A probability problem by Ofiu Sasad

This problem looks a bit difficult at first, but after breaking it down into simple pieces it becomes much more digestible. As is often the case with probability problems, this problem can be broken down into two parts: Counting total possibilities, and counting successful possibilities.

• Part 1: Counting total possibilities

This is simple. We are choosing five distinct numbers between 1 and 20 inclusive. The number of ways to do so is just $\dbinom{20}{5}$ . (We will simplify this later, after solving the other part.)

• Part 2: Counting successful possibilities

This is much less simple. Since often times, there will be multiple pairs of consecutive numbers, direct counting will likely run into the difficulties of PIE. This indicates that we should try complementary counting.

After a bit of consideration, it should seem natural to conjecture that the number of such sets of five is $\dbinom{16}{5}$ . The justification below, while not the first justification I found, is the most elegant and intuitive justification:

Consider choosing five distinct numbers, $a < b < c < d < e$ , between 1 and 16 inclusive. Then we can turn this into five distinct numbers between 1 and 20 inclusive, with no two consecutive, simply by taking $a,b+1,c+2,d+3,e+4$ . These five numbers all clearly differ by more than one (e.g. $b > a \implies b-a > 0 \implies b+1-a > 1$ ), so this is a set with no two consecutive numbers. In addition, since $e \le 16$ , $e+4 \le 20$ , so the numbers do fall between 1 and 20. We can similarly reverse the process, taking $a' < b' < c' < d' < e'$ between 1 and 20 with no two consecutive and turning it into $a',b'-1,c'-2,d'-3,e'-4$ . There is thus a bijection between sets of five numbers from 1 to 16, and sets of five numbers with no two consecutive from 1 to 20.

Returning to the problem at hand, we see that this means that the number of successful possibilities is $\dbinom{20}{5}-\dbinom{16}{5}$ , so the probability is $\dfrac{\binom{20}{5}-\binom{16}{5}}{\binom{20}{5}} = 1 - \dfrac{16*15*14*13*12}{20*19*18*17*16} = 1 - \dfrac{91}{323} = \boxed{\dfrac{232}{323}}$ . (I skipped a few simple computation steps, but they are all straightforward and doable with a calculator)