# What is the probability that when they leave that none of them is wearing either of their own shoes?

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Three ladies get together for a bit of fun. When they arrive they all kick off their shoes. Unfortunately, when they leave they are not quite in the same frame of mind so they each grab a left shoe and a right shoe at random, and put them on.
What is the probability that when they leave that none of them is wearing either of their own shoes?

posted Sep 30, 2018

If the shoes are named
L1 R1 ==> Women 1 (W1)
L2 R2 ==> Women 2 (W2)
L3 R3 ==> Women 3 (W3)

Now all the possible favourable events for the given probability is
[L2 R2 (W1)]×[L3 R3 (W2)]×[L1 R1 (W3)] = 1/9 × 1 × 1/4 = 1/36
[L3 R3 (W1)]×[L1 R1 (W2)]×[L2 R2 (W3)] = 1/9 × 1/4 × 1 = 1/36
[L2 R3 (W1)]×[L3 R1 (W2)]×[L1 R2 (W3)] = 1/9 × 1/2 × 1/2 = 1/36
[L3 R2 (W1)]×[L1 R3 (W2)]×[L2 R1 (W3)] = 1/9 × 1/2 × 1/2 = 1/36

Required Probability = 1/36 + 1/36 + 1/36 + 1/36 = 1/9.

1/9
.
Let's look at the left foot first.
The chance that none has their own shoe is 2/6.
1/2/3 - at least 1 correct
1/3/2 - at least 1 correct
2/1/3 - at least 1 correct
2/3/1 - all wrong
3/1/2 - all wrong
3/2/1 - at least 1 correct
.
The same holds for the right shoe.
Thus the change that none has either shoe correct is 2/6 * 2/6 = 4/36 = 1/9

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