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[Cheddarius]

Okay, this is the discussion from http://www.bay12games.com/forum/index.php?topic=46711.135 .

I'll try to mull this over in my head and get the idea right...

[RedWarrior0]

No other discussion from the game. This is purely for numbers and the probability, as it will help avoid clutter. And ched, I'd suggest moving to Mafia

[Cheddarius]

I'll try to start with grinch probability...
You are one person.
There are 11 other people. Two are grinches.
Suppose you mail one letter. You have a 9/11 chance of not getting a grinch.
Suppose you mail two letters. You have a 9/11 chance of not getting a grinch with your first letter. If this happens, there are 10 targets, again 2 of whom are grinches. You have a 9/11 * 8/10 chance of getting no grinch.
Suppose you mail three letters. You have a 9/11*8/10 chance of no grinches with your first two letters. If this happens, there are 9 targets, 2 of whom are grinches. You have a 9/11*8/10*7/9 chance of getting no grinch.

Ahah, I've figured it out. Red, my friend, you had it right in the end. Chance of getting no grinches with N letters is... how do I set this up...
(9!/11!)/((9-n)!/(11-n)!)
I think.
n=1 gives me (9!/11!)/(8!/10!), which is of course 9/11, and that is correct.
n=2 gives me (9!/11!)/(7!/9!), which is 9*8/11*10, and that is correct.

Okay, I think the chance of getting a grinch is (9!/11!)/((9-n)!/(11-n)!) and santa is 1-(9!/11!)/((9-n)!/(11-n)!), and clean ability is them multiplied. Great!

[Cheddarius]

All right. How do I move this thread?

Also, I'll start work constructing a new probability table...

[RedWarrior0]

Bottom left. And it's now basically wasted bandwith

[Cheddarius]

Here is the table I've generated:

Spoiler (click to show/hide)

Number of Letters	Chance Of No Grinch	Chance of Santa	Chance of "clean" Santa hit
1	0.818	0.182	0.149
2	0.655	0.345	0.226
3	0.509	0.401	0.204
4	0.382	0.618	0.236
5	0.273	0.727	0.198
6	0.182	0.818	0.149
7	0.109	0.801	0.087
8	0.055	0.945	0.052
9	0.018	0.982	0.018
10	0	1	0
11	0	1	0

[Cheddarius]

You mean because the day has ended?

Well, it was an interesting problem, and it might help me in future probability problems and such. Nice working with you, Red.