Saturday, 24 January 2015

Two balls, one urn

Suppose I filled an urn with 100 balls and, having not let my friend see this process, ask her to take out the balls one by one - at random.  After 99 balls, I ask her to stop.  Now, suppose further that she had taken out 99 white balls.  I ask her what the likelihood is that the ball still in the urn is non-white.

It's possible that my friend will say it's pretty unlikely, after all she's just taken 99 balls out at random and they were all white.  She's sort of right, the likelihood is remote at 1%, but it's possible that she might think it's even more improbable.  She'd be wrong though, wouldn't she?

One way we could think of it is "what is the likelihood of my friend picking the only non-white ball in the 100th position, rather than one of the other positions?"  Well, there are 100 positions in a random sequence of 100 balls in which a lone non-white ball could lie, so the likelihood of it lying in the 100th position is the same as it lying in the first, or any other position namely 1/100. Therefore, either the last ball still in the urn is non-white (at a probability of 1/100), or it's white just like all the others.

This logic should work the same no matter how I fill the urn and no matter how many balls I put in the urn.  If there are a million balls, the chances of the last ball being the only non-white one after pulling out 999,999 white balls is one in a million.  If there are two balls, then the chances of the second ball being non-white after the first being white is 50%.

But wait a minute, let's look at my process for filling an urn with two balls.

Let's say that my selection process involves a million different balls from which I make my selection, only of which is white, the rest having different colour and pattern combinations.  I select one ball completely at random and put it in the urn.  The ball I removed is replaced by an impartial third party with an identical one, and then I make another entirely random selection and put it in the urn.

What happens to the probability then?  Say that my friend pulls out a ball, perfectly at random, and it is white.  What are the chances that the one left in the urn will be white?

I encourage people to think carefully about this little conundrum, provide an answer and give a brief explanation.


  1. From your friend's point of view, the logic you provide in the above paragraph seems to be the best she has to work with, so a 50% chance that the remaining ball is white would be a safe bet. But from our point of view, we know that the chance of her drawing another white ball is 1 in a million or less. At best, there was a 2 in a million chance that either ball in the urn would be white (as you chose twice with a 1 in a million chance each time). There is a 1 in 10^11 chance of both balls being white (1/1M * 1/1M). But we know that despite all odds, one of the balls was in fact white, as this is the ball your friend drew. The chance that the remaining ball is white then could be seen as 1/1M (the chance of any single ball being white) or even less (closer to the 1/(10^11) chance of both balls being white). That's all I got.


    1. I've tried to simplify things in a revision to the scenario, B. Perhaps it won't be as confusing - or rather, only confusing because the answer is counterintuitive.


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