## Poll

20 votes (46.51%) | |||

14 votes (32.55%) | |||

6 votes (13.95%) | |||

2 votes (4.65%) | |||

12 votes (27.9%) | |||

3 votes (6.97%) | |||

6 votes (13.95%) | |||

5 votes (11.62%) | |||

12 votes (27.9%) | |||

9 votes (20.93%) |

**43 members have voted**

June 29th, 2021 at 2:29:35 PM
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Extra credit: What's the expected number of rolls to resolve Alan and Bob's dice bet?

I guess no one can solve or no one cares or both. Here's how I solved:

Let y = [{(x/6) + (x/6)^2/2 + (x/6)^3/3! + (x/6)^4/4! + (x/6)^5/5!} * 1/e^(x/6)]

Let z = 1/e^(x/6)

Take the integral over all time of:

6*y^5*z + 15*y^4*z^2 + 20*y^3*z^3 + 15*y^2*z^4 + 6*y*z^5 + z^6 dx

to get 1,981,795,937,732,780 / 152,339,935,002,624 =~ 13.01 rolls.

I did not confirm this with a simulation but that figure seems reasonable since the expected number of rolls to hit all six numbers at least once is easy calculated as 14.7, so we know the answer must be less than that, but probably not by much.

The formula says: sum the probabilities over all time zero to infinity that the bet is in an unresolved state. The first term evaluates the 6 ways of having five numbers with 1-5 hits and one number with zero hits, the next term evaluates the 15 ways of having four numbers with 1-5 hits and two numbers with zero hits and so on: 20*3/3, 15*2/4, 6*1/5, 1*0/6. I realize the formula could be further condensed but I prefer keeping it in a more intuitive format.

I guess no one can solve or no one cares or both. Here's how I solved:

Let y = [{(x/6) + (x/6)^2/2 + (x/6)^3/3! + (x/6)^4/4! + (x/6)^5/5!} * 1/e^(x/6)]

Let z = 1/e^(x/6)

Take the integral over all time of:

6*y^5*z + 15*y^4*z^2 + 20*y^3*z^3 + 15*y^2*z^4 + 6*y*z^5 + z^6 dx

to get 1,981,795,937,732,780 / 152,339,935,002,624 =~ 13.01 rolls.

I did not confirm this with a simulation but that figure seems reasonable since the expected number of rolls to hit all six numbers at least once is easy calculated as 14.7, so we know the answer must be less than that, but probably not by much.

The formula says: sum the probabilities over all time zero to infinity that the bet is in an unresolved state. The first term evaluates the 6 ways of having five numbers with 1-5 hits and one number with zero hits, the next term evaluates the 15 ways of having four numbers with 1-5 hits and two numbers with zero hits and so on: 20*3/3, 15*2/4, 6*1/5, 1*0/6. I realize the formula could be further condensed but I prefer keeping it in a more intuitive format.

Last edited by: Ace2 on Jun 29, 2021

It’s all about making that GTA

June 29th, 2021 at 3:26:06 PM
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Quote:Ace2to get 1,981,795,937,732,780 / 152,339,935,002,624 =~ 13.01 rolls

The last digit of the numerator should be 9.

June 29th, 2021 at 3:31:23 PM
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Source?

It’s all about making that GTA

June 29th, 2021 at 3:53:27 PM
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I modified my earlier calculation to calculate the expectation instead of the probabilities, and I also calculated the integral in Mathematica. Both gave 1981795937732789/152339935002624.

June 29th, 2021 at 4:28:58 PM
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Well it looks like either Mathematica or integral-calculator.com is accurate to "only" 14 digits, which doesn't concern me much.. I've been using the latter for quite some time now and I don't ever recall an error.

It’s all about making that GTA

June 29th, 2021 at 4:54:48 PM
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I would trust Mathematica over integral-calculator. com.

June 29th, 2021 at 9:45:40 PM
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I've seen Excel do this with baccarat calculations, round the final digit to 0 when I import the data. Based on this error, my Spidey sense tells me the int data size on integral-calculator is the same as Excel.Quote:GMI would trust Mathematica over integral-calculator. com.

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June 30th, 2021 at 11:07:44 AM
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Quote:teliotI've seen Excel do this with baccarat calculations, round the final digit to 0 when I import the data. Based on this error, my Spidey sense tells me the int data size on integral-calculator is the same as Excel.

Mathematica can handle arbitrarily large integers in principle, 16 digits is definitely not a problem. LibreOffice Calc can also handle 1981795937732789.

July 1st, 2021 at 11:40:24 AM
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The following question I do not claim to have a correct answer for and perhaps there isn't one. It was inspired the following scene from The Dark Knight. Watching it is not required to understand the problem.

Direct: g3dl32LaOls

Suppose there is a game show with two perfect logicians as players. The game starts with a pot of $1,000,000.

1. Logician A will make a suggestion on how to divide the pot.

2. If Logician B accepts the suggestion, then they split it that way and the game is over.

3. If Logician B rejects the offer, the host will rake 10% out of the remaining pot.

4. Logician B will make a suggestion on how to divide the remaining pot.

5. If Logician A accepts the suggestion, then they split it that way and the game is over.

6. If Logician B rejects the offer, the host will rake 10% out of the remaining pot.

7. Go back to step 1.

The question is how should logician A initially suggest dividing the pot?

Direct: g3dl32LaOls

Suppose there is a game show with two perfect logicians as players. The game starts with a pot of $1,000,000.

1. Logician A will make a suggestion on how to divide the pot.

2. If Logician B accepts the suggestion, then they split it that way and the game is over.

3. If Logician B rejects the offer, the host will rake 10% out of the remaining pot.

4. Logician B will make a suggestion on how to divide the remaining pot.

5. If Logician A accepts the suggestion, then they split it that way and the game is over.

6. If Logician B rejects the offer, the host will rake 10% out of the remaining pot.

7. Go back to step 1.

The question is how should logician A initially suggest dividing the pot?

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan