Baye's Revolver and other conditional probability questions

[GWestphal]

I'm brushing up on Baye's rule and this simple brain teaser has me hung up. I explain it graphically, but I can't explain it using Baye's equation.


You have a revolver and two rounds are put in adjacent chambers. Your friend fires first, then passes it to you. Do you spin and shoot or just shoot?

The probability you survive is 75% if you just shoot, because the fact that he didn't die removes 2 of six possibilities.

BB0000, B0000B are removed

So you have 0000BB, 000BB0, 00BB00 and 0BB000. The latter being a bad situation for you.

But how does that work into the Baye's Theorem? P(B|A)=P(A|B)P(B)/P(A)

That doesn't make 75% to me P(B|A) the probability you survive given your friend didn't die on the first shot.

P(A|B) that you die given that your friend did not (1/4)
P(B) probability that you live, (3/5)
P(A) probability that you die, (2/5)

(1/4*3/5)/(2/5)= 15/40 ~37%

What am a screwing up here?



Next question, you start in a corner of a square, you move to an adjacent corner 50% of the time in either direction. One turn passes. What is the expected number of moves to return to the starting point.


50% it will take 1 move (move back where you came from)
the other 50% you will move away to a place where it will take at least 2 moves

if you are at kitty corner position the chance of moving to the origin in two moves is 2*.5^2

at the other corner you have a 50% of returning to the origin or not.

2/3 of the time you are in a position to return to the origin 50% of the time, and 1/3 of the time you are in a position to return to the origin in two turns 2*.5^2, and .5^3 it will take 3 moves.

I just start getting mixed up here, because it could oscillate back and forth and never get to the origin. What am I doing wrong?

 

[serpretetsky]

Disclaimer: I am not a probability expert, I'm trying to help with the this question in the hope that I help myself get a little better with probability

Question 1: Your events are not very well defined. I'm assuming you want these events:
A: the event that you live playing second
B: the event that your friend lives playing first.

So now the question becomes what is the probability of event A occuring given that event B has occured

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

So.. what's the probability that your friend lived after playing first knowing that you lived after playing second (I'm assuming you keep playing even if your friend died, correct me if im wrong)

Also, what's the probability of you living after picking second (not knowing what your friend picked) ?

Question 2:
Your right, you can oscillate back and forth for ever, however the probability of such an even decreases exponentially.

50% it will take 1 move (move back where you came from)
the other 50% you will move away to a place where it will take at least 2 moves

I think you hit on a pretty key point there. Whenever you are in either of two corners closest to the origin (let's call these "close corners") you have a 50% chance of returning to the origin.

However, consider the other choice, the move to the farthest corner, more carefully. After moving to the farthest corner, there is an illusion of choice: You can move to one of the two "close corners". However, if you really think about it, both of the "closer corners" can actually be merged into a single corner. There is absolutely no difference, for this problem, if you go into one of the "close corners" or the other one. Therefore, as soon as you have moved into the farthest corner, you immediately sacrifice two moves and move straight back into one of the "close corners" again (it doesn't matter which one).

This means you cannot have even moves, all of them are odd. It also makes the probabilities a little easier:

P(X=1) = 50%
P(X=3) = P( X != 1) * 50%
P(X=5) = P( X != 3) * 50%
...
P(X=n) = P(X != n-2) * 50%

where n is a positive odd integer.

Now you should be able to use the definition of expected values (for discrete random variables) to obtain the answer:

Expected Value:
E[X] = Sum from i=1 to i=infinity of x_i * p_i where x_i is the ith random value that the random variable can take on and p_i is the probability of that random value.

 

[Mand]

Chance of a bullet if you spin it is easy: 1/3.

Chance of a bullet if you don't spin is 1/4.

So, don't spin. That part you've established correctly. This bit:

P(A|B) that you die given that your friend did not (1/4)
P(B) probability that you live, (3/5)
P(A) probability that you die, (2/5)

(1/4*3/5)/(2/5)= 15/40 ~37%

What am a screwing up here?

You've defined P(A) and P(B) incorrectly. One of them is you live, the other one is if your friend lives. P(youlive|friendlives) is what you're trying to calculate. Bayes's theorem lets you move from one conditional probability to another - you don't need this, you're just calculating one conditional probability:

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

You've already done this, in the 75% to not get a bullet if you don't spin. What Bayes's Theorem allows you to do is calculate P(friendlives|youlive) once you know P(youlive|friendlives).

Next question, you start in a corner of a square, you move to an adjacent corner 50% of the time in either direction. One turn passes. What is the expected number of moves to return to the starting point.


50% it will take 1 move (move back where you came from)
the other 50% you will move away to a place where it will take at least 2 moves

if you are at kitty corner position the chance of moving to the origin in two moves is 2*.5^2

at the other corner you have a 50% of returning to the origin or not.

2/3 of the time you are in a position to return to the origin 50% of the time, and 1/3 of the time you are in a position to return to the origin in two turns 2*.5^2, and .5^3 it will take 3 moves.

I just start getting mixed up here, because it could oscillate back and forth and never get to the origin. What am I doing wrong?

You're on the right track here. At each corner, there's a 50% branch point you can take, which will take you down another path. You're correct that it's possible to go infinitely without going back to the corner - that's the point. The calculation of this involves an infinite series. You've already established the first few terms, but to do the full calculation you'll have to take the series limit.

 

[TuxDave]

I'm brushing up on Baye's rule and this simple brain teaser has me hung up. I explain it graphically, but I can't explain it using Baye's equation.

You have a revolver and two rounds are put in adjacent chambers. Your friend fires first, then passes it to you. Do you spin and shoot or just shoot?

The probability you survive is 75% if you just shoot, because the fact that he didn't die removes 2 of six possibilities.

BB0000, B0000B are removed

So you have 0000BB, 000BB0, 00BB00 and 0BB000. The latter being a bad situation for you.

But how does that work into the Baye's Theorem? P(B|A)=P(A|B)P(B)/P(A)

That doesn't make 75% to me P(B|A) the probability you survive given your friend didn't die on the first shot.

P(A|B) that you die given that your friend did not (1/4)
P(B) probability that you live, (3/5)
P(A) probability that you die, (2/5)

(1/4*3/5)/(2/5)= 15/40 ~37%

What am a screwing up here?

This is where a little organization goes a long way:
A = You die
B = Your friend lives

So from that:
P(A) = 1/3
P(B) = 2/3

As you manually calculated
P(A|B) = 1/4

Bayes theorem allows you to calculate:
P(B|A) or in other words "if they find your dead body, what's the probably that your friend is alive"

P(B|A) = P(A|B) * P(B) / P(A) = (1/4 * 2/3) / (1/3) = 1/2.

So everything looks fine to me. If you died, that means the revolver was either

BB0000
0BB000

That means it's a 50% chance that your friend didn't die. (even though if I saw my friend shoot himself in the head, I probably wouldn't continue playing )