## Poll

18 votes (51.42%) | |||

13 votes (37.14%) | |||

5 votes (14.28%) | |||

2 votes (5.71%) | |||

11 votes (31.42%) | |||

3 votes (8.57%) | |||

6 votes (17.14%) | |||

5 votes (14.28%) | |||

10 votes (28.57%) | |||

8 votes (22.85%) |

**35 members have voted**

November 26th, 2021 at 8:21:27 AM
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Eureka!

See: OEIS A103277

"Smallest i such that there exists j such that i = x + y + z, j = x*y*z has exactly n solutions in positive integers x <= y <= z".

3, 13, 39, 118, 185, 400, 511, 1022, 1287, 2574, 4279, 8558, 11777, 24377, 23554, 46111, 99085, 165490

Edit: Interesting that it does not monotonically increase. the 15th term, 23554 is smaller than the 14th term.

See: OEIS A103277

"Smallest i such that there exists j such that i = x + y + z, j = x*y*z has exactly n solutions in positive integers x <= y <= z".

3, 13, 39, 118, 185, 400, 511, 1022, 1287, 2574, 4279, 8558, 11777, 24377, 23554, 46111, 99085, 165490

Edit: Interesting that it does not monotonically increase. the 15th term, 23554 is smaller than the 14th term.

Last edited by: gordonm888 on Nov 26, 2021

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

November 27th, 2021 at 8:30:28 AM
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Quote:GialmereQuote:WizardHappy Thanksgiving to you too and all the math geniuses of the forum!

I think everyone has an equal chance to be last.

My solution is I answered this for the 3, 4, and 5-person case and all of them showed, hopefully correctly, that everyone had an equal chance of being last. I can explain how I did those cases on request, but I thought of it like a coin flipping game where the goal was to keep flipping until you won $x or lost $y.

Extending the logic for those three cases, I say regardless of the number of mathematicians, everyone will have an equal chance to be last.

Correct!!

Very good.

Yes. It's a surprisingly fair way to determine who gets stuck with the last helping (if you don't mind spending hours at the table passing food around).

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I am not getting this. What I get is:

If the first person is included in having "touched" the cranberries at the start, then yes, everybody but that person has an equal chance.

If the first person is not included until the cranberries are passed to that person by someone else, then the two people adjacent to the first person are half as likely as the others.

November 27th, 2021 at 12:15:54 PM
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Yes, you serve yourself first. The question is worded poorly. Here is a link to the Riddler page containing the official solve (at the bottom). I would just post it here but the rather fun pie chart animations from Twitter will not embed.

Have you tried 22 tonight? I said 22.

November 27th, 2021 at 6:03:37 PM
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Quote:GialmereHere is a link to the Riddler page containing the official solve (at the bottom). I would just post it here but the rather fun pie chart animations from Twitter will not embed.

link to original post

Thank you. At first I thought that was a "hand waving" kind of proof, but now I can see the logic in it. Here is how I might word it.

Let the starting position be x and y be any other specific person you wish to find the probability of being last.

Eventually, the cranberry sauce will find it's way to someone to either side of y. Then, for y to be last, the sauce must immediately move away from y and eventually get all the way to y's other neighbor, without ever getting to y from the other side. That is 18 steps it must move.

If you have one unit to gamble with on a 50/50 game, your chances of winning n units before losing the one unit is 1/(1+n). I can prove this upon request, but I hope it's obvious.

So, in the case of the sauce moving 18 steps before moving one from the starting point is 1/19.

You can use this logic for anyone at the table.

It's not whether you win or lose; it's whether or not you had a good bet.