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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 ?

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

The expected value is zero

Variance =

2 x C(1,000,000, 0) x 1 / 2^1,000,000 x 1,000,000^2

+ 2 x C(1,000,000, 1) x 1 / 2^1,000,000 x 999,998^2

+ 2 x C(1,000,000, 2) x 1 / 2^1,000,000 x 999,996^2

+ ...

+ 2 x C(1,000,000, 499,999) x 1 / 2^1,000,000 x 2^2

= 8 / 2^1,000,000 x (

C(1,000,000, 0) x 500,000^2

+ C(1,000,000, 1) x 499,999^2

+ ...

+ C(1,000,000, 499,999) x 1^2

)

= 1 / 2^999,997 x (

C(1,000,000, 0) x 500,000^2

+ C(1,000,000, 1) x 499,999^2

+ ...

+ C(1,000,000, 499,999) x 1^2

)

My computer claims the result is exactly 1,000,000; I'll have to see if I can compute that manually

Anyway, the SD = sqrt(variance) = 1000

Expected Value = 0; SD = $1000

I disagree.Quote:ThatDonGuyQuote: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

The expected value is zero

Variance =

2 x C(1,000,000, 0) x 1 / 2^1,000,000 x 1,000,000^2

+ 2 x C(1,000,000, 1) x 1 / 2^1,000,000 x 999,998^2

+ 2 x C(1,000,000, 2) x 1 / 2^1,000,000 x 999,996^2

+ ...

+ 2 x C(1,000,000, 499,999) x 1 / 2^1,000,000 x 2^2

= 8 / 2^1,000,000 x (

C(1,000,000, 0) x 500,000^2

+ C(1,000,000, 1) x 499,999^2

+ ...

+ C(1,000,000, 499,999) x 1^2

)

= 1 / 2^999,997 x (

C(1,000,000, 0) x 500,000^2

+ C(1,000,000, 1) x 499,999^2

+ ...

+ C(1,000,000, 499,999) x 1^2

)

My computer claims the result is exactly 1,000,000; I'll have to see if I can compute that manually

Anyway, the SD = sqrt(variance) = 1000

Expected Value = 0; SD = $1000

link to original post

If, for instance, they flipped three times, then there is a 2/8 chance of a $3 check and a 6/8 chance of a $1 check going either way, for an expected check amount of $1.50.

Yes it’s true than both players have an expected gain/loss of zero (1/8 * 3 + 3/8 * 1 - 3/8 * 1 + 1/8 * 3), but that’s not what’s being asked. Looking for the expected settlement amount.

Also disagree on standard deviation

Quote:Ace2I disagree.Quote:ThatDonGuyQuote: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

The expected value is zero

Variance =

2 x C(1,000,000, 0) x 1 / 2^1,000,000 x 1,000,000^2

+ 2 x C(1,000,000, 1) x 1 / 2^1,000,000 x 999,998^2

+ 2 x C(1,000,000, 2) x 1 / 2^1,000,000 x 999,996^2

+ ...

+ 2 x C(1,000,000, 499,999) x 1 / 2^1,000,000 x 2^2

= 8 / 2^1,000,000 x (

C(1,000,000, 0) x 500,000^2

+ C(1,000,000, 1) x 499,999^2

+ ...

+ C(1,000,000, 499,999) x 1^2

)

= 1 / 2^999,997 x (

C(1,000,000, 0) x 500,000^2

+ C(1,000,000, 1) x 499,999^2

+ ...

+ C(1,000,000, 499,999) x 1^2

)

My computer claims the result is exactly 1,000,000; I'll have to see if I can compute that manually

Anyway, the SD = sqrt(variance) = 1000

Expected Value = 0; SD = $1000

link to original post

If, for instance, they flipped three times, then there is a 2/8 chance of a $3 check and a 6/8 chance of a $1 check going either way, for an expected check amount of $1.50.

Yes it’s true than both players have an expected gain/loss of zero (1/8 * 3 + 3/8 * 1 - 3/8 * 1 + 1/8 * 3), but that’s not what’s being asked. Looking for the expected settlement amount.

Also disagree on standard deviation

link to original post

Understood. You want the mean and the SD of the value of the check, regardless of which player pays it to the other.

For the mean, I get about 797.884361. I do have an exact answer, but it is a fraction with a numerator having 301,029 digits.

Because of this, calculating the SD may take a while...

I ask a related question in Ask the Wizard #358.

Based on the method in that solution, for n flips my estimate is 2^2*sqrt(n/4) * sqrt(2/pi).

The reason for multiplying by 2^2 is with every flip there is a change in money of $2. We square that $2 to get the variance.

Note: Answer updated 7:15 PST 1/25/22, because I didn't correctly account for $2 difference between winning and losing for each flip.

Quote:ThatDonGuy

For the mean, I get about 797.884361. I do have an exact answer, but it is a fraction with a numerator having 301,029 digits.

Because of this, calculating the SD may take a while...

link to original post

Also looking for a formulaic solution for the standard deviation of the check amount, which can also be an accurate estimate

Quote:Ace2Quote:ThatDonGuy

For the mean, I get about 797.884361. I do have an exact answer, but it is a fraction with a numerator having 301,029 digits.

Because of this, calculating the SD may take a while...

link to original postI agree with that answer ($797.88) for the mean, but I’d like to see a formulaic solution for full credit. The solution can be an estimate, albeit a very accurate one (to at least five digits).

Also looking for a formulaic solution for the standard deviation of the check amount, which can also be an accurate estimate

link to original post

I assume by "formulatic solution," you mean something other than the sum of:

C(1,000,000, 0) x (1/2)^1,000,000 x (1/2)^0 x 1,000,000

C(1,000,000, 1) x (1/2)^999,999 x (1/2)^1 x 999,998

C(1,000,000, 2) x (1/2)^999,998 x (1/2)^2 x 999,996

...

C(1,000,000, 499,999) x (1/2)^500,001 x (1/2)^499,999 x 2

C(1,000,000, 500,000) x (1/2)^500,000 x (1/2)^500,000 x 0

C(1,000,000, 500,001) x (1/2)^499,999 x (1/2)^500,001 x 2

...

C(1,000,000, 999,999) x (1/2)^1 x (1/2)^999,999 x 999,998

C(1,000,000, 1,000,000) x (1/2)^0 x (1/2)^1,000,000 x 1,000,000

Yes. I guess I forgot that some people have software to “brute force” things of this magnitude. When I ran some tests in excel, it would not let me calculate combinations too far above 1000.Quote:ThatDonGuyQuote:Ace2Quote:ThatDonGuy

For the mean, I get about 797.884361. I do have an exact answer, but it is a fraction with a numerator having 301,029 digits.

Because of this, calculating the SD may take a while...

link to original postI agree with that answer ($797.88) for the mean, but I’d like to see a formulaic solution for full credit. The solution can be an estimate, albeit a very accurate one (to at least five digits).

Also looking for a formulaic solution for the standard deviation of the check amount, which can also be an accurate estimate

link to original post

I assume by "formulatic solution," you mean something other than the sum of:

C(1,000,000, 0) x (1/2)^1,000,000 x (1/2)^0 x 1,000,000

C(1,000,000, 1) x (1/2)^999,999 x (1/2)^1 x 999,998

C(1,000,000, 2) x (1/2)^999,998 x (1/2)^2 x 999,996

...

C(1,000,000, 499,999) x (1/2)^500,001 x (1/2)^499,999 x 2

C(1,000,000, 500,000) x (1/2)^500,000 x (1/2)^500,000 x 0

C(1,000,000, 500,001) x (1/2)^499,999 x (1/2)^500,001 x 2

...

C(1,000,000, 999,999) x (1/2)^1 x (1/2)^999,999 x 999,998

C(1,000,000, 1,000,000) x (1/2)^0 x (1/2)^1,000,000 x 1,000,000

link to original post

Note that the mean is very close to sqrt(1,000,000) / sqrt(2 PI)

Change the problem so that the mean is 1 / sqrt(2 PI); the SD for 1,000,000 samples is sqrt(M (1 - M) / 1,000,000) = sqrt(M (1 - M)) / 1000

For M = 1 / sqrt(2 PI), SD = sqrt((sqrt(2 PI) - 1) / sqrt(2 PI)) / 1000 = 0.77528 / 1000

If you multiply this by 1,000,000, which is how many tosses were made, you get SD = 775.28

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

Expected value= death by sleep deprivation and kidney failure.

Standard deviation= arthritis and thumb amputation from flipping