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Suppose there are N male and N female students. They are randomly distributed into k groups.

Is it more probable for a male student to find himself in a group with more guys and for female student to find herself in a group with more girls?

The question is motivated by an argument with my mother. She claimed that in the majority of collectives where she was, the number of women besides her was greater than men, while I had the opposite impression, that in most collectives where I was a member there was more men even if not to count myself.

I never was in a majority-girls collective, so I think for a male it is more probable to find oneself in a majority-male collective (even if we exclude oneself).

Anixx
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  • My gut feeling is "no" because after picking one male student, you're choosing from N-1 male students and N female students. Not much difference, but it's there. – El'endia Starman Apr 27 '12 at 06:06
  • @El'endia Starman is not the majority of guys should be in the groups which have more guys and the majority of girls - in the groups with more girls? – Anixx Apr 27 '12 at 06:11
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    Suppose there are three kinds of groups: those with more guys, those with more girls and those which have equal number. In the groups with equal number the number is equal, but among other groups, evidently those with majority of guys have more guys than those with the majority of girls. – Anixx Apr 27 '12 at 06:14
  • Should not the groups with more guys have greater total number of guys than the groups with more girls? – Anixx Apr 27 '12 at 06:18
  • for 2 boys and 2 girls, we can have the following distribution (bb,gg) (bg,bg)(bgg,b)(bggb)(bg,b,g)(b,b,g,g)(bb,g,g)(gg,b,b) so in this case only in 2 conformations i.e (bb,gg) and(bb,g,g) boys outnumber girls, hence the answer should be no – Tomarinator Apr 27 '12 at 06:41
  • I meant the groups are equal in number. – Anixx Apr 27 '12 at 06:46
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    Also I think these groups should be counted several times: (b1b2,g1g2), (b2b1,g1g2), (b1b2,g2g1) etc, which gives more combinations than for a distribution with smaller groups. – Anixx Apr 27 '12 at 06:50
  • but the order of arranging them isnt our concern, so (b1b2,g1g2) are (b2b1,g1g2) both same. and also , in the question are all boys identical? or all girls identical? – Tomarinator Apr 27 '12 at 06:59
  • Arranging is not our concern but we should ascribe different weights to each case. Anyway I was asking about a situation with even groups. – Anixx Apr 27 '12 at 07:22

2 Answers2

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The expected number of males in a randomly-chosen group is exactly half the size of the group.

If a group contains at least one male, then the expected number of males will be slightly larger than half the size (and therefore larger than the expected number of females), so the answer is YES.

There is a bit of a difficulty with the phrasing of the question. For one thing, "randomly distributed" may mean that each group contains exactly $\frac{2 N}{k}$ students, or it may mean that each student chooses a group independently (giving groups of different sizes, some of them perhaps empty). My answer applies to either case. Also, it is not clear whether you are asking if it is more probable for a male student to find himself in a group with more guys than to find himself in a group with more girls, or if you are asking if it is more probable for a male student to find himself in a group with more guys than to find himself in a group with at least as many girls as guys. I am assuming the former interpretation, otherwise (supposing the groups are constrained to be of equal size) if $k = N$, then a male student will find himself partnered with another male with probability $\frac{N-1}{2 N} \lt \frac{1}{2}$, and the answer in this case would be NO.

Dan Brumleve
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  • Well consider the following: suppose there are no groups with equal number of students of both sexes. Each group has either the number of guys or the number of girls exceeding the other by 10. Will the probability to find oneself in a group with greater number of the opposite sex very small compared the the probability to find oneself in a group with the same sex? – Anixx Apr 27 '12 at 06:44
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This problem also gives the answer why you are "always" in the longer queue at the supermarket.

If $k=1$ the answer is trivial: All groups are gender balanced. Therefore we shall assume that $k>1$.

Assume Samuel and Samantha were ill the day the groups were originally formed. If the two Sams are assigned to the groups randomly, the result is just as if there had been no illness.

If Sam and Sam are asigned to the same group (which happens with probability $\frac1k$ if the distribution is uniform, with other distributions, your mileage may vary), we see by symmetry that more guys is exactly as probable as more gals for this group.

If Sam is assigned to a different group than Sam (which happens with probability $1-\frac1k>0$ in the uniform case, but we actually need only assume that the probability is $>0$), then three cases are possible:

  • The group was gender-balanced before with some probability $p$, say (clearly, $p>0$, though the exact value depends on the group size).

  • Or the group had more male members.

  • Or the group had more female members.

By symmetry, the probabilities for the latter two events are equal, hence are $\frac{1- p}2$ each. Then the probability that the group including Sam has more people of the same gender as Sam is at least $p+\frac{1- p}2=\frac{1+ p}2>\frac12$.

In total, it is more probable to find oneself in a group with more people of ones own gender than with less. This holds both for Sam=Samual and Sam=Samantha.

Hagen von Eitzen
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  • Yes but it only evident from your answer that you're more likely to find yourself in a group with slightly more persons of your sex (that is by 1). But I wonder whether the difference is more sufficient if the distribution is far from uniform. – Anixx Oct 27 '12 at 15:39
  • Which distribution is far from uniform? I only assume that the assignment is done in a symmetric fashion (so that the adding of late group members liek Sam&Sam is possible without changing the distribution). The exact value of $p$ above may depend on the exact distribution. Even if the assignment is done by trying to balance the groups, as soon as there are odd-sized groups, the majority in that group will be larger than the minority. – Hagen von Eitzen Oct 27 '12 at 16:21
  • Maybe this is what you are after: In a random group of size $n$, we expect about $\frac n2$ boys/girls. But the typical deviation is $\approx \frac12\sqrt n$, thus about $\frac {n+\sqrt n}2$ people will be of the majority sex and $\frac {n-\sqrt n}2$ peole will be in the minority. – Hagen von Eitzen Oct 27 '12 at 16:28
  • yes. Can you elaborate on this, that is take into account the typical deviation? – Anixx Oct 28 '12 at 07:49