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An elementary school admits students on a basis of a random lottery. There are 50 slots available, and 100 families (students) have applied. Names will be drawn from a hat. Seems pretty straightforward that an individual student has a 1 in 2 chance of getting a slot or 50%. However, the school has a policy that if twins apply, BOTH names are put in the hat, and if and EITHER twin is picked then BOTH twins are accepted. Although each twin has the a same probability of having his/her name picked from the hat as anyone else, intuitively families with twins seem to have an enormous advantage, but I can't figure how to quantify it. This might be a simple problem, and I would think it should be easy to find on the internet, but I'm not getting anywhere with it.

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    $\begingroup$ It would be much fairer, and still accomplish the same social goal, if for twins one name were put in the hat (but if that name is drawn both kids get to go). $\endgroup$ Nov 17 '15 at 6:06
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    $\begingroup$ What happens if one of the twins is drawn on the 50th draw; that is after 49 slots are filled? $\endgroup$
    – DJohnM
    Nov 17 '15 at 6:34
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    $\begingroup$ is the "50 slots" firm in the face of "BOTH twins are accepted"? $\endgroup$
    – DJohnM
    Nov 17 '15 at 7:08
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We make a simplified model. Among the $100$ applicants, there is one pair of twins, named Alpha and Beta (the parents are mathematicians, and Alpha was born first).

The $100$ names are thrown in a hat, and $50$ names are drawn. If one of Alpha or Beta is drawn, but not the other, then one of the $49$ non-twin "winners" is bounced.

What is the probability that neither Alpha's name nor Beta's is chosen? There are $\binom{100}{50}$ equally likely ways to choose $50$ names. There are $\binom{98}{50}$ ways to choose $50$ names none of which is Alpha or Beta. So the probability the twins strike out is $$\frac{\binom{98}{50}}{\binom{100}{50}}.$$ This simplifies to $\frac{49}{198}$, close to $1/4$.

More informally, each name has probability $\frac{1}{2}$ of not being drawn. But the probability neither name is drawn is a little bit smaller than $\left(\frac{1}{2}\right)^2$, because of (mild) dependencies.

So Alpha has probability about $3/4$ of being admitted to the school.

Remark: The addition of a small number of additional twin pairs to the applicant mix does not change the analysis at all: Alpha, and any member of a pair of twins, still has probability close to $3/4$ of getting in, as long as the kids bounced because of application of the twin rule are non-twins.

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I think that symmetry can be used to get an answer here.

Suppose you do the draw as described, settle the edge cases, so that you have divided the 100 students into two lists of fifty names each, with the two twins in the same list.

Then you flip a (fair) coin to see which list is the rejects. Admit the others.

EDIT:

This solution(?) assumes a certain methodology:

Fifty names are chosen at random.

If this list satisfies the "twins together" requirement, the list is used.

If the twins are not together, the list is rejected, and a new set of fifty names is selected.

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