In the current season of reality TV show “Big Brother”, the casts features all returning players and the season is dubbed “All Stars”. At present, two particular players are the vilified target by the larger alliance in the house. For a Head of House competition, the players are asked to make three attempts at performing a task and the winner is the player who scored highest (in the least time in case of a tie). The players were randomly ordered but the twp particular players happened ot land in the final two positions. Some conspiracitoral notions abound about how these two players just happened to be last picked.

     On the onset, the question makes for a short mathematics exercise. Reframing into more mathematical problem form, suppose there are thriteen contestants who are each invited to perform a task. The order in which the contestants are to play was random. What are the chances that any particular pair of players are randomly selected last?

     In (fair) coin flipping, there are is only one outcome, either the coin lands on Heads or lands on Tails. There are two possible outcomes, either Heads of Tails, but only one outcome will happen. The chances of landing either Heads or Tails is 1 out of 2. 1 chance for (either) Heads or Tails out of the two possibilities of Heads and Tails.

     For the roll of a (fair) 6-sided die, there a 1 in 6 chance that the die lands on any particular number, e.g. the 3. There is a 1 out of 2 chance of the die landing on an even number, 2, 4, or 6 because there are six numbers 1, 2, 3, 4, 5, & 6 but only three of these numbers are even, 2, 4, & 6 making 3 chances out of 6 3/6 which reduces to ½.

     So, in math, the probability of someting occurring out of the total number of possible outcome is represented by the fraction: 

     Second, there is a notion in mathematics of counting the number of ways a number of objects can be arranged.

     Take an apple, an orange, and a banana. There are six possilbe ways we can arrange them.

apple, orange, banana

apple, banana, orange

orange, apple, banana

orange, banana, apple

banana, apple, orange

banana, orange, apple

      Counting the number of ways arranging things is called a permutation. (Not to be confused with combination, a similar concept but the difference is order matters.) We care about the order in which things are arranged, in particular two things must be positioned last.

      Let us look again at three things being permuted. To make things easier, let us work with numbers. Let’s suppose we have three contestants, 1, 2, and 3. Now, let’s ask what is the possibility if randomly arranged, that contestant #3 is selected last. Well, following our previous example with the apple, orange, and the banana, we have

1, 2, 3

1, 3, 2

2, 1, 3

2, 3, 1

3, 1, 2

3, 2, 1

     We can see that 3 comes in last in two instances, 1, 2, 3 and 2, 1, 3.

     So, there are two possible chances for 3 to come last out of total of six arrangements, or permutations. Let me note here, we can represent the number of permutations of n things by the factorial (!), thus permutation on three things is 3!

     Therefore, the chances of 3 coming last out of our three contestants is 

     Now, our problem asks us to determine the probability that two players from thirteen are selected last. Well, 13! is a rather large number, to be precise, 13! = 6227020800. I do not know about you, but I have no desire to write out all 6227020800 possible permutations of 13 things, even if were to use numbers 1, 2, 3, 4 etc..

     Instead, I think there must be an easier way to derive at our answer.

     Let us think small, like we had done with the three contestants, let us do similar with say, four contestants.

     Given four contestants, 1, 2, 3, & 4, their possible permutations are 1234, . . . ., OK, I shall stop right here. 4! = 24 and although writing down 24 permutations is far easier than doing 6227020800 permutations, I am not inclined to list them all. Instead, I shall look at the particular cases that I am interested in and count only those. In other words, when does 3 & 4 come last in permutations of all four? Note: Here, for the sake of clarity and ease, let me add the lax condition of not caring whether 3 is before 4 or 4 coming before 3. In other words, xxx34 is treated the same as xxx43.

     That beng the case, need only count the instances which 3 & 4 come last. Those are 1, 2, 3, 4 or 2, 1, 3, 4 and 1, 2, 4, 3 or 2, 1, 4, 3. Four instances — occurrances out of what we know are 24 possible permutations. Therefore, our answer is 

     Ok, that seems fair.

     Now, let’s step up to there being five contestants. How many permutations on 5? 5! = 120. Again, even though 120 < 13!, I am not about to write all 120 possible permutations. But I can go with the idea as did before of noting those instances when 4 & 5 come last position in the lit of permutations, either xxx45 or xxx54. I shall stop right here and make an observation.

     We are interested in all cases in which 1, 2, 3, 4, & 5 are arranged while fixing 4 & 5 in last position. So in other words, want to count the permutaitons of the first three 1, 2, 3 — well, that we have done already and the answer is 3!. But we want to include regardless if 4 comig before 5 and 5 before 4, thus that adds to 3! x 2 = 12. Giving answer of 12 / 5!.

      From the previous smaller step, I think I can go out on a limb and suggest working toward the final answer as involving 11! becasue we are counting all the permutations of 13 contestants while fixing two in the final position. And since do not care which order the final two players come, that means two times the possibilities.

     Thus giving a final answer of 

     This may seem daunting. But observe, 13! = 11! x 12 x 13, so we can reduce the fraction 

     If I had done my arithemtic correct, then the answer is, there is a 1 out of 78 chance that two particular contestants are last when randomly selected out of thirteen players.

     Now, who says reality TV show cannot be educational?

With the exception of looking up the value for 13!, I never had to use a calculator. Damnit. I miss using a calculator.