Six shoes

Probability Level 3

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?

Please provide your answer to three decimal places.

The answer is 0.111.

2 solutions

Geoff Pilling
Sep 20, 2018

There are six ways they could have grabbed their left shoes. Of these, only for two of them do none of them end up with the same left shoe they started with. Therefore, there is a $\frac{1}{3}$ chance that they each are wearing a different left shoe.

Similarly for their right shoe, so when all is said and done, the probably that none of them have either of the same shoes they started with is:

$\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9} =~ \boxed{0.111}$

Sorry for the misreporting, (the notice of your response came after I had deleted the report, so I never saw what you said, which is probably a good thing as I was apparently temporarily occupying a universe in which 3! = 3 :/). Anyway, in an attempt to redeem myself, as this is a double-derangement problem the general solution for $n$ women is $\left( \dfrac{D_n}{n!} \right)^{2}$ , which of course goes to $\dfrac{1}{e^{2}} \approx 0.135$ as $n \to \infty$ .

Brian Charlesworth - 2 years, 8 months ago

Skullslide Gaming
Sep 25, 2018

2/3 "chance to grab another left shoe than yours" * 2/3 "chance to grab another right shoe than yours" * 1/2 "chance for the second person to grab another left shoe. Had 3 at the beginning but the first person grabbed one left shoe" * 1/2 "chance for the second person to grab another right shoe. Had 3 at the beginning but the first person grabbed one shoe" * 1/1 * 1/1 "for the third person.

2/3 2/3 1/2*1/2=1/9

Six shoes

The answer is 0.111.

2 solutions

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