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Quote:Ace2Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?

link to original post

A related question (it is easy if you look at it at the right way): show that the expected value of the check is the same for n flips and for n-1 flips, if n is even.

The average check amount is (1,000,000 * 2 / π)^.5 = $797.88

I described this method back in April (link below). At that time the Wizard’s reply was that I was “overthinking this”, but since then he published a pdf and Ask the Wizard problem using this method

https://wizardofvegas.com/forum/questions-and-answers/gambling/35821-sports-bet-size/#post802206

The average settlement amount is a very useful number. For instance, let’s say you’re going to Vegas for the weekend and expect to play 600 blackjack hands at $100 per hand. Your expectation is to lose $300 +/- 2,800, but that -$300 is basically meaningless…it’s a very long term average of results. But take $2,800 * (2/π)^.5 to get an average settlement amount of about $2,200 and that is much more useful information…it tells you that your result for the weekend will probably resemble something like -$2,500 or +$1,900. As mentioned, I originally posted about the average settlement amount earlier to get a practical figure like this

Quote:GMQuote:Ace2Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?

link to original postIf we use the normal approximation to the binomial distribution, we get that the expected value of the check is about sqrt(2n/π)≈0.7978845*sqrt(n) and the standard deviation is about sqrt((π-2)n/π)≈0.6028103*sqrt(n). For n=1000000, this gives expected value $797.8845 and standard deviation $602.8103. The values I calculated without approximation are $797.8843 and $602.8105.

A related question (it is easy if you look at it at the right way): show that the expected value of the check is the same for n flips and for n-1 flips, if n is even.

link to original post

Shows you how much I know, at least when it comes to normal distributions.

Is there an online source that shows how you came up with those mean and SD values?

The normal distribution has a height of 1/(2π)^.5 at x=0, so that’s it’s mean value. For the coin flip problem, we are only looking at one side of the distribution (since we don’t care if it’s + or -) but we are now saying that one side is the entire distribution. It still has to total 1, so the mean becomes 2 * 1/(2π)^.5 = (2/π)^.5Quote:ThatDonGuyQuote:GMQuote:Ace2Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?

link to original postIf we use the normal approximation to the binomial distribution, we get that the expected value of the check is about sqrt(2n/π)≈0.7978845*sqrt(n) and the standard deviation is about sqrt((π-2)n/π)≈0.6028103*sqrt(n). For n=1000000, this gives expected value $797.8845 and standard deviation $602.8103. The values I calculated without approximation are $797.8843 and $602.8105.

A related question (it is easy if you look at it at the right way): show that the expected value of the check is the same for n flips and for n-1 flips, if n is even.

link to original post

Shows you how much I know, at least when it comes to normal distributions.

Is there an online source that shows how you came up with those mean and SD values?

link to original post

(2π)^.5 also appears in Stirling’s formula for factorials :

(2πn)^.5 * (n/e)^n. Probably appears in many places

Quote:Ace2Quote:ThatDonGuy

Shows you how much I know, at least when it comes to normal distributions.

Is there an online source that shows how you came up with those mean and SD values?

The normal distribution has a height of 1/(2π)^.5 at x=0, so that’s it’s mean value.

Er...huh? Isn't the mean of a normal distribution zero?

I can get sqrt(2/π) as the mean by dividing the integral over 0 to positive infinity of x f(x) dx by the integral over 0 to positive infinity of f(x) dx, where f(x) = 1 / sqrt(2π) * e^(-x%2/2), but I am looking for an easier way to get that number.

Oh, and while I'm in pedantic mode: "it's mean value" means "it is mean value." The word for "something belonging to it" - "its" - does not have an apostrophe.

Are you kidding me?Quote:ThatDonGuy

Oh, and while I'm in pedantic mode: "it's mean value" means "it is mean value." The word for "something belonging to it" - "its" - does not have an apostrophe.

link to original post

Though I’ll admit that language skills are not my strong point, I do know the difference between its and it’s. My iPhone’s “autocorrect” probably did that.

I’m quite surprised you nitpicked that so allow me to make a constructive criticism.

Before I posted this basic coin flip problem, I triple-checked the wording, ensuring there was absolutely no ambiguity to what was being asked. The problem ends with:

“What is the expected value and standard deviation of the check amount ?”

I thought there was no possible way someone could misunderstand that question. But, I was wrong…you did not comprehend it and you answered that the expected value is zero (which is the answer to a different question). You didn’t actually “get it” until after I made a second post with a detailed example.

So your reading comprehension skills have lots of room for improvement

Quote:Ace2

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?

link to original post

It seems this is exactly the same problem as the one-dimensional random walk problem on Wikipedia. To find the exact solution, you need the Pascal's triangle to solve it, but usually the number N is large enough to guarantee the use of its limit. Is this right?

Standard deviation hasn't been considered very much. what is the S.D. limit when N approaches infinity? It seems a previous post says it is sqrt((π-2)n/π).

Quote:Ace2So your reading comprehension skills have lots of room for improvement

link to original post

I have tremendous respect for both of you, especially your math skills. There was obviously some miscommunication on this problem.

I'd like to suggest we pass the peace pipe.

Quote:Wizard

I'd like to suggest we pass the peace pipe.

link to original post

Is that a Futurama reference there ? :)

https://youtu.be/SCFkr7Wj9TQ?t=66

Quote:Ace2Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?

In case anyone is interested, here is how the EV and SD are found via integration

The normal distribution is represented by the function (2π)^(-.5) * e^(-.5x^2). To find the mean value of x on one side of the distribution, we must take the product of all probabilities (of one side) and their corresponding x values and then sum them. This is given by the integral from zero to infinity of: (2π)^(-.5) * e^(-.5x^2) * x dx

There is an antiderivative for this integral as follows:

Let u = (-.5x^2) and substitute into above integral

du/dx = -x

-du = x dx

Now we can rewrite (2π)^(-.5) * e^u * x dx as:

-(2π)^(-.5) * e^u du

For which the antiderivative is simply: -(2π)^(-.5) * e^u

Undoing the substitution gives us the antiderivative of:

-(2π)^(-.5) * e^(-.5x^2)

We are evaluating this from zero to infinity, which is:

0 - (-(2π)^(-.5)) = (2π)^(-.5)

Since this is for half of the distribution, multiply by 2 to get the mean value of: (2/π)^(.5) =~ 0.79788. Multiply by 1,000,000^.5 to get the expected check value of $797.88

Now that we know the mean value of x (on one side) is (2/π)^(.5), we can calculate the variance by taking the integral from zero to infinity of:

(2π)^(-.5) * e^(-.5x^2) * (x - (2/π)^(.5))^2 dx

I don't believe there's a way to get an exact antiderivative for this function, so I used an integral calculator to get the answer of: (π - 2) / (2π)

That is the variance for half the distribution, so multiply by 2 to get the full variance of: (π - 2) / π

Therefore the standard deviation is ((π - 2) / π)^.5 =~ 0.60281. Multiply by 1,000,000^.5 to get the check’s SD of $602.81