# Many urns with colored balls

This is 1.7 from Grimmett's Probability book:

There are n urns of which the r th contains r − 1 red balls and n − r magenta balls. You pick an urn at random and remove two balls at random without replacement. Find probability the second ball is magenta, given that the first is magenta.

This answer is given here: Find probability of specific ball getting selected on second turn

My question: Why is the following wrong (i.e. what I'm doing):

Denote $C_i$ the event that magenta is chosen on the the ith trial. Then: $$P(C_2 = 1|C_1 = 1) = \sum^{n}_{r=1}P(C_2 = 1|C_1 = 1,Urn=r)P(Urn=r)$$ $$= \sum^{n}_{r=1}\frac{n-r-1}{n-2}\frac{1}{n}$$

For $P(C_2 = 1|C_1 = 1,Urn=r)$, I work direcctly in the conditioned sample space, so I know I'm in the $r^{th}$ urn and the first ball drawn is magenta. So I can just write down the probability: $\frac{n-r-1}{n-2}$

• Conditional probabilities don’t combine this way. Here’s an example I think will convince you. There are two coins. One has two heads, and the other has two tails. You choose a coin at random and flip it twice. What is the probability that the second flip is heads, given that the first flip was heads? The answer is $1$, but the method you are trying to use says it is $1\cdot\frac{1}{2}+\frac{1}{2}\cdot$[not really sure, because the first flip can’t be heads if the second coin was chosen, and by the way, your formula has a negative value for the urn where the first ball can’t be magenta]. – Steve Kass May 26 '16 at 22:42
• Or put another way, while there is a “Law of Total Probability” and a “Law of Total Expectation” that use a partition of the event space, there is no “Law of Total Conditional Probability” that is as general. You can only do what you’re trying to do if the probability that the first ball is magenta is the same for each urn, which it is not. A related question on MSE is here: math.stackexchange.com/questions/1209050/… – Steve Kass May 26 '16 at 22:47
• Oh hm, after thinking about what you said I think I just used LOTP wrong. Perhaps, it should be: $P(C_2 = 1|C_1 = 1) = \sum^{n}_{r=1}P(C_2 = 1|C_1 = 1,Urn=r)P(C_1 = 1|Urn=r)P(Urn=r)$ – yoshi May 26 '16 at 22:50

## 1 Answer

Your expression looks not quite right.

If $r=n$ then that term for the probability is negative.

If $n<3$ then the whole thing is nonsensical.

One can see for $n=r$ and $n=r-1$ the term for the conditional probability will be zero, because there aren't enough magenta balls.

So I'd pick a reasonable number ($n=6$, say) and write out the probabilities explicitly to get the form for the term.

• these observations are all true, but I worked directly from definitions. so what, in particular did I miss? In fact I know what the missing term is too -- I just dont know how to reconcile what's missing with how I'm thinking. – yoshi May 26 '16 at 22:15
• I think you just answered your own question in a comment to the original post. Yes, you were applying the Law of Total Probability incorrectly (as if it could be used without modification for conditional probabilities). – Steve Kass May 26 '16 at 23:10