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Sunday Times Teaser 2805 – Greek Urns

by Michael Fletcher

Published: 26 June 2016 (link)

I have three Greek urns. I took some balls (consisting of an odd number of red balls and some black balls) and I placed one or more balls in the first urn, one or more in the second, and the rest in the third. If you chose an urn at random and then a ball at random from that urn, then overall there would be a 50 per cent chance of getting a red ball.

Then I moved some black balls from one of the urns to another. In this new situation, if you chose an urn and then a ball there was a 75 per cent chance of getting a red. In fact, with this set of balls and urns it would be impossible to get a higher percentage than that.

How many red balls and how many black balls were there?

One Comment Leave one →
  1. Brian Gladman permalink

    Here is a manual solution.

    To make the probability of withdrawiing a red ball 3/4, \(r+2\) red balls and \(3r\) black balls must be placed in the three urns in the following way: red \((r,1,1)\) and black \((3r,0,0)\).

    When \(x\) black balls are moved from urn one to urn two, the distribution of black balls becomes: \((3r-x,x,0)\), making the probability of drawing a red ball: \[(r/(4r-x)+1/(1+x)+1/0))/3\] which must be 1/2. This gives the following quadratic for \(x\):\[x^2-(2r+1)x+6r=0\]with the solution:\[\frac{(2r+1\pm\sqrt{(2r-5)^2-24})}{2}\] For integer \(x\) the expression in the square root must be a perfect square (say \(c\)), giving: \[(2r-5)^2-c^2 = 24\]If \(f\) is a factor of 24, we have:\[\begin{array}{lcl}2r-5+c &=&24 \\2r-5-c&=&24/f\end{array}\]which gives the solutions for \(r\) and \(c\) in terms of \(f\) as:\[\begin{array}{lcl}r&=&(24/f+f+10)/4 \\c&=&(24/f-f)/2\end{array}\]Trying the factors of 24 gives only one solution: \(r=5\) and \(c=1\) with \(x=5\) or \(x=6\).

    Hence there are 7 red balls and 15 black balls.

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