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I'm reading Thinking fast and slow and there's this problem I cannot get a correct answer. Here it is:

Half the marbles are red, half are white. Next imagine a very patient person (or a robot) who blindly draws 4 marbles from the urn, and does it all again, many times. If you summarize the results, you will find that the outcome "2 red, 2 white" occurs (almost exactly) 6 times as often as the outcome "4 red" or "4 white".

Could you please explain and show why it's 6 times?

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    $\begingroup$ en.wikipedia.org/wiki/Binomial_distribution $\endgroup$
    – Adrian
    Commented Nov 9, 2014 at 22:05
  • $\begingroup$ @Adrian binomial is for sampling with replacement. It would give an approximate calculation for very large numbers of both kinds of marble in the urn ... but how many marbles are we talking about? $\endgroup$
    – Glen_b
    Commented Nov 9, 2014 at 22:17
  • $\begingroup$ Ah, I found a copy of the problem ... which starts with "Imagine a large urn filled with marbles..." ... user60343, that information is important to answering the question. Is it in your copy? $\endgroup$
    – Glen_b
    Commented Nov 9, 2014 at 22:23
  • $\begingroup$ Doesn't say. But it may be a problem when marbles are returned to the urn, although the the problem implies otherwise, if so I can solve it. Nevertheless, I'm not sure about the if I am right about what the problem implies. Yes, @Glen_b $\endgroup$
    – user60343
    Commented Nov 9, 2014 at 22:30
  • $\begingroup$ Here's an article with Kahneman discussing the same problem. It contains the additional clarification I mentioned. $\endgroup$
    – Glen_b
    Commented Nov 9, 2014 at 22:47

1 Answer 1

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Imagine there are $r$ red and $w$ white marbles in the urn. Let $\text{R}$ be the event "a red marble is drawn" and $\text{W}$ be the event "a white marble is drawn". 4 balls are drawn, without replacement between the draws.

Then $P(\text{RRRR}) = \frac{r}{r+w}\cdot \frac{r-1}{r+w-1}\cdot \frac{r-2}{r+w-2}\cdot \frac{r-3}{r+w-3}$.

However, if $r$ and $w$ are both very large (much, much larger than 4 ... perhaps something more like 400 each, say), then each of those fractions is very close to $\frac{1}{2}$, and the product is about $\frac{1}{16}$.

By symmetry, the calculation for $\text{WWWW}$ is similar.

Indeed, so would it be for WWWR, or RWWR or any other specific order of R and W over 4 marbles.

e.g. $P(RRRW)=\frac{r}{r+w}\cdot \frac{r-1}{r+w-1}\cdot \frac{r-2}{r+w-2}\cdot \frac{w}{r+w-3}$.

But $P(\text{two R, two W, in any order}) = P(RRWW)+P(RWRW)+ P(RWWR)\\ \hspace{6.5cm}+P(WRRW)+P(WRWR)+P(WWRR)$

When $r$ and $w$ are large, each of those 6 arrangements has about the same chance as $P(RRRR)$ and $P(WWWW)$, close to $\frac{1}{16}$.

So - roughly speaking, because there are lots of both kinds, in equal numbers, and there are 6 ways of arranging two red and two white marbles in order, there's about six times the chance of observing two red and two white (about 6/16) in any order as four red (about 1/16) or four white (about 1/16).

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  • $\begingroup$ That seems fine, thank you! I still have one more question which might have caused a huge error in the way I was thinking. It's a word "or" in "4 whites" or "4 reds". Doesn't it mean that the probability of these two have to be summed up? As a result of both of them possible? $\endgroup$
    – user60343
    Commented Nov 9, 2014 at 22:49
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    $\begingroup$ The quotation should be read as "... you will find that the outcome "2 red, 2 white" occurs (almost exactly) 6 times as often as the outcome "4 red" occurs or the outcome "4 white" occurs." $\endgroup$
    – whuber ♦
    Commented Nov 10, 2014 at 0:13

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