Conditional Probability and Slaying Werewolves

[Louisistan]

Q1: P(wolf) = 5/22
P(Def) = 2/22
P(seer) = 1/22

Q2:
Independent: Me being randomly selected as a wolf in ww50 is independent from me being randomly selected as a wolf in ww51, assuming that the host does of 51 not look at the previous game's assignments.

Dependent: me being selected as wolf and the seer identifying me as a wolf are dependent. If I'm not a wolf, I'm not going to be identified as one.

Q3:
P(seer) = 1/15
P(lynch) = 1/15

P(Not(seer) AND lynch) = (1 - 1/15) * (1/15) = (14/15) * (1/15) = 14/225



D1:
I would argue that since the two events are independent (which was the assumption in the example), Aerso being elected mayor does not factor into his trustworthiness. Only the probability of whether or not he is then wolf should be looked at.
However​, there is a case to be made about the two not actually being independent. If he was a wolf, the other wolves might elect him into the position. Or they might deliberately vote for someone else so as to not put him on the spot. But that wasn't part of the example

[Palman]

May 7, 2017 5:19:34 GMT -5 Louisistan said:

D1:
I would argue that since the two events are independent (which was the assumption in the example), Aerso being elected mayor does not factor into his trustworthiness. Only the probability of whether or not he is then wolf should be looked at.
However​, there is a case to be made about the two not actually being independent. If he was a wolf, the other wolves might elect him into the position. Or they might deliberately vote for someone else so as to not put him on the spot. But that wasn't part of the example

I think you make a very interesting point here, but perhaps one which becomes elaborate beyond the scope of the course. To accurately judge that you would have to figure the odds that someone votes for Aerso given they are each of the roles, including whether or not a wolf would vote for or against a fellow wolf, and that depends on the individual players and who the other choices are, and who the wolves are, and if they discuss it before hand, and so on and so on. That asks for not just probability, but a probabilistic behavior model for all 15 people playing for each of the roles they could hold, and a massive conditional probability computation.

[Ater Nox]

Homework 1.

Control:
Good work on questions 1 and 3. If you own a taco farm on the world map, it means you are registered on our forum which probably means you are more likely to be a defender. That being said, I see where you were going.

Sorya:
Good work on the homework questions (despite a typo where you said dependent instead of independent )

Hah:
Excellent work defining your events. Your dependent events look good, but for the independent ones I assume that if you binge Star Trek, you might be more likely to stay up later which means the events are related. Good work on question 3, a slight typo there with 14/225 = 0.062 not 0.055.

Flag Flayers:
Excellent work defining your events. Question 2 is well done (Unless you want to come up with a sequence of chaos theory events in which Tanzoria only plays football when you drink water ) Question 3 is good, you missed a few A and Bs in the maths but I could see where you were going.

Alkasia:
All good

Palman:
All good

Gunny:
All good, though don't forget to define your events

Louis:
All good

Overall the homework questions were very well done, A+ for everyone!

Discussion 1.

Everyone brought a lot of good points to the table. Sorya pointed out that while there was a 5% chance of being elected Mayor and not being a wolf, the remaining 95% chance was split between being elected and a wolf, not elected and a wolf, and not elected and not a wolf. That's right, from the 5% value alone we can't narrow the three other options down and have to consider them as just the remaining 95% chance. Hah then calculated those percentages for each combination, which much better showed the possibilities. Flag Flayers and Alkasia made some good comparisons with the different probabilities. Palman spotted that the two events were independent, and pointed out that regardless of the mayor position the probability is the same. Louis made the excellent point that that assumption might not be the case, which is something we'll think about in a later class.

As Gunny pointed out, 'it is impossible to trust him completely until there is concrete evidence for or against him'. If we make the assumption that the two events were independent, then Aerso being elected mayor tells us nothing about whether he is a wolf or not and we can't trust him any more then usual. Which, because it is Aerso, is a questionable amount.



Overall the discussion was well thought out, A+ for everyone!

[Palman]

But let's be fair. It's always safe to assume Aerso is a wolf.

[Ater Nox]

Lesson 2. Conditional probability


Welcome everyone to the second lesson of Conditional Probability and Slaying Werewolves. In this lesson, we'll be looking at how to deal with dependent events using conditional probabilities.

As usual, if you have any questions, please post them in the thread. The next class will be held in a week.

Conditional probability

Conditional probability is the probability of an event occurring given that another event already has occurred. Take the following events as an example:

Let B be the event that Barry is a werewolf.
Let M be the event that Barry was howling at the moon.

The probability of B alone, P(B), might be quite low. However, if on a stroll last night I spotted Barry howling a the moon, I know have additional information and consider Barry's chances of being a werewolf as quite high.

In mathematical notation, conditional probability is written like so:

P(B | M)

In English, we would say the probability of B given M.

Order of Events

It is very easy to mix up the order of the events you are using, and it is important to remember that the probability of A given B does not equal the probability of B given A.

For example, given that you are a citizen of the United States, there is a very, very small chance you are Donald Trump. However, given you are Donald Trump, there is a very, very high chance you are a citizen of the United States.

Calculating conditional probability

Conditional probability is calculated by dividing the probability of both events occurring by the probability of the event that has occurred. Mathematically:

P(B | M) = P(B and M) / P(M)

For example, if I think the probability of someone being a wolf and howling at the moon is 0.5 and the probability of someone howling at the moon is 0.55, then:

P(B | M) = 0.5 / 0.55
= 0.91

Note, the probability of two events both occurring (P(A and B)) must be equal to, or less than, the probability of either one of the events occurring (both P(A) and P(B)). If 55% of all people howl at the moon, then the proportion of all people who howl at the moon and do something else can't be higher than that 55%.

Independence

Consider two events that are independent:

Let S be the event that Barry is sunbathing at the beach in the Bahamas.
Let L be the event that Ater Nox has leftovers for dinner.

The probability of Barry sunbathing Ater Nox has leftovers for dinner is equal to the probability that Barry is sunbathing.

P(S | L) = P(S)

Because these two events are independent, one event occurring (leftovers) does not effect the probability of the other (sunbathing).

Other Probability Rules

Other probability rules from before hold assuming all values have the same conditions (or events given). For example:

P(Not A|B) = 1 - P(A|B)

Assuming A and B are independent:

P(A and B | C) = P(A|C) * P(B | C)

Another handy rule is:

P(A) = P(A | B)P(B) + P(A | Not B)P(Not B)

You might recognise this as being equivalent to:

P(A) = P(A and B) + P(A and Not B)

Thinking about it, A can occur either with B or without B, those are the only options. So if we add those two probabilities together, we will get the overall probability of A regardless of whether B is happening or not.

Example

There is a werewolf game in which there are 19 other players including 5 wolves. You are the only seer. On day 1, Barry announces to the village that he is the seer. What is the probability he is telling the truth?

Let B be the event that Barry is the seer.
Let Y be the event that you are the seer.

P(B | Y) = P(B and Y) / P(Y)

You know you are the seer, so P(Y) = 1.
You also know that there is only one seer, meaning both you and Barry can't both be the seer, so P(B and Y) = 0.

Substituting those values in:
P(B | Y) = 0/1 = 0

Surprise surprise! Barry is lying!

Homework

Please post your answers to the homework questions in the thread, including any working you do. Everyone may participate in the discussion between now and the next class.

Question 1.
The probability that Barry is a werewolf and hasn't paid his bar tab is 30%. The probability that Barry hasn't paid his bar tab is 90%. What is the probability that Barry is a werewolf given he hasn't paid his bar tab?

Question 2.
There is a werewolf game in which there are 14 other players including 3 wolves. You are the only seer and Barry has just been elected mayor. Using the equation P(A) = P(A | B)P(B) + P(A | Not B)P(Not B), what is the probability that Barry was elected mayor considering the cases where he is or isn't a wolf? List any assumptions you have made, and any probabilities you have chosen.

Question 3.
Continuing from question 2, you then scan Barry and find out he is a wolf. What is the probability that he was elected mayor now?


Discussion 1.
What is the difference between P(A and B) and P(A | B)? When would you use one or the other? Which one is more useful?

[Ater Nox]

Tanzoria, Dominion of Compassion, Woonsocket, Sorya, Louisistan, Hahiha / USSR, Control, Alkasia, Sauron, Flag Flayers, Palman, Elite, Free Guns For All, SweetHaven

If you didn't participate in the last class, you can still join in at any time

[Dominion of Compassion]

May 8, 2017 23:42:36 GMT -5 Ater Nox said:

Tanzoria, Dominion of Compassion, Woonsocket, Sorya, Louisistan, Hahiha / USSR, Control, Alkasia, Sauron, Flag Flayers, Palman, Elite, Free Guns For All, SweetHaven

If you didn't participate in the last class, you can still join in at any time

I've been wrapping up my university work irl

[Flag Flayers]

May 8, 2017 21:40:45 GMT -5 Ater Nox said:

...you missed a few A and Bs in the maths but I could see where you were going.

Wooops I put them in square brackets so it made them all do forum code stuff ._.

Anyways.....

SPOILER: Click to show

Question 1.
The probability that Barry is a werewolf and hasn't paid his bar tab is 30%. The probability that Barry hasn't paid his bar tab is 90%. What is the probability that Barry is a werewolf given he hasn't paid his bar tab?

First of all event naming:

'A' is the probability Barry is a werewolf
'B' is the probability Barry paid his tab

It is known from the question that:

1. P(A And Not B)=0.3
2. P(Not B)=0.9

It is asked to find what P(A|Not B) equals to

Since the probability of Not B is known, to calculate the answer only P(A) is needed.

Expanding the first statement it is found that

P(A)*P(Not B)=0.3
P(A)=0.3/P(Not B)

And since it is known P(Not B)=0.9

P(A)=0.3/0.9
P(A)=0.33...
P(A)=33.33...%

Since the two events are independent (Unless crazy chaos theory stuff)*^{see bonus question 2} it is known that

P(A|Not B)=P(A)

Therefore:

P(A|Not B)=33.33...%

Question 2.
There is a werewolf game in which there are 14 other players including 3 wolves. You are the only seer and Barry has just been elected mayor. Using the equation P(A) = P(A | B)P(B) + P(A | Not B)P(Not B), what is the probability that Barry was elected mayor considering the cases where he is or isn't a wolf? List any assumptions you have made, and any probabilities you have chosen.

The overall player count is 15, 14 if not including myself, there are 3 wolves, I'm the seer.

Event naming time!
'A' is the probability of being elected mayor
'B' is the probability of being a wolf

Jumping right into the calculation,

P(A)=1/15
P(B)=3/14

P(A)=P(A|B)*P(B)+(PA|Not B)*P(Not B)
P(A)=P(A)*P(B)+P(A)*(1-P(B))
P(A)=3/210(Probability of being elected mayor and being a wolf)+11/210(Probability of being elected mayor and not being a wolf)

And since the 2 events are independent even without doing the addition it is known that

P(A)=1/15

This is similar (if not identical) to the first discussion, where the two probabilities are independent, there's a lower probability of Barry being both a wolf and elected mayor, there could be a higher or lower probability of him being elected mayor depending on wolf, player, seer army etc. strategy, and no assumptions could be made on if Barry is innocent or not. While this could be about the question not stating whether being elected as mayor or being a wolf is random, or if the two events are somehow dependent, according to my assumptions, there's no way to tell if Barry is innocent or not.

Question 3.
Continuing from question 2, you then scan Barry and find out he is a wolf. What is the probability that he was elected mayor now?

The probability stated in my previous calculations, 3/210, 0.014, or 1.43%

Bonus question 1.
Come up with a sequence of chaos theory events in which Tanzoria only plays football by you drinking water (Or vice versa since it is an AND statement)

I'll go the other way around for this one:

Tanzoria is playing football, and he's sweating. The sweat evaporates and creates a domino effect that causes an extreme heat wave in Flaglandia (Where I live). Because it is so hot, I decide to drink a glass of water and therefore, the probability of Tanzoria playing football increased the probability of me drinking water. This also proves that my example was wrong and that nealy anything can be related to anything by chaos theory.

Bonus question 2.
Come up with a sequence of chaos theory events in which Barry being a werewolf affects the probability of him paying his tab (Or vice versa since it is an AND statement)

Well since Barry is a werewolf, he can't pay his tab at night, decreasing the time he can pay his tab, and therefore, decreasing the probability of him paying his tab.

Discussion 2.
What is the difference between P(A and B) and P(A | B)? When would you use one or the other? Which one is more useful?

The main difference between P(A and B) and P(A | B) is that P(A and B) takes two events regardless of dependence and gives the probability of both happening. P(A | B) on the other hand, is only useful in dependent events, where the result of one event effects the other, using it on independent events is useless since in that case P(A | B)=P(A)

When it comes to which one it more useful, it depends. While the probability of 2 or more events happening can be very useful for some statistics, as a probability intiate, when it comes to dependent events I feel that P(A | B) would be more useful. If you're in a werewolves game, and some event that has a certain chance of happening happened, you can use that to find how it changed other event's probabilities given that the event happened, which is much more useful then the probability of that event and another event happening.

[Ater Nox]

Flag Flayers, you can have some extra credit for those bonus questions

[SweetHaven]

Just a note to say, I am withdrawing from class

Sorry

Real life has struck again \0/

You can keep the tuition money BTW

[Alkasia]

SPOILER: Click to show

Question 1


Let A= the probability that Barry is a werewolf and B= the probability that Barry has not paid his bar tab



P(A | B)= P(A and B) / P(B)
P(A | B)= .30/.90 = 33.33%



Question 2



Let A= the probability of being elected mayor and B= the probability of being a wolf



P(A)= 1/15

P(B)= 3/14 (since I'm the seer)


Okay let's put this here:

P(A) = P(A | B)P(B) + P(A | Not B)P(Not B)

P(A) = P(A and B) + P(A and Not B)


P(A | B)= ((1/15)(3/14))/(3/14) = 1/15

P(A | Not B)= (1/15)(1-(3/14))= 1/15

P(Not B)= (1-(3/14))= 11/14

P(A)= (1/15)(3/14) + (1/15)(11/14)= 14/210= 1/15



That's the math, but since they're independent events, this can be stated to begin with since P(A | B)= P(A) when the events are independent.



Question 3



P(A and B)= (1/15)(3/14)= 3/210

Discussion 1

The difference between P(A and B) and P(A | B) is in dependency. P(A and B) disregards dependency and gives the probability of both events happening together.
P(A | B) and the accompanying equation only matter when the events are dependent, because when the events are independent, it's merely P(A | B)= P(A), since they don't rely on each other.
You could use P(A | B) when determining a second event with the first one already known. For instance, if you need to draw two cards of the same suit from a deck in order to win (with 13 cards per suit) and you draw a spade, your knowledge of the event would change the probability from 13/52 to 12/51.

I would consider P(A | B) useful more of the time because usually there will be something to base the probability off of, and it gives a more useful and specific answer than P(A and B). At least in Werewolf.

[Ater Nox]

Homework 2

FF and Alkasia: Good work on the homework

For the discussion, FF made a good point about how P(A|B) was useful when there was a dependence between events, and P(AB) useful when the events were independent. Alkasia, I liked your card example to illustrate how the stuff we know can be used to affect our calculations.

Personally, I think P(A|B) is more useful because we always have some information in a werewolf game (B) that we can use to help us calculate A, but of course, it entirely depends on what you want to do.

A+s all round!

[Ater Nox]

Lesson 3. Using Bayes theorem



Welcome everyone to the thirdlesson of Conditional Probability and Slaying Werewolves. In this lesson, we'll be learning about Bayes theorem.

As usual, if you have any questions, please post them in the thread. The next class will be held in a week.

Bayes Theorem

Bayes theorem describes the relationship between conditional probabilities. It is given as:

P(A|B) = P(B|A)P(A) / P(B)

You might remember from out last class that:

P(A | B) = P(A and B) / P(B)

If we rearrange this, we get:

P(A and B) = P(A | B) P(B)

If we do the same thing for P(B | A), we get:

P(B | A) = P(A and B) / P(A)

P(A and B) = P(B | A) P(A)

And so if P(A and B) is equal to both of those equations, the two equations are equal to each other.

P(A | B) P(B) = P(B | A) P(A)

Which is the same as:

P(A|B) = P(B|A)P(A) / P(B)

Using Bayes Theorem

This comes in useful when we want to find P(A|B), but can't easily determine what it is.

For example, in the following example we might be able to come up with a value for the chance that Aerso would vote for one of his fellow wolves if he was one, and the chance he is a wolf or voted for one easily.

P(Aerso is a wolf|Aerso voted for a wolf) = P(Aerso voted for a wolf|Aerso is a wolf) P(Aerso is a wolf)/P(Aerso voted for a wolf)

Example

There are 15 players in the game with 3 werewolves. In the first phase, all players excluding Aerso and Barry voted to lynch Barry. Barry turned out to be a werewolf. What is the probability that Aerso is a werewolf too?

Let A be the event that Aerso is a wolf
Let V be the event that Aerso voted for a villager

P(A)=3/15

We'll assume for now that Aerso randomly selected his vote from all players other than himself.

P(V)=12/14

If Aerso was a wolf, would he vote for another wolf? I think he would, so I'm going to say the probability that Aerso voted for a villager if he is a wolf is 50%.

P(V|A)=0.5

Therefore,
P(A|V) = P(V|A)P(A)/P(V)
=0.5*(3/15)/(12/14)
=11.67%

We can compare that to the probability that Aerso is a wolf, which is 3/15 or 20%. So, based off my theory on Aerso's likelihood to vote for a villager in this situation being 50%, it seems that he is less likely to be a wolf than by random chance.

Homework

Please post your answers to the homework questions in the thread, including any working you do. Everyone may participate in the discussion between now and the next class.

Question 1.
The probability that Aerso is having tacos for dinner is 20%, but the probability that he is having tacos given that it is Monday is 50%. If Aerso has tacos for dinner tonight, what is the probability that it is Monday?

Question 2.
If events A and B are independent, what are P(A and B) and P(A|B) equal to if P(A)=0.5 and P(B)=0.2?

Question 3.
There are 22 people playing a game of werewolf, including 5 werewolves. No one dies in the first phase. You are the seer. Aerso during the first day phase admits to being a werewolf, and so you want to know if you should scan him. With the events 'Aerso is a werewolf' and 'Aerso admitted to being a werewolf', calculate the probability that Aerso is a werewolf given that he admitted it.

Discussion 1.
Provide one or more examples of when Bayes Theorem could be useful to convert an easy to estimate conditional probability into a more useful one in a game of werewolf. Explain what values you might estimate for your conditional probabilities, and the ones other people have mentioned.

[Ater Nox]

Tanzoria, Dominion of Compassion, Woonsocket, Sorya, Louisistan, Hahiha / USSR, Control, Alkasia, Sauron, Flag Flayers, Palman, Elite, Free Guns For All, SweetHaven

Sorry for the delay everyone, I've been down with a bit of a nasty cold for the last week or so. The next class won't be until next week.