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GM
GM
Joined: Jun 16, 2021
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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
Joined: Oct 2, 2017
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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 
Joined: Jun 22, 2011
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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
Joined: Oct 2, 2017
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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 
Joined: Jun 22, 2011
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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.

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