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General discussionSecret Hitler 3 - Beyond Stardom!(800 posts)(800 posts)
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This is my favourite topic
Posted January 17, 2020
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ZFR: 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.
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RWarehall: 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.
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.
Posted January 17, 2020
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GameRager: What KIND of entertainment are we talking here.....cuz anything over PG13 costs extra. :D

*Flaps cape dramatically and then disappears, then reappears to the still empty food table*

"So Chief, when we gonna get some vittles around here?"
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!
Posted January 17, 2020
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supplementscene: So it's (6/17) * (6/17) = 12.3%

So the chance of the last 2 blues not being in the back of the pack = 88%

@ZFR correct me if I'm wrong
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.
Posted January 17, 2020
@RW Sorry, disregard my answer to your problem. It's wrong.

A rough estimate does show 85% ish to be correct. How did you get this? I don't see a quick general formula way, when I'm home I can do a quick excel calculation, by calculating each government separately.

Still, the 88% from secrethitler.io is like I said the probability that 5 governments will have 5 Ls, and the answer to that is correct.
Posted January 17, 2020
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ZFR: 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.

------------------------------------------------------------

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?
Posted January 17, 2020
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ZFR: @RW Sorry, disregard my answer to your problem. It's wrong.

A rough estimate does show 85% ish to be correct. How did you get this? I don't see a quick general formula way, when I'm home I can do a quick excel calculation, by calculating each government separately.

Still, the 88% from secrethitler.io is like I said the probability that 5 governments will have 5 Ls, and the answer to that is correct.
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.
Posted January 17, 2020
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Microfish_1: 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.
If that happens we won't be going into a reshuffle. We'll be opening a champagne bottle.


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Microfish_1: 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?
(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.
Posted January 17, 2020
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RWarehall: There are only 2 ways for all 5 governments to have an L
5 FFL's
and
4 FFL's and 1 FLL
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.
Posted January 17, 2020
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ZFR: I planned to do it the longer way (starting from first government with all branching paths, killing a branch once it becomes impossible).
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.
Posted January 17, 2020
ok, thank you
Posted January 17, 2020
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ZFR: If 1 L card left (everyone told the truth), there is 40% of drawing FFF, 60% of drawing an L.
Called it!
Posted January 17, 2020
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JoeSapphire: our resident statistician could tell you the chances of drawing 1 out of 5 in three consecutive draws, but his powers of number seem to have deserted him. (Perhaps he committed a blasphemy against algebra?)

But from where I'm standing it looks like you've got a pretty good chance of drawing the L.
Hopefully you're right. :)
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ZFR: 5 cards left. 4F, 1L. Probability is: 60% FFL, 40% FFF.
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. :)

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ZFR: Game (Rager/R) Warehall
Yeah I figured that out from a few new posts before yours, but thanks for the info regardless. :)
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supplementscene: It's general play to freeze the player who passes fascist policy but it can be any 4 of the players who played.
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.

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supplementscene: 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
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).
Posted January 17, 2020
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.
Posted January 17, 2020
Okay, let's see. I'm missing these votes:
- GameRager
- Microfish
- supplementscene
Posted January 17, 2020
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GameRager: there's near a 1:2 chance
*Sigh*

Sure. 1:2.
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This is my favourite topic
Community
General discussionSecret Hitler 3 - Beyond Stardom!(800 posts)(800 posts)
(800 posts)