## Poll

 I love math! 18 votes (51.42%) Math is great. 13 votes (37.14%) My religion is mathology. 5 votes (14.28%) Women didn't speak to me until I was 30. 2 votes (5.71%) Total eclipse reminder -- 04/08/2024 11 votes (31.42%) I steal cutlery from restaurants. 3 votes (8.57%) I should just say what's on my mind. 6 votes (17.14%) Who makes up these awful names for pandas? 5 votes (14.28%) I like to touch my face. 10 votes (28.57%) Pork chops and apple sauce. 8 votes (22.85%)

35 members have voted

Ace2 Joined: Oct 2, 2017
• Posts: 1118
June 29th, 2021 at 2:29:35 PM permalink
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.
Last edited by: Ace2 on Jun 29, 2021
It�s all about making that GTA
GM Joined: Jun 16, 2021
• Posts: 35
Thanks for this post from: June 29th, 2021 at 3:26:06 PM permalink
Quote: Ace2

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

The last digit of the numerator should be 9.
Ace2 Joined: Oct 2, 2017
• Posts: 1118
June 29th, 2021 at 3:31:23 PM permalink
Source?
It�s all about making that GTA
GM Joined: Jun 16, 2021
• Posts: 35
June 29th, 2021 at 3:53:27 PM permalink
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.
Ace2 Joined: Oct 2, 2017
• Posts: 1118
June 29th, 2021 at 4:28:58 PM permalink
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
GM Joined: Jun 16, 2021
• Posts: 35
Thanks for this post from:  June 29th, 2021 at 4:54:48 PM permalink
I would trust Mathematica over integral-calculator. com.
teliot Joined: Oct 19, 2009
• Posts: 2269
June 29th, 2021 at 9:45:40 PM permalink
Quote: GM

I would trust Mathematica over integral-calculator. com.

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.
Poetry website: www.totallydisconnected.com
GM Joined: Jun 16, 2021
• Posts: 35
Thanks for this post from: June 30th, 2021 at 11:07:44 AM permalink
Quote: teliot

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.

Mathematica can handle arbitrarily large integers in principle, 16 digits is definitely not a problem. LibreOffice Calc can also handle 1981795937732789.
Wizard Joined: Oct 14, 2009
• Posts: 23632
July 1st, 2021 at 11:40:24 AM permalink
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?
It's not whether you win or lose; it's whether or not you had a good bet.
unJon Joined: Jul 1, 2018