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GM
GM
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January 26th, 2022 at 8:23:37 AM permalink
Quote: Ace2

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

Itís $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


If 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.
Ace2
Ace2
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January 26th, 2022 at 8:44:57 AM permalink
Thanks GM, that is the correct answer for the check amount. Still reviewing your SD answer, which is very close to mine

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 Wizardís reply was that I was ďoverthinking thisĒ, 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, letís say youíre 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 meaninglessÖitís 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 informationÖit 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
Itís all about making that GTA
ThatDonGuy
ThatDonGuy 
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January 26th, 2022 at 10:51:18 AM permalink
Quote: GM

Quote: Ace2

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

Itís $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


If 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?
Ace2
Ace2
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January 27th, 2022 at 4:32:26 PM permalink
Quote: ThatDonGuy

Quote: GM

Quote: Ace2

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

Itís $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


If 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

The normal distribution has a height of 1/(2π)^.5 at x=0, so thatís itís mean value. For the coin flip problem, we are only looking at one side of the distribution (since we donít care if itís + 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)^.5

(2π)^.5 also appears in Stirlingís formula for factorials :
(2πn)^.5 * (n/e)^n. Probably appears in many places
Last edited by: Ace2 on Jan 27, 2022
Itís all about making that GTA
ThatDonGuy
ThatDonGuy 
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January 27th, 2022 at 5:06:56 PM permalink
Quote: Ace2

Quote: 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 thatís itís 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.
Ace2
Ace2
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January 28th, 2022 at 6:13:39 PM permalink
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

Are you kidding me?

Though Iíll admit that language skills are not my strong point, I do know the difference between its and itís. My iPhoneís ďautocorrectĒ probably did that.

Iím 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:

ďWhat is the expected value and standard deviation of the check amount ?Ē

I thought there was no possible way someone could misunderstand that question. But, I was wrongÖyou did not comprehend it and you answered that the expected value is zero (which is the answer to a different question). You didnít actually ďget itĒ until after I made a second post with a detailed example.

So your reading comprehension skills have lots of room for improvement
Last edited by: Ace2 on Jan 28, 2022
Itís all about making that GTA
aceside
aceside
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January 29th, 2022 at 2:52:53 PM permalink
Quote: Ace2

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

Itís $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).
Wizard
Administrator
Wizard
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January 29th, 2022 at 9:10:31 PM permalink
Quote: Ace2

So 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.
ďExtraordinary claims require extraordinary evidence.Ē -- Carl Sagan
TwelveOr21
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January 29th, 2022 at 9:17:07 PM permalink
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
Ace2
Ace2
Joined: Oct 2, 2017
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February 1st, 2022 at 2:28:41 PM permalink
Quote: Ace2

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

Itís $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 checkís SD of $602.81
Last edited by: Ace2 on Feb 1, 2022
Itís all about making that GTA

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