The cards were shuffled at the start. The cards and odds don't change after cards have been played. If it's 3 Fascist policies, there's a 88% chance someone dropped. Which probably means 2 fascist players in play together

You're talking about there being 2 Ls left in the pack, but me and ZFR both know there's only 1, as we've seen 4 Ls on the board and one go to the discard (Well, ZFR might have put another L in the discard, and he might think the same about me and GRW or RWG might have put the 6th L in the discard we don't know but anyway)

Unless you think me and zfr are both lying you must calculate on the premise that there is maximum 1 L left.

You're wrong.

Here is a hint: Assume everyone was telling the truth. Suppose we open 3 more cards and they are F, F and F. What is the probability of the remaining two cards being LL. Is it 100%? Or is it 12.3% because that's the chance that two L cards were at the back of the deck?

5 cards left. 4F, 1L. Probability is: 60% FFL, 40% FFF.

Well, nice. I'd be willing to try to play the last L, if RFG hands me one. But let's be real: a government of two unknowns isn't going through at this point. Let's circle back to GR+RW. That's a tested government.

Game (Rager/R) Warehall

There's 6 Liberal policies in the deck, not 5

So odds change as events happen, okay makes sense. From the observer position though the odds remain that 88% chance that a 5 Liberal Policy happened and there's an assumption that someone dropped a card. It's general play to freeze the player who passes fascist policy but it can be any 4 of the players who played.

I imagine the odds of this happening is influenced by the odds of 4 Liberal players playing in tandem occuring. But given the latter is quite a low probability we can stick to the 88% chance I feel.

First off your odds are greatly off...
At the start of the game, neglecting any draws, the odds of FFF are 24.3%...(11/17)*(10/16)*(9/15)=33/136

Those odds would apply to any set of 3 cards if you are ignoring reported draws. But given that 4 Liberal policies have been passed, reported data should be reliable. Worse case, a Fascist President drew 2 L's and underreported it, meaning there are no L's left in the deck, not the other way around.

If ZFR were Fascist or Hitler, he likely still passes LF so he can underreport and pass shade on the next government. We have accounted for 5 Ls. There is at most 1 left in the deck with 5 cards remaining.

On Secrethitler.io the general consensus is that the chance of a 5 liberal policy deck is 88%. Do you dispute this? Because most players believe it

And experiencing several games daily I will state most games are 5 liberal policy decks.Some are 6 policies, some are 4, occasionally 3

So I think you're wrong.

I gave you the odds...
I showed you exactly how to derive them yourself...
For a set of 3 to lack a Liberal policy they have to all be Fascist.
The deck starts with 11 Fascist policies and 6 Liberal policies out of 17 total.
For the first card to be Fascist, odds are 11 out of the total of 17
Once that is drawn there are now 10 Fascist policies and 6 Liberal policies out of 16 total.
For the second card to be Fascist, odds are 10 out of the total of 16
Once that is drawn there are now 9 Fascist policies and 6 Liberal policies out of 15 total.
For the second card to be Fascist, odds are 9 out of the total of 15
Hence (11/17)*(10/16)*(9/15)

Taking the 103/136 odds of each set of 3 having at least 1 Liberal policy makes the odds of 5 available Liberal policies just 24.9% in all 5 governments to start the game. (103/136)^5

No we can't. 88% is just an arbitrary number that is wrong.

Even if we assume everyone was lying we *saw* 4 policy cards. So there can be only 0, 1 or 2 cards left.

If 0 L cards left (1 or more player lied), there is 100% of drawing FFF, 0% of drawing an L.
If 1 L card left (everyone told the truth), there is 40% of drawing FFF, 40% of drawing an L.
If 2 L cards left (only ZFR lied), there is 10% of drawing FFF, 90% of drawing at least an L.

Your 88%/12% split Is. Wrong. In. Any. Situation.

Since there is absolutely no reason for me to lie about 1 L being gone, even if I'm F, we can assume it's either 60% chance of drawing L, if everyone told the truth and 0% if someone lied and hid that final L.

Got ninja'ed by RW.

^typo. Should be:

If 1 L card left (everyone told the truth), there is 40% of drawing FFF, 60% of drawing an L.

Scene and mathematics are never going to agree with each other on the topic of statistics! ;-)

And in further consideration, that (103/136)^5 is also wrong because it does not consider that each previous one has been guaranteed to contain a L. That 88% number may be correct, but those might be the odds AGAINST Liberals having any chance to win before the shuffle.

No. To 88% is the answer to this question:

What's the probability that the first 15 cards will have at least 5L?

It doesn't guarantee that each of the first 5 governments has 1L, but it guarantees that there are enough L for it to be so.

For the record, barring math errors, I just worked it out. 86.3% chance that at least one of the first 5 governments will fail by cards. That is probably where Scene is getting that number from, but it is backwards.