Non-independent two consecutive draws from two urns

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In summary, the conversation discusses an experiment where balls are drawn from two urns and the number of red balls in the second draw is a random variable. The formula for the probability of this variable is given, and there is a question about the mean and variance of the variable. It is suggested to use binomial and Gaussian distributions for different scenarios.
  • #1
hwangii
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Suppose there are two urns: in urn A, there are r red balls and w white balls. In urn B, there are b black balls.

Suppose we do the following experiment: draw k balls from urn A. Among those k balls, put only the red balls in urn B, and draw n balls from urn B. Then the number of red balls from the second draw is a random variable.

Call the random variable $\tilde{y}$. Then
\begin{align*}
Pr(\tilde{y}=y)=\sum\limits_{x=\max\{y,k-w\}}^{\min\{r,k\}}\frac{{r\choose x}{w\choose k-x}}{{r+w \choose k}}\frac{{x\choose y}{b\choose l-y}}{{x+b\choose l}}
\end{align*}

Does anyone know what the mean and the variance of this random variable are? If you do not know the exact form, what about the asymptotic mean and variance when r, w and b go to infinty with the ratio amongst them constant.

Thanks a lot!
 
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  • #2
For r,w,b -> infinity with constant ratio, with constant n, the probability goes to 0 as we have many black balls and nearly no red balls in the second step.

For r,w -> infinity with constant n,b, the first drawing becomes a binomial distribution as function of k.

In general: Expand your (n choose k) as factorials, simplify, approximate them with the Stirling formula, simplify, and see what you get.

For r,w,b,n -> infinity, gaussian distributions are good.
 

FAQ: Non-independent two consecutive draws from two urns

What is the difference between independent and non-independent two consecutive draws from two urns?

Independent draws from two urns means that the outcomes of the first draw do not affect the probabilities of the outcomes of the second draw. Non-independent draws, on the other hand, means that the outcomes of the first draw do affect the probabilities of the outcomes of the second draw.

How is the probability calculated for non-independent two consecutive draws from two urns?

The probability for non-independent draws is calculated by considering the outcomes of the first draw and then adjusting the probabilities for the second draw based on those outcomes. This adjustment can be done using conditional probabilities.

Can you provide an example of non-independent two consecutive draws from two urns?

One example is drawing two marbles from two urns, where the first urn contains 3 red marbles and 7 blue marbles, and the second urn contains 5 red marbles and 5 blue marbles. The probability of getting a red marble on the second draw is higher if a red marble was drawn on the first draw (non-independent) compared to if a blue marble was drawn on the first draw (independent).

How does the number of draws affect the probability for non-independent two consecutive draws from two urns?

The more draws that are done, the more the probabilities of each draw will be affected by the outcomes of the previous draws. This means that the probability for non-independent draws will change with each subsequent draw, unlike independent draws where the probability remains constant.

Is there a way to make non-independent draws behave like independent draws?

Yes, by increasing the number of draws and ensuring that each draw is done with replacement (putting the drawn item back into the urn before the next draw), the probabilities for non-independent draws can approach the probabilities for independent draws.

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