Wednesday, May 1, 2013

Wednesday Brain Teaser 5-1-13

Did you know Lewis Carroll was a mathematician?  Here's one of his favorite ones:

On return from the battlefield, the regiment is badly battle-scarred.  If 70% of the soldiers have lost an eye, 75% have lost an ear, 85% have lost a leg and 80% have lost an arm, what percentage at least must have lost all four?


  1. At least 45% exhibit all four types of wounds.

    More accurately: the intersection of all four classes of wounding is in the range 45% ≤ x ≤ 70%.

    I vaguely remember that Lewis Carrol was a mathematician, but didn't know much more.

    1. ...and I was wrong. If we use the following,

      Set A: lost an eye (70%).
      Set B: lost an ear (75%).
      Set C: lost an arm (80%).
      Set D: lost a leg (85%).

      Then the smallest possible intersection of sets A and B is 45% of the regiment. (Thus the wrong answer I gave above. I got it by filling in a bar-chart of 100% from both ends. The zone in the middle, covered by both sets, is 45%...)

      The smallest possible intersection of sets A,B, and C is 25% of the regiment.

      The smallest possible intersection of sets A,B,C, and D is 10% of the regiment.

      Counter-intuitive that it can be so small.

  2. SJ - I also envisioned bar graphs from both ends. Interesting. I started with the two highest percentages, though - don't know why. The minimum 85-80 intersection (your set D and C) is 65. The intersection with 75% (set B), would be 40. The minimum intersection of that with 70% (set A) is 10%.

    There's another way: Minimizing the number of people with 4 wounds is the same as maximising the number of people with 3 wounds. A+B+C+D = 310. So after we have stuffed as many people as possible into the three wounds category, 100 people with 3 wounds each, there are still 10 left over - the unfortunates who got all four.

    It makes it less counterintuitive to me when I frame it as maximising the threes.

  3. I ALSO envisioned bar graphs from both ends. I knew, intuitively, that it must be wrong, because I was missing the other two sets, but I didn't know how to factor that in. AVI's second paragraph helped.