Brainteasers & Puzzles

Three Children, One Apple

Two tosses give four equally likely pairs. Assign HH, HT, and TH to the three children and re-toss on TT, so conditional on stopping each child wins with probability exactly 1/3 and the number of rounds is geometric with mean 4/3.

solvedeasy1 min

You have three children but only one apple, and you want a fair coin to decide who gets it, with each child equally likely. A single coin gives only two outcomes, so how do you split the apple three ways?

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#Make four equal outcomes

One toss gives two outcomes, but a pair of tosses gives four equally likely ones, HH, HT, TH, and TT. Hand three of them to the three children, one apiece, and set the fourth aside.

#Discard and re-toss

When TT comes up, ignore it and toss the pair again. Conditioned on stopping, each of HH, HT, and TH is equally likely, so every child receives the apple with probability

P(a given child wins)=13.(1)\PP(\text{a given child wins}) = \frac{1}{3}. \tag{1}
HHchild 1HTchild 2THchild 3TTre-toss
Two tosses give four equally likely outcomes. Hand HH, HT, and TH to the three children and throw the pair again on TT. Conditioned on stopping, each child wins with probability one third.

#A biased coin for two

The same discard trick even tames a biased coin. For two children, toss the coin in pairs and read HT for one child and TH for the other, throwing away HH and TT. Both mixed pairs carry the same probability p(1−p)p(1-p) whatever the bias, so each child wins exactly half the time.

#Thirds from halves

Toss twice, give HH, HT, and TH to the three children, and re-toss on TT. The discard restores perfect thirds from a coin that can only ever speak in halves.