## Poll

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**46 members have voted**

Quote:unJonNot sure this is optimal as feels like the kind of thing that should go to 1/e as bikes go to infinity but

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

Quote:ThatDonGuyQuote:unJonNot sure this is optimal as feels like the kind of thing that should go to 1/e as bikes go to infinity but

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

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

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?

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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.

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

100 * 2436559 / 720720 =~ 338.07

This formula would be useful for a high number of motorcycles

FYI your first answer is expressed in miles and your solution is a mix of miles and km.Quote:WizardQuote: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.

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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 nQuote:unJonQuote:ThatDonGuyQuote:unJonNot sure this is optimal as feels like the kind of thing that should go to 1/e as bikes go to infinity but

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

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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)

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What’s the standard deviation ?Quote:WizardThe 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/

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Quote:Ace2What’s the standard deviation ?Quote:WizardThe 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/

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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.

What is the standard deviation?

Closed-form solutions only

Quote:Ace2The 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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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.

I confirmed the answer using a Markov chain

Does anyone want more time or should I post the solution?Quote:Ace2The 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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Quote:Ace2Does anyone want more time or should I post the solution?Quote:Ace2The 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

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

Initial roll | # ways | E rolls | Diff to average (*165) | squared | ||

2,3,7,11,12 | 12 | 1 | 12 | -392 | 153664 | 1843968 |

4 or 10 | 6 | 5 | 30 | 268 | 71824 | 430944 |

5 or 9 | 8 | 4.6 | 36.8 | 202 | 40804 | 326432 |

6 or 8 | 10 | 4.27272727272727 | 42.7272727272727 | 148 | 21904 | 219040 |

36 | 3.37575757575758 | 2820384 | ||||

165 | 2.8776492194674 | |||||

557 | 1.69636352809986 |

Disagree. Did you confirm that number with anything (Markov, simulation)?Quote:ThatDonGuyQuote:Ace2Does anyone want more time or should I post the solution?Quote:Ace2

What is the standard deviation?

Closed-form solutions only

link to original post

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

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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

Quote:Ace2Disagree. Did you confirm that number with anything (Markov, simulation)?Quote:ThatDonGuyQuote:Ace2Quote:Ace2

What is the standard deviation?

Closed-form solutions only

link to original post

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

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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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No, I didn't. I'll wait to see your solution before trying to figure out where I went wrong on this.

Quote:Ace2

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

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

Very good. Next round of beers is on meQuote:ThatDonGuyI 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

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