Marilyn Vos Savant (self-proclaimed owner of the "world's highest IQ", as confirmed by Guiness Book of World Records) in the May 5, 2002 issue of Parade Magazine addresses the following question from a reader (colored and underlined words are my doing, and are not in the original.):
 

Reader's Question:     "...inexpensive tickets for a hit play are distributed before every show by lottery.  In the morning, 65 numbered wristbands are offered at the theater on a first-come, first-served basis.  Two hours before showtime, 20 of the numbers are chosen at random.  The people with those wristbands may purchase the tickets.  If my daughter gets a wristband, the chances of her number being chosen are 20 out of 65.     Here's what we can't figure out:  If she and I each get a wristband, what are the chances that at least one of us will have our number called?  In theory, because we each have a 20/65 chance, you'd think our collective chances would go up to 40/65, but that can't be right.  Help!" [End of question.]

    Marilyn gave the reply listed below, which is fairly obviously (to a math teacher) incorrect.  Marilyn is very smart, and usually limits herself to mathematically related questions that are amenable to careful, though relatively basic, logical analysis--she is very good at logic--but that don't require more complicated, specialized quantitative mathematical processes.  (I don't know how much actual math training she has.)  Unfortunately, even simple-sounding probability problems are notoriously tricky to conceptualize and formulate correctly, and often require deceptively complex and mathematically involved reasoning.
    Marilyn has every reason to be confident about her ability to handle such problems, as she successfully handled the (in)famous "Monty Hall" 3-door problem several years ago, which does lend itself to basic logical analysis, while being potentially deceptive to even (some) mathematicians, many of whom publicly claimed she was wrong about it (which she was not!)  But for the problem above I think she suffers from a tad of overconfidence that a simple approach will work.  Of course it's possible that she's correct and that I'm the overconfident one!
 

Marilyn's Answer:     "Here's one way to look at it:  If one person has a 20/65 chance alone, then two people together (such as you and your daughter) have a 20/64 collective chance of winning at least one ticket opportunity.  Three people would have a 20/63 chance.     Stretch this point to consider the case of your daughter and 45 more people.  With 46 people together in this group, they would have a 20/20 collective chance (100%) of winning at least one ticket opportunity among them.  That's because 65 people have wristbands, 20 of those people will win, and 45 of them will go without a ticket opportunity.  So with any group of 46 people, at least one will win for sure."  [End of answer.]

    This answer struck me immediately as weird, and intuitively wrong.  By counting the two people together as "one" person, thus lowering the number of "total entrants" to 64 instead of 65, she ignores various detailed ways in which either person or both could win when their separate numbers are drawn.  The reader's first claim (in red in her question above, and with which Marilyn agreed in her answer) about having a single wristband is easily seen to be correct, and Marilyn's last claim (in green in her answer above) about 46 wristbands is also obviously correct.  But the logic and math in the claims about 2 wristbands or 3 wristbands, and in fact, for all numbers of wristbands from 2 to 45, is in error.
    After some scratchwork, I sent an e-mail to Marilyn, a slightly edited version of which follows.  (Everything below up to the purple "In fact" can be understood by a non-mathematically trained reader, while the parts that follow the "In fact" are for the mathematically inclined!)
 

My e-mail:         "Hi, Marilyn,         Your answer to the question of the chances of 2 people getting tickets when 20 wristbands are chosen from 65 isn't correct.  To see this simply, imagine there were only 4 wristbands given out, and that 2 of them were chosen to receive tickets.  The chance of one person (me) winning--say I have wristband #2--would indeed be 2/4:   I will win if any of the following three combinations are chosen:  (2, 1) or (2, 3) or (2, 4).  And I will lose if (1, 3) or (1,4) or (3,4) are chosen. My chances of winning are thus 3 winning combo's out of 6 total, or 3/6, which also equals 2/4.         But if my friend Al also has a wristband, and my number is #2 and his is #3, at least one of us will win if the following number combinations are picked:  (2, 1) (2, 3) (2, 4) (1, 3) or (3, 4).  The only way at least one of us won't win is if (1,4) is chosen.  Thus our chances of winning are 5 winning combinations out of 6 total, or 5/6 = 83.3 %.  Yet by the logic in your answer, going from 1 wristband to 2 would only have increased our chances from 2/4 to 2/3, or 66.7%, which is incorrect.         In fact, as you may know, the correct way to calculate these probabilities is by use of the binomial coefficients, i.e. the C(n,r) formula for "Combinations of n things taken r at a time":  C(n,r) =  n! / [ (n-r)! r! ] .         For the original problem, the probability that at least one of two wristbands wins when 20 out of 65 are chosen is [100% minus P], where P is the probability that NEITHER of the two wristbands will be chosen among the 20.  The denominator of P is C(65, 20)--the total ways to draw any 20 wristbands from any of the 65 choices.  The numerator of P is the number of ways to have neither wristband chosen, or C(63, 20) , reflecting the fact that any of the 63 other numbers (besides those of the 2 specified wristbands) can be picked to be among the 20.         So the answer to the question your reader posed is:  100% - [ C(63, 20) / C(65,20) ] or 1 - (99/208) or 109/208 or 52.4%.  Your answer of 20/64 or 31.25% is thus significantly off the mark.         For, say, 17 people, the probability that at least one of them would win is similarly [100% minus Q], where Q is the probability that NONE of the 17 numbers would be selected among the 20.  In this case, with 48 numbers other than those of the 17 people, Q would equal   C( 48 , 20 ) / C (65, 20)  = ...whatever.  Then subtract from 1 to get the final answer, which would NOT be the answer you implied of 20 / (65-16) or  20/49.        Of course, with 46 people, the correct answer is indeed (as you stated) 100%, as the pigeonhole principle replaces the (now impossible to formulate) C(n,r) process, and the probability that none of the 46 numbers would be chosen is zero.  But all of the other answers postulated (for from 2 to 45 people) in your original reply are incorrect.         [Further math stuff in e-mail omitted here-P.B.]         Sincerely,         Paul Boisvert   Chicago, IL ps:  Loved your Monty Hall analysis from a while ago, which is indubitably correct!"   [End of my e-mail.]

    In fact, immediately below is a simpler, more direct way to see the correct probabilities, without all the C(n,r)'s above.  The fact that both methods give the same answer is further confirmation that Marilyn's approach is flawed.
 

Simpler, More Direct Analysis, Without All the C(n,r) Combinatorics:     The best way to think of this problem is to imagine that the wristband holders don't look at the number on their wristbands! They remain in ignorance of their  number until long after the "winning" 20 numbers have been selected.  Imagine those winning numbers, once selected, being displayed on a public sign in the following way:  the numbers from 1 to 65 are listed in order, one underneath the next, and next to each of the 20 winning numbers is put a "W".  (The other 45 losing numbers have nothing listed next to them.)     Before looking at the sign, and still without knowing what number is on their own wristband, each person can then ask what the chances are that their wristband number will turn out to have a "W" next to it when they look at the list.     a.)   For a single person (me), their are 65 possible positions where my (still unknown) number could appear on the list, and 20 of those possible positions have a "w" next to them, so my probability of winning is indeed 20/65, as both the reader and Marilyn agreed is correct--they are right about that.    b.)   But for two people (me and Al), the probability that at least one of us wins is still 100% minus [P] , where P = the probability that neither of us wins.    c.)   It's now easy to figure out how neither of us could win, by multiplying two probabilities together:           Prob. [neither of us wins]   =  [my probability of not winning] times [Al's probability of also not winning, given that I didn't win].    d.)  My probability of not winning is calculated by the method in step a.):  there are still 65 positions on the list my number could be, and 45 of them are not winning.  So this probability is 45/65    e.)  For Al also not to win (assuming I haven't won either), his number cannot be in my (already occupied) position on the list, so there are 64 possible positions left where his number might be.  Of those, 44 are non-winning--my number must be already occupying one of the 45 non-winning position, leaving 44 possibilities for him.   So his probability is 44/64.    f.)   So the probability that at least one of us wins is  100% - [ (45/65) x (44/64) ]  =  1 - [99/208]  =  109/208 =  52.4% as in my e-mail to Marilyn.     The beauty of this method is that it instantly generalizes to all other numbers of wristbands:  for example, take 5 people (A,B,C,D,E) with wristbands:     Prob[at least one of the 5 wins]  =  1 - Prob[none of the 5 win] .     For none of the 5 to win, first A must lose with prob.  45/65.      Then B has prob. 44/64 of also losing, as in step e.) above.     Then C's number can't be in A's or B's position, leaving 63 possible positions for it.  And 43 of those are non-winnning, since two non-winning positions are already occupied by A's and B's numbers.  So C's prob. = 43/63.      Similarly D's prob. is 42/62, and E's is 41/61, etc.  This simple process works for any number of wristbands.      So for 5 people, the probability that at least one wins is  1 - [ 45/65 x 44/64 x 43/63 x 42/62 x 41/61 ]  =  335,149 / 393,328  = 85.2%      Marilyn's answer would have been 20/61 = 32.8%, a far cry from the correct answer.      Finally,      Does the combinatoric method in the e-mail, using C(n,r), give the same answer for 5 people (85.2%) as this direct method?      Yes, indeed!    With 65 - 5 = 60 , the C(n,r) method yields :     1 - [ C( 60 , 20 ) / C (65, 20) ]  =  85.2%

    Looks like Marilyn is human after all.  Of course, with this, her first humiliating public error, she's now a mere 46,655 humiliating public errors behind me, so she still has a lot of catching up to do.  (And believe me, I won't be sitting around idly failing to screw up, either, so I doubt she'll ever overtake me!)
 

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