Cricket gloves

An interesting interview puzzle asked in trader round at Da Vinci Derivatives

Indian cricket team has 5 pairs of gloves that each have a distinct number 1−5. The 10 gloves are randomly paired up. Find the probability that the gloves are paired up such that the values of any pair differ by at most 1.

Try to solve this problem yourself before moving on to the solution below

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Solution

We are going to solve the more general case with n pairs of gloves labelled 1 to n. Note that there are (2n)!/(2^n * n!) ways to pair up the 2n gloves, as there are (2n)! total arrangements, n! ways to re-label the pairs, and then 2 ways to switch around the order in each pair.

Let q_{n} be the number of arrangements that satisfy our condition. Consider the two gloves labelled n, say n1 and n2. Decide the partner for n1 first. We either have that n2 is paired with n1, in which case, we go back to the same problem but with n-1 pairs of gloves instead of n. Otherwise, n1 is paired with one of the gloves labelled n-1, of which there are 2 ways n-1 to pick that glove. Afterwards, we know that n2 is paired with the other glove labelled and that becomes fixed. Then, this goes back to the same problem but with n-2 pairs of gloves instead. Therefore, we get the recurrence relation

\( q_{n}=q_{n-1}+2q_{n-2}\)

Remember that the 2 in front of q_{n-2} represents the fact that we have 2 options of the glove labelled (n-1) to match with (n1) in that sub-case. We now need some initial conditions. Note that (q_{1}=1) , as there is clearly only one pair. Furthermore, we have (q_{2}=3.) as we can pick the partner for any one of the gloves in 3 ways, and that fixes the other pair immediately. The characteristic equation of this recurrence relation is (r^2 -r-2=0,) of which the solutions are (r=2,-1.) Therefore,

\((q_{n}=c_{0}\cdot2^{n}+c_{1}\cdot(-1)^{n}) \)

Plugging in the initial conditions yields that (1=2*c_0 -c_1) and (3 = 4*c_0 +c_1) Solving these yields that (c_0=2/3) and

c_1= 1/3. Therefore,

\( q_{n}=\frac{2^{n+1}+(-1)^{n}}{3}\)

Therefore, the probability of this event occurring with n pairs is given by

\( p_{n}=\frac{q_{n}}{\frac{(2n)!}{2^{n}\cdot n!}}=\frac{(2^{n+1}+(-1)^{n})\cdot2^{n}\cdot n!}{3(2n)!} \)

Substituting in n=5, we get that p_5 = 1/45

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😃 Over to you: Were you able to solve this?

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