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Re:Counterintuitive problem: People in a room keep giving do... by gragas on Sunday July 09, 2017 @11:36AM (#14730183)
>The system only has 25^5 states

Not true. The problem is in one-to-one correspondence with the non-negative solutions to

a + b + c + d + e = 25

which is in one-to-one correspondence with the positive solutions to

a + b + c + d + e = 30

which is a simple counting problem---(29 choose 4) = 23751.

In my opinion, you should really try to prove your math to yourself before answering with such an authoritative tone.

Re:Counterintuitive problem: People in a room keep giving do... by jordigh on Sunday July 09, 2017 @11:39AM (#14730209)
I guess what sounds authoritative is that I'm using jargon while trying to explain the jargon at the same time, as if I knew what I was talking about. I've since edited my comment. I know it's stars and bars, but I keep messing up the count. Anyway, it's a finite amount, it's not huge, and your count is wrong too by a sister comment. :-)
Re:Counterintuitive problem: People in a room keep giving do... by gragas on Sunday July 09, 2017 @11:44AM (#14730246)
I agree with your point. :-)

Why is my solution incorrect, though?

* Oh, I see why my solution is wrong. It's never the case that one person has all the money. I should take some of my own advice.

Re:Counterintuitive problem: People in a room keep giving do... by jensv on Sunday July 09, 2017 @12:15PM (#14730430)
Counting is just very difficult. You can easily be off by orders of magnitude and not even recognize it.
Re:Counterintuitive problem: People in a room keep giving do... by Retric on Monday July 10, 2017 @11:53AM (#14737048)
One player can't get all the money as they have to hand out a dollar every turn and they can't receive all the money on the same turn.

23751 - 5 = 23746