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?

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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.

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…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.

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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.

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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.

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