, which states that as the number of trials increases, the observed results will approach the expected probability. In this case, the expected probability of heads is 50% and tails is also 50%, so after 10 flips, we would expect to see an equal number of each.

## 3. How does the law of probability apply to coin flips?

The law of probability states that the more times an experiment is repeated, the closer the observed results will be to the expected probability. In the case of coin flips, this means that as the number of flips increases, the number of heads and tails will approach a 50/50 split.

## 4. What is the probability of getting 3 heads in a row on a coin flip?

The probability of getting 3 heads in a row on a coin flip is 12.5%, or 1 in 8. This can be calculated by multiplying the individual probabilities together - in this case, 0.5 x 0.5 x 0.5 = 0.125. This is because each coin flip is an independent event, so the probability of getting heads on each flip does not change.

## 5. How does the concept of probability relate to real-life situations?

Probability is used in many real-life situations, such as predicting the weather, medical diagnoses, and gambling. It helps us understand the likelihood of certain outcomes and make informed decisions based on that information. In the case of coin flips, understanding probability can help us make better predictions and manage risk.

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