# Picking teams for a game

This was a question that came up at work, and we ended up with several different answers!

In the game, there are two teams of four, and each person gets a randomly assigned number from one to eight. This is done by picking the numbers out of a bag.

The question is: if the order of people within the team doesn't matter - that is, if the team draws 1,2,5,8 or 8,2,5,1 it's the same - then what are the odds of either team drawing 1,2,3,4 or 5,6,7,8?

My answer was that if one team gets either of those groups, the other team will always have the other group, so the fact that there are two teams doesn't matter. Thus, it comes down to: what's the chance of a team picking [1,2,3,4] or [5,6,7,8], and I make that 1/35.

Am I correct? Does it make any difference how the numbers are drawn - for example, if one team draws four and then the other does, or if they draw alternately?

• You are correct that it's 1/35. However the real question is what combinations humans think of as "un-random". If one team drew 2,3,4,5, would you still be here asking the same question? What if one team drew 1,3,5,7? Oct 11, 2013 at 13:40
• The reason for the significance is that the players stand in a line with the positions given by their numbers. So the original question started as: "what's the chance that you'll end up with team A and team B standing AAAABBBB or BBBBAAAA?" Oct 11, 2013 at 13:44
• You are correct that the order of draws doesn't matter. Any process that randomly picks four out of the eight will give $\frac 1{35}$ Oct 11, 2013 at 14:09

There are ${8 \choose 4} = 70$ ways of selecting teams.

Only two of those selections are of the form 1234|5678. Therefore

$$P = \frac{2}{70} = \frac{1}{35}$$

Alteratively

There are $\frac{8 \choose 4}{2} = 35$ ways of selecting teams as Team 1 is indistinct from Team 2 and creates symmetry.

There is now only one configuration that is of the form 1234|5678.

$$P = \frac{1}{35}$$

• More structural than my answer (+1). Oct 12, 2013 at 7:06

Let team 1 start with drawing 4 numbers. Some number is drawn by the first drawing person. Let's say he draws 4. The second one must draw 1,2 or 3 and the probability that that occurs is $\frac{3}{7}$. Going on like that you get a probability of $\frac{3}{7}\times\frac{2}{6}\times\frac{1}{5}$ that you end up with [1,2,3,4] and [5,6,7,8].

• There are $7$ numbers left after the first chose his ;) Oct 11, 2013 at 14:07
• Which agrees with the $\frac 1{35}$ that OP got. Oct 11, 2013 at 14:08

Number of ways of forming the team such that one team gets the numbers 1, 2,3 ,4 is the number of permutations of the numbers 1 to 8, in which the numbers 1 to 4 occupy the first four positions or the last four positions. This is 2.4!.4! .

Number of possible ways of forming the teams is the number of permutations of the numbers 1 to 8, which is 8!.

Hence, probability that one team gets the numbers 1,2,3,4 is 2.4!.4!/8!. This is 1/35.

I think it matters whether teams pick alternately or not. If they do then aren't the odds

TeamA (4/8 * 3/6 * 2/4 * 1/2) * TeamB(4/7 * 3/5 * 2/3 * 1) = 1/105

• I think your arithmetic is wrong here. This evaluates to 1/70. The reason it's half of 1/35 is that your calculation is for the probability that Team A pick [1,2,3,4] and Team B pick [5,6,7,8], whereas the reverse is also allowed, giving you 1/35. Oct 14, 2013 at 8:31