# Magic Urn Problem

ADDENDUM: First off, thank you all for your insightful answers to this weekend, fun problem, and especially Mark L Stone for his initial comment. I'm probably missing the point in some of the answers, in which case, please change them slightly so that I see it and can happily accept one. But, I guess the issue that I'm interested in is not so much how to solve the problem, or how some answers are nonsensical, but How can it be that even though the equations that I used to solve the problem (and remember I didn't know the answer was going to be 3R/3B at that point) yield the correct answer even with the inclusion of an impossible scenario (2 pairs of red)?

The Wall Street Journal has a new section with math questions. Today's question was open for discussion with posts by readers. So, in this regard, I presume it is fair to also discuss it here among a generally more expert audience. If this reasoning is faulty, just let me know and I'll erase the question.

Your friend tells you: This is a remarkable urn. It contains only red and black balls. If you reach in and take one ball at random, there’s an equal chance of drawing red or black. So you might think that if you instead take two at random, it’s 50/50 that your balls will match. Well, you’d be wrong—but if they do match, and then you reach your other hand in and take two more, your chances of matching a second time are 50/50! How many balls in your friend’s urn (before you take any out)?

Like a good number of people who still remember the basics of combinations and permutations, I followed this reasoning:

We start off with equal numbers of red and black balls ("equal chance of drawing red or black"), let's name this number $n$. After the first round you may have gotten two red or two black("but if they DO match, and then..."). Let's go with red, i.e. you have drawn two reds. You started with equal numbers or red and black or $2n$ balls in the urn, and after drawing the first pair, you have $2n-2$ with $n-2$ red balls remaining in the urn, and $n$ black balls.

They tell us that the chance of repeating the experiment and getting a second pair remain $0.5$ ("...your chances of matching a second time are 50/50"). So $0.5$ should be the probability of drawing a second pair of red balls, plus the probability of getting a pair of black balls:

$$0.5 = p(2R|\text{after Red pair})+p(2B|\text{after Red pair})= \frac{{n-2\choose 2}}{{2n-2\choose2}} + \frac{{n\choose 2}}{{2n-2\choose2}}$$

The answer turns out to be $n=3$ for a total of $6$ balls in the urn ($3$ Red and $3$ Black). Yet this makes it physically impossible for us to draw two red balls on the second round.

What am I missing?

In other words, we get the right result (see comments); however, we can't go back and "verify" that we got the correct answer, because when you go back to the equation above and plug in $3$ into the $n$'s, we get:

$0.5 = = \frac{{1 \choose 2}}{{4\choose 2}} + \frac{{3\choose 2}}{{4\choose 2}}$, and ${1 \choose 2}$ makes no sense. It's probably the way math is telling us that we can't draw two pairs of red balls with only 3 red balls available. If this is true, it is really cool in a wonderfully circular fashion: we got the right answer including an impossible premise in the equation, in such a way that the result prompts us to modify the equation that generated the answer in the first place!

We would have gotten the same result with:

$0.5 = p(2B|\text{after Red pair})= \frac{{n\choose 2}}{{2n-2\choose2}}$ despite the fact that we didn't know at first that the scenario of drawing two pairs of red balls was impossible.

• "Yet this makes it physically impossible for us to draw two Red balls on the second round." So what? With one red and 3 black, ${3\choose 2}$ out of ${4\choose 2}$ possibilities are both black, i.e., the same color, which is 50/50. I think the question is meant to be interpreted as the 2nd pair match each other, not that they match the first pair - perhaps this is where you are "going wrong". Sep 20, 2015 at 1:07
• @MarkL.Stone It is just puzzling to get the right answer (6 balls) even when I have included in the equation the probability calculus of a scenario that is physically impossible: drawing a second pair of red balls after having drawn a first pair of red balls. Sep 20, 2015 at 1:28
• You missed out the case when the first fair is a black ball. So the left side should have a factor of $\frac{1}{4}$ instead of $\frac{1}{2}$ Sep 20, 2015 at 5:55
• using the 6 balls (3 red and 3 black) what is the probability of drawing two of the same color (i.e., the first step in the question). Words + math is helpful. thx.
– user89957
Sep 20, 2015 at 22:11
• we don't know the result is 6 until we get it. but we can get it using a formula that includes a physical impossibility as you suggest. that makes it interesting to me. Sep 20, 2015 at 22:25

What am I missing? How can it be that even though the equations that I used to solve the problem (and remember I didn't know the answer was going to be 3R/3B at that point) yield the correct answer even with the inclusion of an impossible scenario (2 pairs of red)?

$$0.5 = p\,(2\,R\,|after\,Red\,pair)+p\,(2\,B\,|after\,Red\,pair)= \frac{{n-2\choose 2}}{{2n-2\choose2}} + \frac{{n\choose 2}}{{2n-2\choose2}}$$

holds only for $n-2 \geq 2$ because $n\choose r$ is a non zero for $n \geq r$, so the solution $n=3$ to this equation is bounded by this constraint: $n \geq4$, which implies there is no solution.

## Solution

P1, P2 represent the two pairs drawn.

Scenario1: {P1: RR, P2: BB}

Scenario2: {P1: RR, P2: RR}

Scenario3: {P1: BB, P2: RR}

Scenario4: {P1: BB, P2: BB}

$$0.5 = P(P2=BB|P1=RR) + P(P2=RR|P1=RR) + P(P2=RR|P1=BB) + P(P2=BB|P1=BB)$$ $\implies$ $$0.5 = 2*\big(P(P2=BB|P1=RR) + P(P2=RR|P1=RR) \big)$$ $\implies$ $$\frac{1}{4} = \frac{n\choose 2}{{2n-2}\choose 2} + \frac{{n-2}\choose 2}{{2n-2}\choose 2}$$ $\implies$ $2n^2-7n+9 =0$

I get no solution for $n$

• You have to check the special cases where $n-2<2$. E.g. $n=3$ is a trivial solution. Your argument shows that there are no other solutions. Sep 20, 2015 at 12:11
• I meant n=3, as in 3 red balls and 3 black ones. Sep 20, 2015 at 18:22
• But, $n=3$ is not a solution for the corrected version of equation. OP's equation ignores Scenario3,4 Sep 20, 2015 at 19:27
• I guess the way that you would solve this problem is: 1) Derive the solution OP has derived 2) Verify that it doesn't work for all cases, e.g. $n=2$ and $n=3$. Try it for the trivial cases and see that you get a solution. Your argument serves the purpose of excluding other solutions. I am not sure if (and how) you would create a general solution that covers all cases by using combinations in your argument, though it may well be possible. Sep 20, 2015 at 19:38

You miss that for n<4 this: $$0.5 = p\,(2\,R\,|after\,Red\,pair)+p\,(2\,B\,|after\,Red\,pair)= \frac{{n-2\choose 2}}{{2n-2\choose2}} + \frac{{n\choose 2}}{{2n-2\choose2}}$$ does not hold.

First, for $n=1$ the problem makes no sense, because you cannot take 2 balls and then 2 more balls.

For $n=2$ or $n=3$, $p\,(2\,R\,|after\,Red\,pair)=0$ rather than $\frac{{n-2\choose 2}}{{2n-2\choose2}}$, so your formula should be

$$0.5 = p\,(2\,R\,|after\,Red\,pair)+p\,(2\,B\,|after\,Red\,pair)=\frac{{n\choose 2}}{{2n-2\choose2}}$$

which holds for n=3 but not for n=2.

This may not be formal but I figured: \begin{align} \frac{(n-2)(n-3)}{(2n-2)(2n-3)} + \frac{(n)(n-1)}{(2n-2)(2n-3)} &= .50 \\[10pt] \frac{(n-2)(n-3)+(n)(n-1)}{(2n-2)(2n-3)} &= .50 \end{align} Solving for $n$, QED $n=3$.

So it starts with 3 of each color, i.e., 6 balls total.

• the OP might benefit more from your answer if you explain the steps Sep 20, 2015 at 17:27
• I believe this is the motivation: Assume 1st 2 balls red. Prob of the next two being red is the expression to the left of the + sign. Prob of the next two being black is the expression to the right of the + sign. Sep 26, 2015 at 2:57

Let $n$ be the number of balls. Then solve: $$\left[\frac{\frac n 2 }{n-2} \right] \left[\frac{\frac n 2 - 1}{n-3} \right] = \frac 1 2$$