# Probability of a certain outcome when drawing balls from an urn with 2 colors without replacement

Consider an urn containing $$i$$ red and $$j$$ blue balls. We draw $$n$$ balls from the urn without replacement. If $$(x_1,\dots,x_n)$$ denotes an outcome, how should one canonically define the probability of this outcome?

I think the probability depends on the order on which red (or blue) balls are drawn. In other words, $$RBBBBBBB$$ will have a different probability from $$BBBBBBBR$$ since it seems intuitively obvious that the chances of getting a red in the beginning would be less and in the end should be more (since later there are lesser blue balls).

I looked at hypergeometric distribution but that is not what I really want.

• Have you looked at examples? $2$ red and $1$ blue seems like the first non-trivial case.
– lulu
Oct 30, 2019 at 13:25

For this post, I will use $$n\frac{r}{~}$$ to denote the falling factorial: $$n\cdot(n-1)\cdot(n-2)\cdots(n-r+1)=\frac{n!}{(n-r)!}$$

I will make a few changes to notation. It should be clear why I did so in a moment. Let us instead talk about the total number of red balls as $$I$$ and the total number of blue balls as $$J$$. Further, let us talk about the total number of balls as $$I+J = N$$. We will then be wanting to draw $$n$$ balls total, and let $$i$$ instead represent the total number of red balls that happened to have been drawn (rather than the total number available) and similarly $$j$$ be the number of blue balls that were drawn.

Imagine that the balls are all uniquely labeled. Recognize then that each of the $$N\frac{n}{~}$$ ways of selecting $$n$$ balls in sequence from the $$N$$ available balls are equally likely to have occurred.

Let us consider a specific sequence of colors of balls that contains $$i$$ red balls and $$j$$ blue balls. Let us count how many ordered sequences of labeled balls result in this sequence of colors.

From left to right, decide which specific red ball occupies a space intended for a red ball to go. The first time there will be $$I$$ options for the specific red ball, then $$I-1$$ for the next, and so on... resulting in $$I\frac{i}{~}$$ ways in which we may select which red ball happened to go in which spot.

Similarly, from left to right, decide which specific blue ball occupied a space intended for a blue ball to go. As before, this will result in $$J\frac{j}{~}$$ ways in which this can be done.

We get then a probability of:

$$\dfrac{I\frac{i}{~}\cdot J\frac{j}{~}}{N\frac{n}{~}}$$

Notice, this does not change based on the order in which the colors of the balls occurred. It is exactly as probable to have gotten a sequence RBBBBB as it is to have gotten a sequence BBBBBR. While yes the probability that the first ball is red is less than the probability that the $$n$$'th ball is red given that the first n-1 balls were blue, that is irrelevant and what we should have been asking is what the probability is that the first ball was red compared to the probability the $$n$$'th ball was red where this second probability is not conditioned on anything. This is similar to how it is equally likely to have drawn a queen on the first draw of a deck as it is to have drawn a queen on the second draw from a deck or indeed the $$n$$'th draw from a deck for any $$n$$.

From the above observations and derived formula, we can then further derive the formula for the hypergeometric distribution by accounting for all of the $$\binom{n}{i}$$ orders in which we could have seen red vs blue balls.

• Can you expand on this: "This is similar to how it is equally likely to have drawn a queen on the first draw of a deck as it is to have drawn a queen on the second draw from a deck". Oct 30, 2019 at 13:53
• @Shahab math.stackexchange.com/questions/1287393/… The probability the first card in a deck is a queen is $\frac{4}{52}=\frac{1}{13}$. The probability that the second card is a queen (with no condition on what the first card is) is also $\frac{4}{52}=\frac{1}{13}$, not $\frac{4}{51}$ like is a common mistake. Oct 30, 2019 at 13:59
• @JMoravitz would you be able to help me on this question?: math.stackexchange.com/questions/4481749/… it's similar to this one Jun 28 at 18:15