Probability - Independent Events - Help Plox I'm braindead

[Frog]

Question:

You roll 3 numerical dice.

One is a 6 sided die (1-6)

One is a 4 sided die (1-4)

One is a 3 sided die (1-3)

If they are each rolled once, what is the probability at least one rolls a number 1.

I feel like the simple answer is gamblers fallacy, each event is independent.

But I also feel like there is a compound probability proof I'm completely forgetting.

Logically I'm compelled to think it's as simple as:

Probability of an event happening = Number of ways it can happen / Total number of outcomes

E.g. 3/13 or 23% of one die rolling a number 1

Anyone chime in?

[Cryptonic]

Nope, cause they're non-mutable exclusive.

P(A or B or C) = P(D) + P(C) - P(D and C)
P(D) = P(A) + P(B) - P(A and B)

[Cryptonic]

Or maybe since they're all 1s, it is mutually exclusive lol. Been like 5 years since stats, IDR all the rules.
Ask Orpz lol


Edit: nvm since it's all just 1, it is mutually exclusive. Multiple dice tricked me. Just tried the formula and it was 0% lol probably is just the number of positive outcomes/number of total outcomes

[Frog]

Yeah See? It's deceptive! Lol.

I think the other way of measuring probability which results in a number close to 0 is from using permutations formula.

E.g.
D1 - 1
D2 - 1,2,3,4
D3 - 1,2,3

D1 - 2
D2 - 1,2,3,4
D3 - 1,2,3

But even factorials should be easy.

I'll try it this way and see if I get zero again, but I think it's more reliable.
I think I'd need to measure the (1-P[of not one 1]).

[Helz]

40.5-43.3% chance

[Frog]

40.5-43.3% chance

Plox how? Lol. It's just something that's been bugging me in the back of my mind.

[Cryptonic]

1/6 OR 1/4 OR 1/3

wait no 1 sec, that made 0 sense


Ok I think it'd be easier to do..

1-(5/6 x 3/4 x 2/3)
1-(30/72) = 58.3%?

Does that make sense? I haven't probability'd in so long.

[Frog]

1/6 OR 1/4 OR 1/3

wait no 1 sec

Lol.

(3-[5/6+3/4+2/3])/3 = 25%

Whaaat????

[Cryptonic]

Or you can do it the more complex way:

But, you have to include all outcomes
like
1,1,1
1,2,1
1,3,1
1,4,1
1,1,2
1,1,3
1,2,2
1,2,3
1,3,2
1,3,3
1,4,2
1,4,3
2,1,1
2,1,2
2,1,3
3,1,1
3,1,2
3,1,3
4,1,1
4,1,2
4,1,3
5,1,1
5,1,2
5,1,3
6,1,1
6,1,2
6,1,3
2,2,1
2,3,1
2,4,1
3,2,1
3,3,1
3,4,1
4,2,1
4,3,1
4,4,1
5,2,1
5,3,1
5,4,1
6,2,1
6,3,1
6,4,1

Possible Positive Outcomes = 42

That is 72 possible outcomes (6*4*3)

42/72 = 58.3%

[Frog]

Or you can do it the more complex way:

But, you have to include all outcomes
like
1,1,1
1,2,1
1,3,1
1,4,1
1,1,2
1,1,3
1,2,2
1,2,3
1,3,2
1,3,3
1,4,2
1,4,3
2,1,1
2,1,2
2,1,3
3,1,1
3,1,2
3,1,3
4,1,1
4,1,2
4,1,3
5,1,1
5,1,2
5,1,3
6,1,1
6,1,2
6,1,3
2,2,1
2,3,1
2,4,1
3,2,1
3,3,1
3,4,1
4,2,1
4,3,1
4,4,1
5,2,1
5,3,1
5,4,1
6,2,1
6,3,1
6,4,1

Possible Positive Outcomes = 42

That is 72 possible outcomes (6*4*3)

42/72 = 58.3%

^This! I just forgot how to solve for that mathematically instead of long form! :-P

Obviously the reason I'm asking this here is because it pertains to Mafia/games.

The only other permutations formulas that are relevant would be something along the lines of:

---

In a pool of 16 players, 7 are scum. If 2 players are picked at random what is the probability they are all scum?

There are C(7,2) ways to pick a group of 2 scum out of 7.

There are C(16,2) ways to pick any group of 2 players out of 16.

So the probability is [C(7,2)] / [C(16,2)] = [(21)] / [(120)] = 0.175
or 2 in 12 chances that a towny would pick 2 scum at random

c(x,y)= (x!/[(X!-Y!)*y])

What is the probability of at least 1 being scum?
We find the complement event and subtract from 1.

The complement event of at least one scum is having no scum, which means all town. (assuming you are Town)

There are C(9,2) ways to pick a group of 2 towns out of the 9.

There are C(16,2) ways to pick any group of 2 players out of the 12.

So the probability is [C(9,2)] / [C(16,2)] = 36/ 120
or 3 out of 10 chances that a towny would pick at 2 town at random out of 2 picks.

or

100%-30%=

70% chances a towny would pick 1 scum out of 2 random picks.

The combined probabilities of choosing scum in 2 picks suggests the strategy should be everyone randomly picks 2 players to town block with until mechanical patterns present themselves.

]`§-----

There are C(3,2) ways to pick a group of 2 scum out of 3.

There are C(12,2) ways to pick any group of 2 players out of 12.

So the probability is [C(3,2)] / [C(12,2)] = [(6/2)] / [(479,001,600/7,257,600)] = 3/66
or 1 in 22 chances that a towny would pick 2 scum at random

c(x,y)= (x!/[(X!-Y!)*y])

What is the probability of at least 1 being scum?
We find the complement event and subtract from 1.

The complement event of at least one scum is having no scum, which means all town. (assuming you are Town)

There are C(9,2) ways to pick a group of 2 towns out of the 9.

There are C(12,2) ways to pick any group of 2 players out of the 12.

So the probability is [C(9,2)] / [C(12,2)] = [(362880/10080
)] / [(479,001,600/7,257,600
)] = 36/ 66
or 55% chances that a towny would pick at 2 town at random out of 2 picks.

or

100%-55%=

45% chances a towny would pick 1 scum out of 2 random picks.

The combined probabilities of choosing scum in 2 picks suggests the strategy should be everyone randomly picks 2 players to town block with until mechanical patterns present themselves.

[Frog]

The conclusions aren't necessarily correct by the way, especially if everyone (including scum) knows that this strategy is being used.

The simple answer is WIFOM.

The longer answer has to do with playing off of knowledge of rationality, and knowledge of knowledge of rationality, and so on.

[Cryptonic]

Using probability is the best way to mislead as scum.

[Frog]

Using probability is the best way to mislead as scum.

I thought using tables of claims and interactions was the best way. :-P

[Cryptonic]

Lol yea that can help, but using words is easy to find where someone is being soft. Numbers are absolute

You say the probability of randomly lynching a scum, add variables that make someone more/less likely, but lynch on someone you know it town.

Town has a 0% to lynch scum if they follow your lead, but they'll be like "65% chance this person is scum?! That's probably our best move."

[Frog]

Lol yea that can help, but using words is easy to find where someone is being soft. Numbers are absolute

You say the probability of randomly lynching a scum, add variables that make someone more/less likely, but lynch on someone you know it town.

Town has a 0% to lynch scum if they follow your lead, but they'll be like "65% chance this person is scum?! That's probably our best move."

Lol. Yeah. 3 words for lies. Lies, Damned Lies, and Statistics.

The risk of strategy being scum manipulated is almost always too high to disclose said strategies as town. That's why I normally (odds rolling town vs. odds rolling scum - normally town) keep it to myself. :-P

Anyway, it's just one tool for the toolbox when defending an action you might take.

I agree that there is an art to Mafia/Games because you rarely ever get one strictly dominant strategy (the most common exceptions being lynch everyday as town, or keeping town size at odd numbers for particular games). It's choosing between the weakly dominated strategies by sizing up your opponents that gets you the best strategy for that particular game.

[Orpz]

The best way is to use complement probabilities.
Let A = event of the 6-headed die not rolling a one, B = event of the 4-headed die not rolling a one, C = event of the 3-headed die not rolling a one, and furthermore P(A), P(B), P(C) denote their respective probabilities.

Then our desired probability is calculated as:
1 - P(A and B and C), and by independence
1 - P(A)*P(B)*P(C) =
1 - (5/6)*(3/4)*(2/3) =

7/12

[Orpz]

Since there's C[16,2] total ways to pick two players out, the probability of picking at least 1 scum out of 2 is equal to the probability of picking 1 scum out of 2 plus probability of picking 2 scum out of 2.

Assuming there are 7 scums and 9 towns,

P(At least 1 scum) = P(exactly 1 scum) + P(exactly 2 scums) = C[7,1]*C[9,1] / C[16,2] + C[7,2]/C[16,2] = 0.525 + 0.175 = 0.7

You can also use complement probability here but I like to count scums instead of I know the probabilities of each amount of scums

[Frog]

The best way is to use complement probabilities.
Let A = event of the 6-headed die not rolling a one, B = event of the 4-headed die not rolling a one, C = event of the 3-headed die not rolling a one, and furthermore P(A), P(B), P(C) denote their respective probabilities.

Then our desired probability is calculated as:
1 - P(A and B and C), and by independence
1 - P(A)*P(B)*P(C) =
1 - (5/6)*(3/4)*(2/3) =

7/12

TY! Elegant simple solution.

[Frog]

Since there's C[16,2] total ways to pick two players out, the probability of picking at least 1 scum out of 2 is equal to the probability of picking 1 scum out of 2 plus probability of picking 2 scum out of 2.

Assuming there are 7 scums and 9 towns,

P(At least 1 scum) = P(exactly 1 scum) + P(exactly 2 scums) = C[7,1]*C[9,1] / C[16,2] + C[7,2]/C[16,2] = 0.525 + 0.175 = 0.7

You can also use complement probability here but I like to count scums instead of I know the probabilities of each amount of scums

Lol yeah, my conclusions for those in written form aren't correct. I replaced the numbers but forgot to change the conclusion to match 70% chance of hitting at least one scum out of 2 random picks in that scenario.

[Frog]

The main purpose of this is checking for balance in games. Not necessarily confirmed strategy to be used every time going forward. It doesn't account for power in roles. Power can be adjusted based on probability of random success.

[Cryptonic]

1/6 OR 1/4 OR 1/3

wait no 1 sec, that made 0 sense


Ok I think it'd be easier to do..

1-(5/6 x 3/4 x 2/3)
1-(30/72) = 58.3%?

Does that make sense? I haven't probability'd in so long.

yay math genius orpz confirms I'm not crazy

[Cryptonic]

The main purpose of this is checking for balance in games. Not necessarily confirmed strategy to be used every time going forward. It doesn't account for power in roles. Power can be adjusted based on probability of random success.

You can't really balance a game with probability, though, cause no game is played thru random acts.
Easy way to balance is

Number of TPR = Number of Citizens +1 = Number of Non-Town +1
Number of possible deaths = Number of possible ways for someone to survive.

[MattZed]

Ugh Orpz beat me to it.

[yzb25]

You can't really balance a game with probability, though, cause no game is played thru random acts.
Easy way to balance is

Number of TPR = Number of Citizens +1 = Number of Non-Town +1
Number of possible deaths = Number of possible ways for someone to survive.

Is there some kind of reasoning behind that, or is that from experience from loads of games?

[Cryptonic]

Is there some kind of reasoning behind that, or is that from experience from loads of games?

You can't balance that way because Mafia is not a game of randomness and probability.
You can't make a game expecting everyone to play a certain way, because that won't happen.

If your question is about the reasoning behind the formula, it's mainly from experience. If you throw in custom roles, then obviously that will skew balance, ect. My logic behind using that as a base for games is that non-town roles need room to hide, and Citizens give them that room. So, equal amounts of Citizens & Non-Town. Town would also make up 2/3 of the game while Non-Town makes up 1/3. So, TPR would need to usually have 1 more player. Unless, of course, it's mathematically impossible, then # TPR can = # Cits