

So this has been stuck in my head, and i'm sure I could open up some math tutorials on it, but don't want to relearn high school stats to get my answer.
So assume there are four cards, one of them is a traitor.
Obviously there's a 25% chance of drawing the traitor in the first draw.
Then a 33% chance of drawing the traitor in the second draw.
But that is assuming that the traitor still exists after that first draw.
What I am asking: Is the chance to draw the traitor different for the person that first drew the card than the person that drew the last card or is it the same chance no matter what.

Dan Blum
United States Wilmington Massachusetts

You mean there are four cards and four people each draw one? If so, everyone has the same 25% chance of getting the traitor card.



tool wrote: You mean there are four cards and four people each draw one? If so, everyone has the same 25% chance of getting the traitor card.
Yeah draws one, one after another. Gotcha.



Maybe I miss something, but I would think that...
1. In the first draw the probability is 25% for everyone.
2. In the second draw, the probability is 33% percent for everyone, given that the traitor hasn't been drawn. It is 0% if the traitor has been drawn before.
3. In the third draw, the probability is 50% percent for everyone, given that the traitor hasn't been drawn. It is 0% if the traitor has been drawn before.
4. In the fourth draw, the probability is 100% to draw the traitor, given that the traitor hasn't been drawn. It is 0% if the traitor has been drawn before.
I would say the key is the term "given that..."  we talk about conditioned probabilities like P(traitor in draw itraitor still available in draw i1); so yes, depending on when you draw, the probability changes according to your position i.
Because those are conditioned probabilities that depend on the random event if the traitor has been drawn before or not, it is, however, hard to judge whether this is fair for everyone. The a priori probability (25%) is the sole thing that is equal for everyone, which is how I would justify that the probability changes over the draws.



Teeguru87 wrote: Maybe I miss something, but I would think that...
1. In the first draw the probability is 25% for everyone.
2. In the second draw, the probability is 33% percent for everyone, given that the traitor hasn't been drawn. It is 0% if the traitor has been drawn before.
3. In the third draw, the probability is 50% percent for everyone, given that the traitor hasn't been drawn. It is 0% if the traitor has been drawn before.
4. In the fourth draw, the probability is 100% to draw the traitor, given that the traitor hasn't been drawn. It is 0% if the traitor has been drawn before.
I would say the key is the term "given that..."  we talk about conditioned probabilities like P(traitor in draw itraitor still available in draw i1); so yes, depending on when you draw, the probability changes according to your position i.
Because those are conditioned probabilities that depend on the random event if the traitor has been drawn before or not, it is, however, hard to judge whether this is fair for everyone. The a priori probability (25%) is the sole thing that is equal for everyone, which is how I would justify that the probability changes over the draws.
If the draw is blind, I think the probability is the same as when everyone gets a card at the same time.



Wormaap wrote: Teeguru87 wrote: Maybe I miss something, but I would think that...
1. In the first draw the probability is 25% for everyone.
2. In the second draw, the probability is 33% percent for everyone, given that the traitor hasn't been drawn. It is 0% if the traitor has been drawn before.
3. In the third draw, the probability is 50% percent for everyone, given that the traitor hasn't been drawn. It is 0% if the traitor has been drawn before.
4. In the fourth draw, the probability is 100% to draw the traitor, given that the traitor hasn't been drawn. It is 0% if the traitor has been drawn before.
I would say the key is the term "given that..."  we talk about conditioned probabilities like P(traitor in draw itraitor still available in draw i1); so yes, depending on when you draw, the probability changes according to your position i.
Because those are conditioned probabilities that depend on the random event if the traitor has been drawn before or not, it is, however, hard to judge whether this is fair for everyone. The a priori probability (25%) is the sole thing that is equal for everyone, which is how I would justify that the probability changes over the draws.
If the draw is blind, I think the probability is the same as when everyone gets a card at the same time.
The question isn't that they're drawn at the same time, the question is if they were drawn one after another.



It's still 25%
For the second person drawing we are saying: the chance of drawing the traitor GIVEN it hasn't been drawn before. So this is:
Probability of traitor AND traitor wasn't drawn before
= 33% * 75%
= 25%



tekshi wrote: Wormaap wrote: Teeguru87 wrote: Maybe I miss something, but I would think that...
1. In the first draw the probability is 25% for everyone.
2. In the second draw, the probability is 33% percent for everyone, given that the traitor hasn't been drawn. It is 0% if the traitor has been drawn before.
3. In the third draw, the probability is 50% percent for everyone, given that the traitor hasn't been drawn. It is 0% if the traitor has been drawn before.
4. In the fourth draw, the probability is 100% to draw the traitor, given that the traitor hasn't been drawn. It is 0% if the traitor has been drawn before.
I would say the key is the term "given that..."  we talk about conditioned probabilities like P(traitor in draw itraitor still available in draw i1); so yes, depending on when you draw, the probability changes according to your position i.
Because those are conditioned probabilities that depend on the random event if the traitor has been drawn before or not, it is, however, hard to judge whether this is fair for everyone. The a priori probability (25%) is the sole thing that is equal for everyone, which is how I would justify that the probability changes over the draws.
If the draw is blind, I think the probability is the same as when everyone gets a card at the same time. The question isn't that they're drawn at the same time, the question is if they were drawn one after another. My remark was meant to clarify. Nothing changes between the two situations, provided the draw is blind.



greendayfan333 wrote: It's still 25%
For the second person drawing we are saying: the chance of drawing the traitor GIVEN it hasn't been drawn before. So this is:
Probability of traitor AND traitor wasn't drawn before
= 33% * 75%
= 25%
Excuse me if I got the problem wrong, but I dare say this is not true.
The probability to draw the traitor given it hasn't been drawn before is not the probability of the conjunction but the conditioned probability, which is defined as:
P(AB) = P(A^B)/P(B)
which means:
P(draw traitor in this drawno traitor drawn before) = P(draw traitor in this draw and no traitor drawn before) / P(no traitor drawn before)
The problem here is that A and B are not stochastically independent, which means P(AB) is not the same as P(A), which further means that P(A^B) is not the same as P(A) * P(B) but P(A^B) = P(AB) * P(B).
P(A^B), the probability to draw the traitor in any draw and not to have the traitor drawn any draw before is .25, as you said, because
for the first draw it is .25* 1 = .25 for the second draw it is .33*.75 = .25 for the third draw it is .50*.50 = .25 for the fourth draw it is 1 * .25 = .25
BUT the probability we search for here is not P(A^B) but P(AB), which is changing because P(B) changes over the course of the draws from 1 to .25.
That means for any given draw, the probability P(AB) is .25/P(B) or
P(draw traitor in this drawno traitor drawn before) = 0.25 / P(no traitor drawn before)
That means that
for the first draw it is .25/1 = .25 for the second draw it is .25/.75 = .33 for the third draw it is .25/.50 = .50 for the fourth draw it is .25/.25 = 1

Stephen Miller
United Kingdom Newport Gwent

tekshi wrote: So this has been stuck in my head, and i'm sure I could open up some math tutorials on it, but don't want to relearn high school stats to get my answer.
So assume there are four cards, one of them is a traitor.
Obviously there's a 25% chance of drawing the traitor in the first draw.
Then a 33% chance of drawing the traitor in the second draw.
But that is assuming that the traitor still exists after that first draw.
What I am asking: Is the chance to draw the traitor different for the person that first drew the card than the person that drew the last card or is it the same chance no matter what.
Same chance no matter what.
I worked this out as a child for a game show I was watching, once, to work out if picking first was an advantage or not.
25% first card is a traitor. 75% first card isn't.
If not...
33% chance second card is a traitor... (33% of 75% is 25%) 67% chance second card isn't (67% of 75% is 50%) 50% chance no traitor cards yet.
If neither...
50% chance third card is a traitor (50% of 50% is 25%) 50% chance third card isn't (50% chance of 50% is 25%) 25% chance no traitor cards yet.
If none of above...
100% chance fourth card is a traitor (100% of 25% is 25%)
So, 25% for each.



Gizensha wrote: tekshi wrote: So this has been stuck in my head, and i'm sure I could open up some math tutorials on it, but don't want to relearn high school stats to get my answer.
So assume there are four cards, one of them is a traitor.
Obviously there's a 25% chance of drawing the traitor in the first draw.
Then a 33% chance of drawing the traitor in the second draw.
But that is assuming that the traitor still exists after that first draw.
What I am asking: Is the chance to draw the traitor different for the person that first drew the card than the person that drew the last card or is it the same chance no matter what. Same chance no matter what. I worked this out as a child for a game show I was watching, once, to work out if picking first was an advantage or not. 25% first card is a traitor. 75% first card isn't. If not... 33% chance second card is a traitor... (33% of 75% is 25%) 67% chance second card isn't (67% of 75% is 50%) 50% chance no traitor cards yet. If neither... 50% chance third card is a traitor (50% of 50% is 25%) 50% chance third card isn't (50% chance of 50% is 25%) 25% chance no traitor cards yet. If none of above... 100% chance fourth card is a traitor (100% of 25% is 25%) So, 25% for each.
Okay, I really got the problem wrong. I assumed that the question is about the probability in a particular draw which changes over the draws.
For any player to draw the traitor in any draw the overall probability remains .25 as Gizensha pointed out and which I calculated as P(A^B) which remains constant over the draws.
Thus I have to correct my statement that it is hard to judge if this is a priori fair for everyone.

Philip Kitching
United Kingdom

Part of the problem with "feeling" probabilities is that it is difficult to avoid knowing what each player drew when you step through the process in your head.
In effect, there is a difference between dealing 4 cards and simultaneously revealing vs dealing and revealing one card at a time.
So the second player does have a 33% chance of being the traitor but only because 25% of the time you'd have been playing a different game where the first player was the traitor.
In practical terms there is no impact unless players can act on the knowledge of player one being a traitor or not before player two is revealed.
Consider three forms of Russian roulette. A) one gun with four bullets, one live, three blank. Each player shoots, then passes the gun on. B) four guns, each with one bullet (one live, three blank). Each player shoots in turn. C) four guns, each with one bullet (one live, three blank). All players shoot at once.
The probabilities are all 25%, but the conditional probability of being the fourth player in games B & C is 100%, since you don't play in 75% of the games.

CARL SKUTSCH
United States New York New York
Agricola, Sekigahara, Concordia, Innovation, COOKIE!!! (and Guinness)
SANJURO: You're all tough, then? GAMBLER: What? Kill me if you can! SANJURO: It'll hurt.

Wait, what, seriously?
The only problem here is in framing the question.
If you're saying that after each draw the person drawing the traitor has to reveal it, well then, yeah, the fourth person drawing, if the traitor has not yet been revealed, is 100% going to be the traitor, duh. In the same way that the 2nd person is 33% going to be the traitor.
However, the way I read the OP's post (it was a little confusing) was the each person was drawing in order but that the traitor was not being revealed by anyone. So, before the third person drew, for example, they would have (as far as they could tell) a 25% of drawing the traitor. (Because there are four cards, 2 in the hands of the already drawn players, 2 in the deck. All 4 cards are unknown to current drawer #3 so there is a 1 in 4 chance that they will have the traitor.)
It's all a matter of perspective and knowledge. If nothing is revealed, the chance is always 25% from the pov of the drawer. If we're using someone's pov who has knowledge, the odds shift.


