## 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/

link to original post

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.

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

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.

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

link to original post

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

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.

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

link to original post

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.

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.

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

Dog Hand

Quote:DogHandJoeman,

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

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

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.

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.

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/2

^{n}

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

The result still seems a bit counterintuitive though.

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

Quote:WizardA 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.

Proof by convection is preferred these daysQuote:unJon

There’s likely a proof by induction I could craft in here.

link to original post

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%.

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.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.

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

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

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.

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.

Quote:charliepatrickI'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.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.

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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I mean to say there is at least one head and an average of one more, for a total of 2.

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.

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I think I would accept that wager.

Quote:charliepatrickSuppose 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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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.

Quote:WizardQuote: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.

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I think I would accept that wager.

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

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.

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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.

Quote:ThatDonGuyQuote: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

Quote:unJonQuote:ThatDonGuyQuote: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.

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

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

Overall Result: Alice: 4042 Bob: 994561

Parms: ndx: 1000000 Time:22:33:7:276

((1 + 5^.5)/2)^102 / (5^.5 * 2^100)

Quote:unJonName your stakes.

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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.

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

Quote:charliepatrickWhat's interesting is if they were both paid the same for each win, who would win more matches.

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If there were an infinite number of flips, would the chances approach 50/50?

Overall Result: Alice: 49449 Bob: 49972

Parms: ndx: 100000 Time:6:28:2:968

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

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.