How did you get this?

The probability that at least one of the 5 governments will get FFF is 75.1%

(1 - (1- ((11/17)*(10/16)*(9/15))^5) = 75.1

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88.97% is the probability that the first 5 governments will have 5 or more Ls to share among themselves (though not necessarily each one will get one). This is the number scene and people on secrethitler.io refer to.

after ye learn you the english lan-goo-age and start callin it...grub
Pooka, be entertained!

Also, I appreciate Book and Zeo in the story flavor!

Oh, just saw this. The maths is very wrong here.

It's (6/17) * (5/16) = 11.03%

Close to what you wrote, but not exactly.

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Regardless, at this stage it's all irrelevant. The chances are:

60% if everyone told the truth.
0% if someone lied and hid the last L.

And, in the unlikely event that I'm Fascist and a complete moron, 90% if I got LLL but lied about it.

So, 60 % chance we get an L in this gov, IF both are L--and I hope they are--then we have 5 L 0 F going into the reshuffle with 1?L vs 10F if I recall correctly.

Why (6/17) * (5/16) not (6/17) * (6/17)?
I feel I should know, but math escaped as fast as it could when I didn't use it every day. Also, how do you still remember this stuff?

I did it the long way, going through each consecutive possibility.
The shortest version...
There are only 2 ways for all 5 governments to have an L
5 FFL's
and
4 FFL's and 1 FLL

5 FFL's = (11/17)*(10/16)*(6/15)*3*(9/14)*(8/13)*(5/12)*3*(7/11)*(6/10)*(4/9)*3*(5/8)*(4/7)*(3/6)*3*(3/5)*(2/4)*(2/3)*3 = 243/6188 = ~3.9%
[Note: All the extra (*3's) are because the L can be in any of the 3 positions in the selection]

4 FFL's and 1 FLL = (11/17)*(10/16)*(6/15)*3*(9/14)*(8/13)*(5/12)*3*(7/11)*(6/10)*(4/9)*3*(5/8)*(4/7)*(3/6)*3*(3/5)*(2/4)*(1/3)*3* 5 = 1215/12376 = ~9.8%
[Note: Last (*5) is because the FLL can be in any of the 5 governments]

Added together and reduced gives 243/1768 = ~13.7% chance all 5 governments are good by cards.

But yes, for our game, this doesn't matter. Most games won't pass 4 Ls in a row just from the cards.

If that happens we won't be going into a reshuffle. We'll be opening a champagne bottle.


(6/17) * (6/17) would have been the probability of taking a card, putting it back, shuffling, taking another card and have us pull an L in both cases.

But in our case we take 2 cards out of the deck. i.e. we take one, then out of the 16 remaining take another one.
So the probability of the first card being L is 6/17, but when we draw the second card, one L is out already, so there is only 16 cards left, of which 5 are L.

That's acutally a pretty neat shortcut I didn't think about. I planned to do it the longer way (starting from first government with all branching paths, killing a branch once it becomes impossible).

Your maths seems right. There are 6 possible combinations:

(5 FFLs)
FFL,FFL,FFL,FFL,FFL

(4 FFLs + FLL)
FLL,FFL,FFL,FFL,FFL
FFL,FLL,FFL,FFL,FFL
FFL,FFL,FLL,FFL,FFL
FFL,FFL,FFL,FLL,FFL
FFL,FFL,FFL,FFL,FLL

Nice!

So: there is actually an ~88.9% chance of getting 5+ Ls to share, but only ~13.7% chance of sharing them nicely.

That's exactly what I did, but when I got to the end there were only 6 sets which worked and 5 were variations of the same theme with the same probability. So I was able to offer a more succinct solution.

ok, thank you

Called it!

Hopefully you're right. :)
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If true than there's near a 1:2 chance that the next govt will get 3 F cards. That kinda sucks, tbh....but at least we have several Ls passed already. :)

Yeah I figured that out from a few new posts before yours, but thanks for the info regardless. :)
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What if a liberal gets 3 F cards and they're frozen needlessly? It can happen, you know.

Although, I do like your continuing adherence to the "suspect everyone not proven liberal" policy.

You are basing your stance on a different set of games with a differing play style(fast as opposed to slow, people drawing a certain number of L policies per deck with most if not all games, etc)....to me that is a somewhat bad idea.
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Done till Post 362...will do some more in a bit(gotta do some stuff IRL for a minute).

OK I have a strategy I'd like a discussion on and it will allow us to either confirm whether fascists are in play and give Rager/RWarehall 100% chance of playing the final policy. If we top deck twice - we could turn over a Liberal policy or Rager will have 2 Liberal policies in his hand and will be able to force a liberal policy through. If there is a Liberal in RWarehall and Rager a Liberal policy will be passed

If they don't play a Liberal policy they are pretty much confirmed fascist and we can demand ZFR is the Special Election. An argument will probably ensue as RWarehall will also demand the SE but Liberals will Top Deck to ZFR. I personally find ZFR more proven than Joe as we know he played a Liberal policy on choice and that he forced the 4th Liberal policy. In theory Joe could have received 2 Liberal policies and 1 fascist and discarded a Liberal one.

Ofcourse we can just play as we were as we lose the chance to use the investigation etc if we TD the investigation.

Okay, let's see. I'm missing these votes:
- GameRager
- Microfish
- supplementscene

*Sigh*

Sure. 1:2.