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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)
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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.
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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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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?
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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
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May 23rd, 2024 at 8:33:49 AM permalink
A coin is flipped 100 times.

Every time there are two consecutive heads (HH), Alice gets one point.
Every time there is a heads followed by a tails (HT), Bob gets two points.

Who is more likely to win?

Fair warning this was asked of me outside the forum and I'm not sure my answer is right. If it is, I'm not sure the reason is right.
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May 23rd, 2024 at 11:39:29 AM permalink
If the series of flips starts with some Tails, then ignore these until there's a Head. From this Head (or the first flip if it is a Head), if there are any more Heads, Alice will get one (or more) points for each Head, but eventually (unless one reaches the 100th flip) there will be a Tail and Bob gets two points. This ends what we'll call the current "game" until the next "game" is initiated when there's anoither Head.

For each "game", the chances of Alice getting 0 points (hT) is 1/2, 1 point (i.e. hHT) is 1/4, etc. This averages out to Alice getting one point, whereas Bob always (except the last "game") gets two.

Thus games are repeated every time there is a Head that followed a Tail (or was the first flip).

On average there will be significantly more games than the last game (which doesn't give Bob any points, unless the last flip is a Tail). So overall one expects Bob to get more points than Alice.
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May 23rd, 2024 at 1:32:50 PM permalink
Perhaps I am misunderstanding the question, but I would think Bob would win by an average 2:1 margin. Whenever T comes up, neither scores any points on the next flip. Whenever H comes up, there is an even chance that the next flip is H or T. Since Bob gets 2 points for a T and Alice only gets 1 for an H, Bob should, on average, double Alice's score for each trial of 100 flips.

Out of curiosity, I whipped up a simulation and ran 10,000 trials (1 mil flips). The average result was: Alice: 25.0085 Bob: 50.0458
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May 23rd, 2024 at 1:44:48 PM permalink
Joeman,

You're missing the fact that Alice can win several times on a streak of H, while Bob can win only when the H-streak ends.

Dog Hand
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May 23rd, 2024 at 2:24:08 PM permalink
Quote: DogHand

Joeman,

You're missing the fact that Alice can win several times on a streak of H, while Bob can win only when the H-streak ends.

Dog Hand
link to original post



Yes, but won't that be offset by the times Bob wins multiple times without Alice winning at all when it's choppy?

Here's my code if you're interested.

Alice = 0
Bob = 0

Flip = Rnd
If Flip < 0.5 Then
PrevResult = "H"
Else
PrevResult = "T"
End If

For i = 2 To 100
Flip = Rnd
If Flip < 0.5 Then
Result = "H"
Else
Result = "T"
End If

If PrevResult = "H" Then
If Result = "H" Then
Alice = Alice + 1
Else
Bob = Bob + 2
End If
End If

PrevResult = Result
Next i




I did notice a slight error in my code. Originally, the FOR statement went from 1 to 100, so my original sim numbers upthread are off a bit since I was flipping 101 times per game, and not 100.
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May 23rd, 2024 at 2:24:25 PM permalink
Interesting different way to look at the puzzle. At any given time there's either just been a Head or Tail. When it's been a Tail neither side can win on the next flip. When it's been a Head then one side is bound to win on the next flip, and it's 50:50 which player will win. However Alice only gets 1 while Bob gets 2.

It's true that Alice can win several in a row, while if Bob wins then no-one can win on the next flip.

However after any given flip, the EV for Bob is always twice as much as that for Alice.
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May 23rd, 2024 at 4:20:10 PM permalink

Skip all tosses until you get to the first head.
Then, count the number of heads until the next tail; if there were N heads (not counting the first head), Alice will get N points, and Bob will get 2.
We are now back to the original condition of skipping tosses until another head appears.

The expected length of a run of heads (that came after a tail) followed by a tail is 1/2 x 1 + 1/4 x 2 + 1/8 x 3 + ...
= 1/2 x (1 + 1/2 x 2 + 1/4 x 3 + ...)
= 1/2 x (1 + 1/2 + 1/4 + ...)^2
= 2
Each run of 2 heads followed by a tail earns both Alice and Bob 2 points.
Therefore, both are equally likely to win.

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May 23rd, 2024 at 6:17:36 PM permalink
Quote: ThatDonGuy


Skip all tosses until you get to the first head.
Then, count the number of heads until the next tail; if there were N heads (not counting the first head), Alice will get N points, and Bob will get 2.
We are now back to the original condition of skipping tosses until another head appears.

The expected length of a run of heads (that came after a tail) followed by a tail is 1/2 x 1 + 1/4 x 2 + 1/8 x 3 + ...
= 1/2 x (1 + 1/2 x 2 + 1/4 x 3 + ...)
= 1/2 x (1 + 1/2 + 1/4 + ...)^2
= 2
Each run of 2 heads followed by a tail earns both Alice and Bob 2 points.
Therefore, both are equally likely to win.


link to original post




You write:
1/2 x 1 + 1/4 x 2 + 1/8 x 3 + ...
= 1/2 x (1 + 1/2 x 2 + 1/4 x 3 + ...)
= 1/2 x (1 + 1/2 + 1/4 + ...)^2
= 2
I don't see how you get from the 2nd line to the third line, but the first line is the series:
∑(n=0-∞) n/2n

which does indeed converge to 2, so I agree with your result.

The result still seems a bit counterintuitive though.
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May 23rd, 2024 at 6:28:14 PM permalink
Quote: gordonm888


You write:
1/2 x 1 + 1/4 x 2 + 1/8 x 3 + ...
= 1/2 x (1 + 1/2 x 2 + 1/4 x 3 + ...)
= 1/2 x (1 + 1/2 + 1/4 + ...)^2
= 2
I don't see how you get from the 2nd line to the third line

link to original post



1 + 2 a + 3 a^2 + 4 a^3 + ... = (1 + a + a^2 + a^3 + ...)^2
For -1 < a < 1, this is (1 / (1 - a))^2 = 1 / (1 - a)^2

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May 23rd, 2024 at 6:56:25 PM permalink
Quote: Wizard

A coin is flipped 100 times.

Every time there are two consecutive heads (HH), Alice gets one point.
Every time there is a heads followed by a tails (HT), Bob gets two points.

Who is more likely to win?

Fair warning this was asked of me outside the forum and I'm not sure my answer is right. If it is, I'm not sure the reason is right.
link to original post





Start with 3 flips:

HHH - Alice wins 2-0
HHT - Bob wins 1-2
HTT - Bob wins 0-2
HTH - Bob wins 0-2
THH - Alice wins 1-0
TTH - tie 0-0
THT - Bob wins 0-2
TTT - tie 0-0

Bob wins twice as often.

Looking at 4 flips.

Half of the 3-series end in T so adding another flip won’t change the result. Bob won 3 and one tie in the series that end in T. So far there are 6 Bob wins and 2 ties with 4 flips.

The other half end in H. So the 4th flip will give Alice 1 half the time and Bob 2 half the time. That turns into 3 Alice wins, 4 Bob wins and one tie.

For a total of 10 Bob wins, 3 Alice wins and 3 ties.

Eventually the ties will become a small fraction of the outcomes. And it looks to me like Bob winning more often is stable.

There’s likely a proof by induction I could craft in here.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
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May 23rd, 2024 at 9:58:42 PM permalink
Quote: unJon


There’s likely a proof by induction I could craft in here.
link to original post

Proof by convection is preferred these days
It’s all about making that GTA
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May 23rd, 2024 at 10:39:12 PM permalink

I think with an infinite number of flips, both sides would have the same expected number of points.

Here is why. First, take any series of two or more Tails and replace them with just one Tail. This is allowed because consecutive tails result in no points either way. Then consider the number of heads between any two tails. There must be at least one, giving Bob two points. The expected number of heads between tails is 1/0.5 = 2, giving Alice two points.

The cap on the number of flips cuts Bob's way. Alice relies on long strings of heads sometimes to stay even. A cap on flips hinders that.

So, over 100 flips, I think it would be close, but Bob's chances would be a little better. I would say his chances of winning would be greater than 50% but less than 51%.
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May 24th, 2024 at 2:15:54 AM permalink
Quote: wizard

..Then consider the number of heads between any two tails. There must be at least one, giving Bob two points. The expected number of heads between tails is 1/0.5 = 2, giving Alice two points.

I'm guessing it is correct that the expected number of Heads between two Tails might be 2, however you know there must be at least one Head. Also the series THT results in no payout to Alice. So technically one has to look at how many Heads there are after TH but before T, given there may be none. Alternatively subtract 1 (the first Head) from your 2.

TH T has a probability of 1/2. The subsequent flip after TH is a Tail. Payout to Alice = 0.
TH H T: p=1/4, The subsequent flips after TH are H then T, Alice paid 1.
TH HH T: p=1/8, The subsequent flips after TH are H, H then T; Alice paid 2.
etc
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May 24th, 2024 at 4:02:34 AM permalink
Quote: Wizard


I think with an infinite number of flips, both sides would have the same expected number of points.

Here is why. First, take any series of two or more Tails and replace them with just one Tail. This is allowed because consecutive tails result in no points either way. Then consider the number of heads between any two tails. There must be at least one, giving Bob two points. The expected number of heads between tails is 1/0.5 = 2, giving Alice two points.

The cap on the number of flips cuts Bob's way. Alice relies on long strings of heads sometimes to stay even. A cap on flips hinders that.

So, over 100 flips, I think it would be close, but Bob's chances would be a little better. I would say his chances of winning would be greater than 50% but less than 51%.

link to original post



The expected number of H is 2, but that only gives Alice 1 point. And then B gets 2 points with the T.

I would wager if you simulate out the number of times Alice is ahead of Bob after 100 flips 10,000 times, Alice would be ahead about 0% of the time.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
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May 24th, 2024 at 4:16:52 AM permalink
Suppose you made a long list of the sets of 100 flips. It would then be reasonable to go through a process of counting how many payouts were made on the 2nd flip, 3rd flip etc.

So the process adopted could be for i=2 thru 100 (e.g. i=2 looks at the first two flips, hence the pair of flips ending in position 2) just look at every set and see what payouts there were.

At each stage if there was "HH" then Alice gets 1, "HT" then Bob gets 2, "TT" no payout, "TH" no payout.

Eventually after running the i=2 thru 100, this process finds all the payouts.

Now just considier any particular column (i.e. an i ). The chances of "HH" is the same at "HT", so on average there will be the same number of payouts to Bob as Alice.

Thus overall, when we add up all i, i.e. in total, there will be the same number of payouts to Alice and Bob.

Since Bob receives $2 and Alice $1, Bob will win more.
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May 24th, 2024 at 5:41:27 AM permalink
Quote: charliepatrick

Quote: wizard

..Then consider the number of heads between any two tails. There must be at least one, giving Bob two points. The expected number of heads between tails is 1/0.5 = 2, giving Alice two points.

I'm guessing it is correct that the expected number of Heads between two Tails might be 2, however you know there must be at least one Head. Also the series THT results in no payout to Alice. So technically one has to look at how many Heads there are after TH but before T, given there may be none. Alternatively subtract 1 (the first Head) from your 2.

TH T has a probability of 1/2. The subsequent flip after TH is a Tail. Payout to Alice = 0.
TH H T: p=1/4, The subsequent flips after TH are H then T, Alice paid 1.
TH HH T: p=1/8, The subsequent flips after TH are H, H then T; Alice paid 2.
etc

link to original post




I mean to say there is at least one head and an average of one more, for a total of 2.
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May 24th, 2024 at 5:42:49 AM permalink
Quote: unJon


The expected number of H is 2, but that only gives Alice 1 point. And then B gets 2 points with the T.

I would wager if you simulate out the number of times Alice is ahead of Bob after 100 flips 10,000 times, Alice would be ahead about 0% of the time.

link to original post




I think I would accept that wager.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
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May 24th, 2024 at 5:47:49 AM permalink
Quote: charliepatrick

Suppose you made a long list of the sets of 100 flips. It would then be reasonable to go through a process of counting how many payouts were made on the 2nd flip, 3rd flip etc.

So the process adopted could be for i=2 thru 100 (e.g. i=2 looks at the first two flips, hence the pair of flips ending in position 2) just look at every set and see what payouts there were.

At each stage if there was "HH" then Alice gets 1, "HT" then Bob gets 2, "TT" no payout, "TH" no payout.

Eventually after running the i=2 thru 100, this process finds all the payouts.

Now just considier any particular column (i.e. an i ). The chances of "HH" is the same at "HT", so on average there will be the same number of payouts to Bob as Alice.

Thus overall, when we add up all i, i.e. in total, there will be the same number of payouts to Alice and Bob.

Since Bob receives $2 and Alice $1, Bob will win more.

link to original post




The reason this doesn't work is each scoring is correlated to the one before and after. This fact favors Alice. I contend over infinite flips, Alice will get 2/3 of the payouts.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
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May 24th, 2024 at 6:01:07 AM permalink
Quote: Wizard

Quote: unJon


The expected number of H is 2, but that only gives Alice 1 point. And then B gets 2 points with the T.

I would wager if you simulate out the number of times Alice is ahead of Bob after 100 flips 10,000 times, Alice would be ahead about 0% of the time.

link to original post




I think I would accept that wager.

link to original post



Name your stakes. I would be shocked if as the number of flips goes to infinity, the % Alice wins doesn't go to:

0%
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
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May 24th, 2024 at 6:24:22 AM permalink
Quote: ThatDonGuy


Skip all tosses until you get to the first head.
Then, count the number of heads until the next tail; if there were N heads (not counting the first head), Alice will get N points, and Bob will get 2.
We are now back to the original condition of skipping tosses until another head appears.

The expected length of a run of heads (that came after a tail) followed by a tail is 1/2 x 1 + 1/4 x 2 + 1/8 x 3 + ...
= 1/2 x (1 + 1/2 x 2 + 1/4 x 3 + ...)
= 1/2 x (1 + 1/2 + 1/4 + ...)^2
= 2
Each run of 2 heads followed by a tail earns both Alice and Bob 2 points.
Therefore, both are equally likely to win.


link to original post



While "the expected length of a run of heads followed by a tail" is 2, Alice doesn't get a point for each head, but for each time HH come up. The first head was in TH.
The number of points Alice will get in each run is 1/2 x (0 + 1/2 x 1 + 1/4 x 2 + ...) = 1, after which Bob gets 2 points.
Therefore, Bob is more likely to win.

Simulation bears this out - Bob wins almost every time.

unJon
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May 24th, 2024 at 10:00:08 AM permalink
Quote: ThatDonGuy

Quote: ThatDonGuy


Skip all tosses until you get to the first head.
Then, count the number of heads until the next tail; if there were N heads (not counting the first head), Alice will get N points, and Bob will get 2.
We are now back to the original condition of skipping tosses until another head appears.

The expected length of a run of heads (that came after a tail) followed by a tail is 1/2 x 1 + 1/4 x 2 + 1/8 x 3 + ...
= 1/2 x (1 + 1/2 x 2 + 1/4 x 3 + ...)
= 1/2 x (1 + 1/2 + 1/4 + ...)^2
= 2
Each run of 2 heads followed by a tail earns both Alice and Bob 2 points.
Therefore, both are equally likely to win.


link to original post



While "the expected length of a run of heads followed by a tail" is 2, Alice doesn't get a point for each head, but for each time HH come up. The first head was in TH.
The number of points Alice will get in each run is 1/2 x (0 + 1/2 x 1 + 1/4 x 2 + ...) = 1, after which Bob gets 2 points.
Therefore, Bob is more likely to win.

Simulation bears this out - Bob wins almost every time.


link to original post



I just ran about 100 simulations in Excel and Bob won every one of them.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
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May 24th, 2024 at 11:15:12 AM permalink
Quote: unJon

Quote: ThatDonGuy

Quote: ThatDonGuy


Skip all tosses until you get to the first head.
Then, count the number of heads until the next tail; if there were N heads (not counting the first head), Alice will get N points, and Bob will get 2.
We are now back to the original condition of skipping tosses until another head appears.

The expected length of a run of heads (that came after a tail) followed by a tail is 1/2 x 1 + 1/4 x 2 + 1/8 x 3 + ...
= 1/2 x (1 + 1/2 x 2 + 1/4 x 3 + ...)
= 1/2 x (1 + 1/2 + 1/4 + ...)^2
= 2
Each run of 2 heads followed by a tail earns both Alice and Bob 2 points.
Therefore, both are equally likely to win.


link to original post



While "the expected length of a run of heads followed by a tail" is 2, Alice doesn't get a point for each head, but for each time HH come up. The first head was in TH.
The number of points Alice will get in each run is 1/2 x (0 + 1/2 x 1 + 1/4 x 2 + ...) = 1, after which Bob gets 2 points.
Therefore, Bob is more likely to win.

Simulation bears this out - Bob wins almost every time.


link to original post



I just ran about 100 simulations in Excel and Bob won every one of them.

link to original post




I ran 10 million, and Alice won about 1 out of every 250, with 1 out of every 750 being a tie.

charliepatrick
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May 24th, 2024 at 2:36:12 PM permalink
Yes I got the same idea
Overall Result: Alice: 4042 Bob: 994561
Parms: ndx: 1000000 Time:22:33:7:276
Ace2
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May 24th, 2024 at 4:44:35 PM permalink
All I know is that Alice’s chance of scoring zero points is

((1 + 5^.5)/2)^102 / (5^.5 * 2^100)
It’s all about making that GTA
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May 25th, 2024 at 8:52:07 AM permalink
Quote: unJon

Name your stakes.
link to original post



I made a mistake in equating the number of heads in a streak to the number of points Alice gets. I should have subtracted one.

Agreed that Bob would likely win.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
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May 25th, 2024 at 9:54:33 AM permalink
What's interesting is if they were both paid the same for each win, who would win more matches.
I don't know how to calculate it but I assume they would, on average, have the same total winnings. However the probability curves are slightly different, so ...
Parms: ndx: 100000000 Time:17:16:32:465
Wins: 1 Alice: 1 Bob: 0
Wins: 2 Alice: 19 Bob: 0
Wins: 3 Alice: 89 Bob: 0
Wins: 4 Alice: 360 Bob: 0
Wins: 5 Alice: 1314 Bob: 0
Wins: 6 Alice: 4156 Bob: 0
Wins: 7 Alice: 11544 Bob: 0
Wins: 8 Alice: 28598 Bob: 0
Wins: 9 Alice: 63996 Bob: 0
Wins: 10 Alice: 131164 Bob: 1
Wins: 11 Alice: 248439 Bob: 2
Wins: 12 Alice: 438079 Bob: 37
Wins: 13 Alice: 720444 Bob: 305
Wins: 14 Alice: 1125696 Bob: 1970
Wins: 15 Alice: 1654547 Bob: 10463
Wins: 16 Alice: 2307315 Bob: 46383
Wins: 17 Alice: 3072128 Bob: 172400
Wins: 18 Alice: 3906842 Bob: 539534
Wins: 19 Alice: 4762913 Bob: 1419963
Wins: 20 Alice: 5569211 Bob: 3177943
Wins: 21 Alice: 6264828 Bob: 6018639
Wins: 22 Alice: 6797722 Bob: 9692121
Wins: 23 Alice: 7125695 Bob: 13318741
Wins: 24 Alice: 7216604 Bob: 15605041
Wins: 25 Alice: 7079137 Bob: 15599196
Wins: 26 Alice: 6722226 Bob: 13320260
Wins: 27 Alice: 6212706 Bob: 9693416
Wins: 28 Alice: 5568956 Bob: 6011627
Wins: 29 Alice: 4852210 Bob: 3174704
Wins: 30 Alice: 4113392 Bob: 1425050
Wins: 31 Alice: 3396494 Bob: 539927
Wins: 32 Alice: 2730224 Bob: 173062
Wins: 33 Alice: 2146093 Bob: 46387
Wins: 34 Alice: 1646665 Bob: 10517
Wins: 35 Alice: 1228464 Bob: 1967
Wins: 36 Alice: 899168 Bob: 295
Wins: 37 Alice: 641667 Bob: 43
Wins: 38 Alice: 447728 Bob: 6
Wins: 39 Alice: 306150 Bob: 0
Wins: 40 Alice: 204346 Bob: 0
Wins: 41 Alice: 133685 Bob: 0
Wins: 42 Alice: 85235 Bob: 0
Wins: 43 Alice: 53900 Bob: 0
Wins: 44 Alice: 33129 Bob: 0
Wins: 45 Alice: 19864 Bob: 0
Wins: 46 Alice: 11696 Bob: 0
Wins: 47 Alice: 6798 Bob: 0
Wins: 48 Alice: 3854 Bob: 0
Wins: 49 Alice: 2099 Bob: 0
Wins: 50 Alice: 1177 Bob: 0
Wins: 51 Alice: 595 Bob: 0
Wins: 52 Alice: 324 Bob: 0
Wins: 53 Alice: 151 Bob: 0
Wins: 54 Alice: 83 Bob: 0
Wins: 55 Alice: 32 Bob: 0
Wins: 56 Alice: 26 Bob: 0
Wins: 57 Alice: 9 Bob: 0
Wins: 58 Alice: 9 Bob: 0
Wins: 59 Alice: 3 Bob: 0
Wins: 61 Alice: 1 Bob: 0
Overall Result: Alice: 45744183 Bob: 48579146
Parms: ndx: 100000000 Time:17:36:33:799
Is it obvious that Bob would tend to win matches more as some of Alice's wins depend on streaks, so has a longer tail, whereas Bob has a narrower range of scores.
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May 25th, 2024 at 8:12:43 PM permalink
Quote: charliepatrick

What's interesting is if they were both paid the same for each win, who would win more matches.
link to original post



If there were an infinite number of flips, would the chances approach 50/50?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
charliepatrick
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May 25th, 2024 at 10:46:19 PM permalink
It does look like it - this is 10000 flips
Overall Result: Alice: 49449 Bob: 49972
Parms: ndx: 100000 Time:6:28:2:968
Alice won 2263 thru 2736 flips/10k, Bob won 2390 thru 2604.
Onviously the number of ties tends towards zero.

edit: It looks as if Bob might win more than Alice, sadly I have to leave now so can't look further 
Overall Result: Alice: 49475 Bob: 49991
Parms: ndx: 100000 Time:6:48:2:602
Last edited by: charliepatrick on May 25, 2024
ThatDonGuy
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May 26th, 2024 at 6:53:55 AM permalink
In an infinite number of flips, it would be 50/50.

Define a "head run" as a run of tosses that begins with a head, and a "tail run" as a run of tosses that begins with a tail. Assume the first toss is part of a tail run.
The expected value of a tail run is, Alice = 0, Bob = 0
The expected value of a head run that finishes (i.e. a tail is tossed) is, Alice = 1, Bob = 1.

However, if the number of flips is finite, there is the possibility of the final run ending on a head run, in which case, Alice gets a positive number of points, and Bob gets none, so Alice would be expected to finish ahead of Bob.
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