Expected tries to to remove ball from urn

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Discussion Overview

The discussion revolves around calculating the expected number of attempts required to draw a black ball from an urn containing 3 red balls and 1 black ball. Participants explore various methods for determining this expected value, including the use of probability distributions and the implications of drawing without replacement.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant initially approaches the problem using the expected value formula and hypergeometric distribution but finds inconsistencies in their results.
  • Another participant suggests that the probability of drawing the black ball on any given attempt is uniformly 1/4, but acknowledges the need to adjust for the probabilities of previous draws.
  • Participants clarify that the scenario involves drawing without replacement, which affects the probabilities of subsequent draws.
  • There is a discussion about calculating probabilities for drawing the black ball on the second attempt, considering the outcomes of the first draw.
  • One participant proposes a method of listing possible outcomes (b, rb, rrb, rrrb) to derive the probabilities and expectations, indicating a combinatorial approach.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the expected value, with no consensus reached on a single method or solution. The discussion remains unresolved regarding the most straightforward way to derive the expected number of attempts.

Contextual Notes

Participants highlight the complexities introduced by drawing without replacement, which complicates the probability calculations for each subsequent draw. There are also references to various probability distributions that may apply, but no agreement on their effectiveness in this context.

funnyguy
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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?
 
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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.
 

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