# Conditional Probability. Drawing the ball from the urn.

Reading the paper I have encountered the following example:

The urn that have been filled with black and white balls as follows. First, a ball is placed in the urn according to the outcome of a fair coin toss. If the coin toss for an urn produces heads, the “coin ball” placed in the urn is black; it is white otherwise. In addition to a coin ball, the urn contains n “noncoin balls,” of which exactly n/2 are black and n/2 are white.

The agent is invited to bet on the colour of the two coin balls. Any bet (on a ball of some colour drawn from some urn) pays one dollar if the ball has the desired colour and zero otherwise.

Suppose now that one draw from the urn is observed and that a ball drawn is black. Calculate the conditional probability of a black coin ball.

The correct answer is (n/2+1)/(n+1). But I can't understand how this result was obtained. I kindly ask for the help from the community.

• "Reading the paper..." Huh? Which paper? Commented Sep 9, 2019 at 19:29
• Sorry that I haven't mentioned that. Epstein, Larry G., and Martin Schneider, 2008, “Ambiguity, Information Quality, and Asset Pricing,” Journal of Finance 63 (1), 197–228
– user702441
Commented Sep 9, 2019 at 21:10

To have an easy exposition, i will will denote by $$k$$ the natural number $$k:=n/2$$. We consider the given two-steps experiment with the probabilities given by the tree model.

                                       Black ===> Relevant, event E
/
/ (k+1)/(2k+1)
/
Black coin ball, (k+1)B and kW
/                              \
/                                \ etc.
/ 1/2                              \
/                                    White
/
*
\
\                                    Black ===> Relevant, event F
\ 1/2                              /
\                                / k/(2k+1)
\                              /
White coin ball, kB and (k+1)W
\
\ etc
\
White


Later edit: In the above diagram, we have the following compositions of the urn after the first coin ball:

• if we add a black coin ball, then there are $$(k+1)$$ B(lack) balls and $$k$$ W(hite) balls in the urn,
• if we add a white coin ball, then there are $$k$$ B(lack) balls and $$(k+1)$$ W(hite) balls in the urn.

So we are on the relevant side, and compute the probability to land "up", in the event $$E$$, considering that we already "sit" in the event $$E\cup F$$.

The corresponding probability is: $$\frac{P(E)}{P(E\cup F)} = \frac{P(E)}{P(E)+ P(F)} = \frac {\displaystyle \frac 12\cdot\frac{k+1}{2k+1}} {\displaystyle \frac 12\cdot\frac{k+1}{2k+1} + \frac 12\cdot\frac{k}{2k+1}} = \frac{(k+1)}{(k+1)+k}\ .$$

• dan__fulea, thank you for the answer. Could you please clarify what does it mean: (k+1)B and kW? Why blacks are (k+1) and whites are just k in this case, so they total up to 2k+1?
– user702441
Commented Sep 9, 2019 at 21:28
• i edited the answer, so that the abbreviation used is clear. Commented Sep 9, 2019 at 21:34
• I might completely misunderstand the question but I can't get why the ball is added to the urn. From my understanding the balls are taken out from the urn.
– user702441
Commented Sep 9, 2019 at 21:45
• I am answering a question that does not need the agent and a "two coin ball", things that are hard to figure out. The following scenario is the one i had to isolate somehow, so let us put it in short sentences. The urn has $k$ W(hite) balls and $k$ B(lack) balls. We add one "coin ball", which is either B or W with equal probability. Then we extract one ball from the urn with the $2k+1$ balls and an uncertain composition. We get the supplementary information, that this ball is B. The question is now, which is the probability that the added coin ball is also B. The answer is for this qu'n. Commented Sep 9, 2019 at 22:10