The complete question is

Two bags are to be put altogether with $5$ red and $7$ white balls, neither bags being empty. How must one divide the balls as to give a person who draws one ball from either bag the least chance of drawing a red ball?

Let $a$ be number of red balls and $b$ be number of white balls in in bag 1. Similarly, $c$ and $d$ are the red and white balls in bag 2. I need to minimise

$$p = 0.5 (\frac{a}{a+b} + \frac{c}{c+d})$$

$0.5$ being the probability of picking one of the two bags. Beyond this point I am stuck. More related to optimization than probability I guess. Can anyone explain the approach to this problem?

Thanks in advance


  • $\begingroup$ Hint: $a+c = 5, b+d =7$ so $2p = \frac{a}{a+b}+\frac{5-a}{12-a-b}$ $\endgroup$ Mar 10, 2016 at 4:54
  • $\begingroup$ @GrahamKemp: I don't quite understand your hint. Certainly, one should make those substitutions to reduce it to a problem of two variables, but where does one proceed? Are you suggesting a multivariable optimization from here, or some sort of strategy by inspection? $\endgroup$ Mar 10, 2016 at 5:17

3 Answers 3


Let's see if we can answer this without any strenuous calculation. You correctly noted that the probability is given by

$$p = .5\left(\frac{a}{a+b} + \frac{c}{c+d}\right)$$

We can see that it suffices to minimize the expression inside. Well, we know that neither of the two bags are empty. This means that $1 \leqslant (a+b), (c+d) \leqslant 11$. In particular, we have

$$\frac{a}{a+b} + \frac{c}{c+d} \geqslant \frac{a}{11}+\frac{c}{11} = \frac{a+c}{11} = \frac{5}{11}$$

This gives us a lower bound on the probability, so if we can find a configuration which achieves this minimum, we're done. But this is easy to do; simply put $a = 5, c = 0, b = 6$ and $d = 1$; in other words, segregate all of the red balls and all but one white ball into one bag, and put one white ball into the remaining bag so that it is not empty.

  • $\begingroup$ @ Alex, Even logically your solution makes sense. Both bags have equal probability of being picked up. So make probability equal to zero in one and as low as possible in the other. Can that be argued? $\endgroup$
    – Bootstrap
    Mar 10, 2016 at 6:35
  • $\begingroup$ @Bootstrap: yep, that sounds about right! I think my explanation is merely a formal argument for the heuristic you just made. It's also expressed nicely in another answer below. Cheers! $\endgroup$ Mar 10, 2016 at 18:07

I remember a puzzle that went something like this: There are $500$ gold coins and $500$ silver coins. They are to be distributed to two bags, and you are allowed to choose one bag at random, and pick one coin. How will you maximize the chance of getting a gold coin ?

The fairly obvious answer was $499$ gold + $500$ silver in one, and $1$ gold in the other, almost raising P(gold) to $\frac34$

Here you are to minimize P(red) which is the same as maximize P(white), and the same logic as in the puzzle would divide as $5$ red +$6$ white in one, and $1$ white in the other.

  • $\begingroup$ Thanks for the answer anil. Learning much faster here.I guess learning does in happen in groups. $\endgroup$
    – Bootstrap
    Mar 11, 2016 at 8:30
  • $\begingroup$ You're welcome ! $\endgroup$ Mar 11, 2016 at 8:47


let bag 1 be $B_1$, we have$$B_1 = P(red) = \frac{a}{b}$$

let bag 2 be $B_2$, we have$$B_2 = P(red) = \frac{5-a}{12-b}$$

then, $$P = 0.5 (\frac{a}{b} + \frac{5-a}{12-b})$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .