Who would you want investigated if that occurs? I'd like to see Joe or ZFR or maybe Agent as we wouldn't be able to trust Lift with The Presidency if he plays a Fascist policy.

I tend to dislike the use of humour at the expense of reason in these games as I see it as deflection. Although I think you've used it in prior games, so I'm not sure I see it as out of character.

A) You're assuming a scum president would always select Hitler as their chancellor. Would they? I don't think so. Why risk FFF and casting bad light on themselves and Hitler, practically screwing up their game..
You're also assuming a liberal won't accidentally select Hitler.

B) Math is wrong. By your math there is a 110% chance of selecting Hitler after the first 3 governments.

Assuming a scum president would always select Hitler as their chancellor, the probability of Hitler being in government is
As a chancellor, selectted by a Fascist president: (1 - (7/10)*(7/10))
As a president: 2/10
Total: (1 - (7/10)*(7/10)) + 2/10 = 71%

C) Given the assumptions this is probably very far from the truth.

@Lift, you've played a lot of Secret Hitler in RL, right? Out of curiosity, can you comment on how many games had Hitler in the first 2 governments? Ballpark figure.

@All, incidentally, how much experience you have with Secret Hitler.

It doesn't follow. In your calculations you assumed everyone has an equal chance of being Hitler (hence 20% of him being a president). If then you claim that some players are less/more likely to be Hitler (due to being too aggressive (?)), then obviously the calculations are wrong.
Not commenting about you/Trent being likely or less likely to be Hitler, but you can't reach your conclusion from your premises.

Joe was #10. I chose him as my chancellor (based random.org).

I was lied to by the president himself. I would definitely have voted for Captain Sapphire as he is a man of strength and that is the only attribute that commands my respect.

I don't think this can be right. I've been trying to wrap my head round the maths for it and not doing too well.

But you have LLL at just under 3% and LLF at 24%ish, which can't really be right as the difference is multiplying ((6/17)(5/14)) by 11/15 instead of 4/15. So I'd expect the chance of drawing LLF to be just under three times larger than the chance of drawing LLL, so like 8%?

I calculated it as 11/136. Which looks close enough to 8/100 to satisfy me.

The math in blotunga's attachment is right.

but how??

aha! I see it!

I've been doing the chance of drawing LLF, but of course you might draw LFL or FLL... which means I calculate those ones too and... take the average? Or something like that.

(11/17)(6/16)(5/15)
(6/17)(11/16)(5/15)
(6/17)(5/16)(11/15)

are all going to give different results...

now I have to wait two minutes...

oh buggerit I edited instead of posting how did I manage that? You can't quote yourself captain!!


also I forgot to say but I've never played this game before

https://www.dcode.fr/picking-probabilities
If you don't like my image. Enter 17 (total), 11 (no of fascist), 3 (picked), 3 (FFF).
Do the same with 17, 6, 3, 2 (LLF)

turns out they don't give different results... but three eights are 24 so I guess in that situation you just add the averages together and fate says "hm seems legit"

And alas here we have a LLF with the same odds.

The math is explained at the bottom of the page. Basically you can have FFF, LFF, FLF, FFL, LLF, LFL, FLL, LLL. You probably missed a few of those in your odds :).

You think, having first distrusted your beautifully colour coded and heavily detailed grid, that I'm just going to trust some website that you direct me to??

but I get it now. There's 8ish% chance you'll draw LLF, 8ish% chance that you'll draw LFL, and 8ish% chance that you'll draw FLL, so out of you 100 draws, 24ish are devoted to draws that give you two Ls and one F.

Boy, maths is fun.

https://en.wikipedia.org/wiki/Combination
http://mathworld.wolfram.com/BinomialCoefficient.html

(I'll be using the "choose" nCk notation (http://mathworld.wolfram.com/Choose.html) since it's easier to write on forums).

We are choosing 3 cards out of 17

First we look at sample space
Total combinations = 17C3 = 680


There are 11F cards. Our event space is the number of ways to choose FFF
Number of FFF combinations = 11C3 = 165

Therefore P(FFF) = 165/680 = 24.26

For LLF, the even space is the number of ways to choose LL AND (i.e. multiplied by) the number of ways to choose F
Number of LLF combinations = 6C2 * 11C1 = 15 * 11 = 165

Therefore P(LLF) = 165/680 = 24.26

I meant to write "event space"

How nice. So out of 680 ways of doing it, there are 55 ways of picking FLL, 55 ways of picking LFL and 55 ways of picking LLF.

Thanks for explanation. It's been nearly a decade since I barely passed the statistics section of my maths AS level.

So did they choose 11 and 6 precisely because of that effect, or would it be the same if you had 13 fascist and 7 liberal for example?.... hm I'll think about that some other time.

This concludes Sapphematics with Captain Sapphire. Remember; numbers are everybody's friend!