Expected tries to to remove ball from urn

  • Thread starter Thread starter funnyguy
  • Start date Start date
  • Tags Tags
    Ball
AI Thread Summary
The discussion revolves around calculating the expected number of tries to remove a black ball from an urn containing 3 red and 1 black ball, specifically without replacement. The initial approach using expected value formulas and hypergeometric distribution led to confusion, prompting a reevaluation of the probabilities involved. Participants noted that the probability of drawing a black ball changes with each attempt, starting at 1/4 and adjusting based on the remaining balls. A simpler method was suggested, focusing on the direct probabilities of outcomes for each possible number of draws. The conversation highlights the complexity of calculating expectations in scenarios involving dependent probabilities.
funnyguy
Messages
16
Reaction score
0
I have a problem dealing with expected values. I'll jump right into the problem:

There are 3 red and 1 black balls in an urn. What is the expected number of times a ball must be removed before a black ball is removed.

I originally thought of this using

E[X]=1*P(X=1) + 2*P(X=2) + ...

I filled in P's using hyper geometric distribution. But didn't get the correct result.

Trying a uniform for each trial I did this
1*1/4 + 2*1/3 + 3*1/2 + 4*1/1

I found the following as a solution from a LONG derivation of formulas:
k(r+b+1)/(b+1) where k=1 in this case.
From a problem setup the same but using negative hyper geometric.

This makes no sense to me. Is there some more simple way of thinking of this?
 
Last edited:
Physics news on Phys.org
The probability that the ball is picked in the nth attempt is just 1/4, as all balls are equal. This allows to calculate the expected value easily.

Trying a uniform for each trial I did this
1*1/4 + 2*1/3 + 3*1/2 + 4*1/1
You would have to consider the probability that the ball was not drawn before in each summand.
 
mfb said:
The probability that the ball is picked in the nth attempt is just 1/4, as all balls are equal. This allows to calculate the expected value easily.You would have to consider the probability that the ball was not drawn before in each summand.

Sorry, but I forgot to mention this is without replacement.

So to start the black ball probability is 1/4, but after the 1st attempt the black ball is 1 of 3 remaining, etc.
 
funnyguy said:
Sorry, but I forgot to mention this is without replacement.
I know.
So to start the black ball probability is 1/4, but after the 1st attempt the black ball is 1 of 3 remaining, etc.
Sure, but the probability that you draw a second ball is just 3/4. If you multiply 1/3 by 3/4, you get 1/4 again.
 
mfb said:
I know.
Sure, but the probability that you draw a second ball is just 3/4. If you multiply 1/3 by 3/4, you get 1/4 again.

Maybe I see what you're saying.

P(X=1) = 1/4
P(X=2) implies P(X=2) AND P(X!=1) = 1/3 * (1- P(X=1))

?
 
Right.
It is possible to extend this for P(X=3) and P(X=4), but that is unhandy. You can get the probabilities in a direct way as well.
 
b, rb, rrb and rrrb are the 4 possible cases corresponding to your X=0,1,2,3. Find the later three probabilities and hence the expectation.
 
Back
Top