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ThatDonGuy
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May 9th, 2024 at 3:32:08 PM permalink
Quote: unJon

Not sure this is optimal as feels like the kind of thing that should go to 1/e as bikes go to infinity but
link to original post



Since the total is 100 x a sum of reciprocals, the distance diverges as the number of cars approaches infinity

unJon
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May 9th, 2024 at 3:44:58 PM permalink
Quote: ThatDonGuy

Quote: unJon

Not sure this is optimal as feels like the kind of thing that should go to 1/e as bikes go to infinity but
link to original post



Since the total is 100 x a sum of reciprocals, the distance diverges as the number of cars approaches infinity


link to original post



Yeah thought that was neat. I also did find an e in there.

Since it’s really just the harmonic series, an estimate of the distance with N motorcycles is 100 * ( ln(N) + Euler’s Constant)
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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May 9th, 2024 at 4:42:45 PM permalink
Quote: Gialmere


You have 16 motorbikes gathered together at the edge of a desert plain. Each bike has a rider and a full tank of gas. A bike can travel 100 kilometers on a full tank.

Using all 16 bikes as a team, how far can you get one of the bikes to travel across the desert?
link to original post




I get 338.0728993 miles.

I'll explain my answer if I'm right and nobody else already did.



Congratulations to UnJon for beating me on this one.

Here is my thinking.

Everyone should drive as far as they can until they get to some point where one biker can top off the other 15 bikers and be left with nothing. When will that be?

Divide his gas into 16 parts. 15 of those parts will eventually be divided to the other bikers. One part will be used up on the journey. The distance traveled on 1/16 of a tank is 100/16 km = 6.25 km. At that point, everybody will have 1-6.25/100 = 93.75% of a tank. If you take the remaining gas of one biker and divide it up among the other 15, each other biker gets 0.9375/15 = 6.25% of a tank, exactly enough to top off.

Then they repeat this, but divide a tank into 15 parts. That will get them an extra 6 2/3 miles. Then they split up the gas of one biker to top off.

In the end, the distance of the furthest biker is 100*(1/16 + 1/15 + 1/14 + ... + 1/1 ) = 338.0729 km.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Ace2
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May 9th, 2024 at 4:55:48 PM permalink
The answer can also be found by taking 100 times the integral from zero to infinity of:

(1 - (1 - 1/e^(x/16))^16) / 16 =

100 * 2436559 / 720720 =~ 338.07

This formula would be useful for a high number of motorcycles
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Ace2
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May 9th, 2024 at 5:05:45 PM permalink
Quote: Wizard

Quote: Gialmere


You have 16 motorbikes gathered together at the edge of a desert plain. Each bike has a rider and a full tank of gas. A bike can travel 100 kilometers on a full tank.

Using all 16 bikes as a team, how far can you get one of the bikes to travel across the desert?
link to original post




I get 338.0728993 miles.

I'll explain my answer if I'm right and nobody else already did.



Congratulations to UnJon for beating me on this one.

Here is my thinking.

Everyone should drive as far as they can until they get to some point where one biker can top off the other 15 bikers and be left with nothing. When will that be?

Divide his gas into 16 parts. 15 of those parts will eventually be divided to the other bikers. One part will be used up on the journey. The distance traveled on 1/16 of a tank is 100/16 km = 6.25 km. At that point, everybody will have 1-6.25/100 = 93.75% of a tank. If you take the remaining gas of one biker and divide it up among the other 15, each other biker gets 0.9375/15 = 6.25% of a tank, exactly enough to top off.

Then they repeat this, but divide a tank into 15 parts. That will get them an extra 6 2/3 miles. Then they split up the gas of one biker to top off.

In the end, the distance of the furthest biker is 100*(1/16 + 1/15 + 1/14 + ... + 1/1 ) = 338.0729 km.

link to original post

FYI your first answer is expressed in miles and your solution is a mix of miles and km.
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Gialmere
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May 9th, 2024 at 5:30:01 PM permalink
It's nice to see Ace2 posting again, and it's not even a craps puzzle.
Have you tried 22 tonight? I said 22.
Ace2
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May 9th, 2024 at 5:42:36 PM permalink
Quote: unJon

Quote: ThatDonGuy

Quote: unJon

Not sure this is optimal as feels like the kind of thing that should go to 1/e as bikes go to infinity but
link to original post



Since the total is 100 x a sum of reciprocals, the distance diverges as the number of cars approaches infinity


link to original post



Yeah thought that was neat. I also did find an e in there.

Since it’s really just the harmonic series, an estimate of the distance with N motorcycles is 100 * ( ln(N) + Euler’s Constant)

link to original post

I never thought of this before, but an intuitive way to define/calculate the Euler–Mascheroni constant is to take the integral from zero to infinity of (1 - (1 - 1/e^(x/n))^n) / n dx minus the integral from one to n of 1/x dx. It will equal the EM constant as n approaches infinity and will be quite accurate at lower n
Last edited by: Ace2 on May 9, 2024
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Ace2
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May 10th, 2024 at 12:35:43 PM permalink
Quote: Wizard

The question was asked in this post on the average number of spins to see every number appear at least twice in double-zero roulette?


234.8326629288898



Integrate from 0 to infinity of 1-(1-exp(-x/38)*(1+x/38))^38 dx

Recommended integral calculator: https://www.integral-calculator.com/

link to original post

What’s the standard deviation ?
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Ace2
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Wizard
May 12th, 2024 at 11:39:57 AM permalink
Quote: Ace2

Quote: Wizard

The question was asked in this post on the average number of spins to see every number appear at least twice in double-zero roulette?


234.8326629288898



Integrate from 0 to infinity of 1-(1-exp(-x/38)*(1+x/38))^38 dx

Recommended integral calculator: https://www.integral-calculator.com/

link to original post

What’s the standard deviation ?
link to original post


If anyone cares, it’s the square root of the integral from zero to infinity of

(1 - ((x/38) + 1) / e^(x/38))^37 * (x/38) / e^(x/38) * (x - v)^2 dx

where v is the expected value of ~234.8 spins

This evaluates to a standard deviation of ~55.65 spins.
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Ace2
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May 14th, 2024 at 3:57:16 PM permalink
The expected number of rolls for a craps passline bet to be resolved is 557/165 =~ 3.38

What is the standard deviation?

Closed-form solutions only
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ThatDonGuy
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May 14th, 2024 at 4:34:12 PM permalink
Quote: Ace2

The expected number of rolls for a craps passline bet to be resolved is 557/165 =~ 3.38

What is the standard deviation?

Closed-form solutions only
link to original post


Do you already have a solution for this?

I was working on this, but it's a bit of a mess.

If m is the mean (in this case, 557/165), the variance is the sum of:
1/3 x (1 - m)^2
+ 1/6 x 1/4 x ((2 - m)^2 + 3/4 x (3 - m)^2 + (3/4)^2 x (4 - m)^2 + ...)
+ 2/9 x 5/18 x ((2 - m)^2 + 13/18 x (3 - m)^2 + (13/18)^2 x (4 - m)^2 + ...)
+ 5/18 x 11/36 x ((2 - m)^2 + 25/36 x (3 - m)^2 + (25/36)^2 x (4 - m)^2 + ...)

Now, for |a| < 1, (2 - m)^2 + a (3 - m)^2 + a^2 x (4 - m)^2 + ...
= (4 + 9 a + 16 a^2 + 25 a^3 + ...) - 2m (2 + 3 a + 4 a^2 + ...) + m^2 (1 + a + a^2 + a^3 + ...)
= (4 - 3 a + a^2) / (1 - a)^3 - 2m (2 - a) / (1 - a)^2 + m^2 / (1 - a)^2
so just plug that back into the first equation three times, with a = 3/4, 13/18, and 25/36 respectively, then calculate the sum and take the square root to get the SD.
unJon
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May 14th, 2024 at 4:36:29 PM permalink
If I know Ace2 it’ll be an integral from zero to infinity of an equation with a bunch of e to the power of stuff equation.
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Ace2
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May 14th, 2024 at 4:49:30 PM permalink
Yes I do have an exact algebraic solution and it does not use calculus

I confirmed the answer using a Markov chain
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Ace2
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May 15th, 2024 at 1:36:36 PM permalink
Quote: Ace2

The expected number of rolls for a craps passline bet to be resolved is 557/165 =~ 3.38

What is the standard deviation?

Closed-form solutions only
link to original post

Does anyone want more time or should I post the solution?
It’s all about making that GTA
ThatDonGuy
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May 15th, 2024 at 3:44:21 PM permalink
Quote: Ace2

Quote: Ace2

The expected number of rolls for a craps passline bet to be resolved is 557/165 =~ 3.38

What is the standard deviation?

Closed-form solutions only
link to original post

Does anyone want more time or should I post the solution?
link to original post




If m is the mean (in this case, 557/165), the variance is the sum of:

1/3 x (1 - m)^2
+ 1/6 x 1/4 x ((2 - m)^2 + 3/4 x (3 - m)^2 + (3/4)^2 x (4 - m)^2 + ...)
+ 2/9 x 5/18 x ((2 - m)^2 + 13/18 x (3 - m)^2 + (13/18)^2 x (4 - m)^2 + ...)
+ 5/18 x 11/36 x ((2 - m)^2 + 25/36 x (3 - m)^2 + (25/36)^2 x (4 - m)^2 + ...)

Lemma:
(2 - m)^2 + a (3 - m)^2 + a^2 x (4 - m)^2 + a^3 (5 - m)^2 + ...
= 4 + 9 a + 16 a^2 + 25 a^3 + ...
- 2m (2 + 3 a + 4 a^2 + 5 a^3 + ...)
+ m^2 (1 + 2 a + 3 a^2 + 4 a^3 + ...)

= -1/a + 1/a (1 + 4 a + 9 a^2 + 16 a^3 + ...)
- 2m ((1 + a + a^2 + a^3 + ...) + (1 + 2a + 3 a^3 + 4 a^3 + ...))
+ m^2 (1 + 2 a + 3 a^2 + 4 a^3 + ...)

= -1/a + 1/a (1 + a) / (1 - a)^3
- 2m (1 / (1 - a) + 1 / (1 - a)^2)
+ m^2 / (1 - a)^2

= -1/a + 1/a (1 + a) / (1 - a)^3 + (m^2 + 2 a m - 4 m) / (1 - a)^2

Substitute 557/165 for m and 3/4, 13/18, and 25/36 for a, gives a variance of 68,448,497 / 2,940,300

Standard Deviation = sqrt(variance) = sqrt(205,345,491) / 2,970 = about 4.8249

Last edited by: ThatDonGuy on May 15, 2024
charliepatrick
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May 15th, 2024 at 4:24:07 PM permalink
Just a quick look gets SQRT (2820384/165/165/36), but I haven't checked it yet.
Initial roll# waysE rollsDiff to average (*165)squared
2,3,7,11,1212112-3921536641843968
4 or 10653026871824430944
5 or 984.636.820240804326432
6 or 8104.2727272727272742.727272727272714821904219040
363.375757575757582820384
1652.8776492194674
5571.69636352809986
Ace2
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May 15th, 2024 at 4:26:59 PM permalink
Quote: ThatDonGuy

Quote: Ace2

Quote: Ace2

The expected number of rolls for a craps passline bet to be resolved is 557/165 =~ 3.38

What is the standard deviation?

Closed-form solutions only
link to original post

Does anyone want more time or should I post the solution?
link to original post




If m is the mean (in this case, 557/165), the variance is the sum of:

1/3 x (1 - m)^2
+ 1/6 x 1/4 x ((2 - m)^2 + 3/4 x (3 - m)^2 + (3/4)^2 x (4 - m)^2 + ...)
+ 2/9 x 5/18 x ((2 - m)^2 + 13/18 x (3 - m)^2 + (13/18)^2 x (4 - m)^2 + ...)
+ 5/18 x 11/36 x ((2 - m)^2 + 25/36 x (3 - m)^2 + (25/36)^2 x (4 - m)^2 + ...)

Lemma:
(2 - m)^2 + a (3 - m)^2 + a^2 x (4 - m)^2 + a^3 (5 - m)^2 + ...
= 4 + 9 a + 16 a^2 + 25 a^3 + ...
- 2m (2 + 3 a + 4 a^2 + 5 a^3 + ...)
+ m^2 (1 + 2 a + 3 a^2 + 4 a^3 + ...)

= -1/a + 1/a (1 + 4 a + 9 a^2 + 16 a^3 + ...)
- 2m ((1 + a + a^2 + a^3 + ...) + (1 + 2a + 3 a^3 + 4 a^3 + ...))
+ m^2 (1 + 2 a + 3 a^2 + 4 a^3 + ...)

= -1/a + 1/a (1 + a) / (1 - a)^3
- 2m (1 / (1 - a) + 1 / (1 - a)^2)
+ m^2 / (1 - a)^2

= -1/a + 1/a (1 + a) / (1 - a)^3 + (m^2 + 2 a m - 4 m) / (1 - a)^2

Substitute 557/165 for m and 3/4, 13/18, and 25/36 for a, gives a variance of 68,448,497 / 2,940,300

Standard Deviation = sqrt(variance) = sqrt(205,345,491) / 2,970 = about 4.8249


link to original post

Disagree. Did you confirm that number with anything (Markov, simulation)?

You can set up a very simple Markov chain (only four states/columns) and get the answer. My formulaic solution agrees to the Markov chain to ten digits
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ThatDonGuy
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May 15th, 2024 at 4:53:01 PM permalink
Quote: Ace2

Quote: ThatDonGuy

Quote: Ace2

Quote: Ace2

The expected number of rolls for a craps passline bet to be resolved is 557/165 =~ 3.38

What is the standard deviation?

Closed-form solutions only
link to original post

Does anyone want more time or should I post the solution?
link to original post




If m is the mean (in this case, 557/165), the variance is the sum of:

1/3 x (1 - m)^2
+ 1/6 x 1/4 x ((2 - m)^2 + 3/4 x (3 - m)^2 + (3/4)^2 x (4 - m)^2 + ...)
+ 2/9 x 5/18 x ((2 - m)^2 + 13/18 x (3 - m)^2 + (13/18)^2 x (4 - m)^2 + ...)
+ 5/18 x 11/36 x ((2 - m)^2 + 25/36 x (3 - m)^2 + (25/36)^2 x (4 - m)^2 + ...)

Lemma:
(2 - m)^2 + a (3 - m)^2 + a^2 x (4 - m)^2 + a^3 (5 - m)^2 + ...
= 4 + 9 a + 16 a^2 + 25 a^3 + ...
- 2m (2 + 3 a + 4 a^2 + 5 a^3 + ...)
+ m^2 (1 + 2 a + 3 a^2 + 4 a^3 + ...)

= -1/a + 1/a (1 + 4 a + 9 a^2 + 16 a^3 + ...)
- 2m ((1 + a + a^2 + a^3 + ...) + (1 + 2a + 3 a^3 + 4 a^3 + ...))
+ m^2 (1 + 2 a + 3 a^2 + 4 a^3 + ...)

= -1/a + 1/a (1 + a) / (1 - a)^3
- 2m (1 / (1 - a) + 1 / (1 - a)^2)
+ m^2 / (1 - a)^2

= -1/a + 1/a (1 + a) / (1 - a)^3 + (m^2 + 2 a m - 4 m) / (1 - a)^2

Substitute 557/165 for m and 3/4, 13/18, and 25/36 for a, gives a variance of 68,448,497 / 2,940,300

Standard Deviation = sqrt(variance) = sqrt(205,345,491) / 2,970 = about 4.8249


link to original post

Disagree. Did you confirm that number with anything (Markov, simulation)?

You can set up a very simple Markov chain (only four states/columns) and get the answer. My formulaic solution agrees to the Markov chain to ten digits
link to original post


No, I didn't. I'll wait to see your solution before trying to figure out where I went wrong on this.
Ace2
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May 15th, 2024 at 5:22:53 PM permalink
Quote: Ace2

The expected number of rolls for a craps passline bet to be resolved is 557/165 =~ 3.38

What is the standard deviation?

Closed-form solutions only
link to original post


After the first roll (2/3 probability that a point was rolled) the potential outcome/variance then follows a geometric distribution based on which point was rolled. If, for instance, the point is 8 then there is an 11/36 chance the bet will be resolved on each subsequent roll with an expectation of 36/11 additional rolls to resolve and variance of 36/11 * (36/11 - 1). But that variance is for the distribution starting on the first roll so you must adjust for it starting on the second roll and having different overall EV (557/165 instead of 36/11).

So you get to the combined variance as follows (u=557/165):

1/3*(1-u)^2 + 2/3*
{3/12*[(4*3) + (4-u)^2 - 9/36(1-u)^2]*36/27 +
4/12*[(3.6*2.6) + (3.6-u)^2 - 10/36(1-u)^2]*36/26 +
5/12*[(36/11*25/11) + (36/11-u)^2 - 11/36(1-u)^2]*36/25}

=~ 9.0237649219, which I confirmed with a Markov chain

Take the square root of that to get the SD of ~3.003958

No integrals required this time. Just fourth-grade arithmetic and fifth-grade English
Last edited by: Ace2 on May 15, 2024
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ThatDonGuy
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May 15th, 2024 at 7:55:55 PM permalink
I found the error


If m is the mean (in this case, 557/165), the variance is the sum of:

1/3 x (1 - m)^2
+ 1/6 x 1/4 x ((2 - m)^2 + 3/4 x (3 - m)^2 + (3/4)^2 x (4 - m)^2 + ...)
+ 2/9 x 5/18 x ((2 - m)^2 + 13/18 x (3 - m)^2 + (13/18)^2 x (4 - m)^2 + ...)
+ 5/18 x 11/36 x ((2 - m)^2 + 25/36 x (3 - m)^2 + (25/36)^2 x (4 - m)^2 + ...)

Lemma:
(2 - m)^2 + a (3 - m)^2 + a^2 x (4 - m)^2 + a^3 (5 - m)^2 + ...
= 4 + 9 a + 16 a^2 + 25 a^3 + ...
- 2m (2 + 3 a + 4 a^2 + 5 a^3 + ...)
+ m^2 (1 + a + a^2 + a^3 + ...)

= -1/a + 1/a (1 + 4 a + 9 a^2 + 16 a^3 + ...)
- 2m ((1 + a + a^2 + a^3 + ...) + (1 + 2a + 3 a^3 + 4 a^3 + ...))
+ m^2 (1 + a + a^2 + a^3 + ...)

= -1/a + 1/a (1 + a) / (1 - a)^3
- 2m (1 / (1 - a) + 1 / (1 - a)^2)
+ m^2 / (1 - a)

= -1/a + 1/a (1 + a) / (1 - a)^3 + (2 a m - 4 m) / (1 - a)^2 + m^2 / (1 - a)

Substitute 557/165 for m and 3/4, 13/18, and 25/36 for a, gives a variance of 245,672 / 27,225
SD = sqrt(245,672) / 165, or about 3.0039582

Ace2
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May 15th, 2024 at 10:04:41 PM permalink
Quote: ThatDonGuy

I found the error


If m is the mean (in this case, 557/165), the variance is the sum of:

1/3 x (1 - m)^2
+ 1/6 x 1/4 x ((2 - m)^2 + 3/4 x (3 - m)^2 + (3/4)^2 x (4 - m)^2 + ...)
+ 2/9 x 5/18 x ((2 - m)^2 + 13/18 x (3 - m)^2 + (13/18)^2 x (4 - m)^2 + ...)
+ 5/18 x 11/36 x ((2 - m)^2 + 25/36 x (3 - m)^2 + (25/36)^2 x (4 - m)^2 + ...)

Lemma:
(2 - m)^2 + a (3 - m)^2 + a^2 x (4 - m)^2 + a^3 (5 - m)^2 + ...
= 4 + 9 a + 16 a^2 + 25 a^3 + ...
- 2m (2 + 3 a + 4 a^2 + 5 a^3 + ...)
+ m^2 (1 + a + a^2 + a^3 + ...)

= -1/a + 1/a (1 + 4 a + 9 a^2 + 16 a^3 + ...)
- 2m ((1 + a + a^2 + a^3 + ...) + (1 + 2a + 3 a^3 + 4 a^3 + ...))
+ m^2 (1 + a + a^2 + a^3 + ...)

= -1/a + 1/a (1 + a) / (1 - a)^3
- 2m (1 / (1 - a) + 1 / (1 - a)^2)
+ m^2 / (1 - a)

= -1/a + 1/a (1 + a) / (1 - a)^3 + (2 a m - 4 m) / (1 - a)^2 + m^2 / (1 - a)

Substitute 557/165 for m and 3/4, 13/18, and 25/36 for a, gives a variance of 245,672 / 27,225
SD = sqrt(245,672) / 165, or about 3.0039582


link to original post

Very good. Next round of beers is on me
It’s all about making that GTA
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