Marilyn vos Savant

Savant was asked the following question in her September 9, 1990 column:[18]

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

This question is called the Monty Hall problem due to its resembling scenarios on the game show Let's Make a Deal; its answer existed before it was used in "Ask Marilyn". She said the selection should be switched to door #2 because it has a ​²⁄₃ chance of success, while door #1 has just ​¹⁄₃. To summarize, ​²⁄₃ of the time the opened door #3 will indicate the location of the door with the car (the door you had not picked and the one not opened by the host). Only ​¹⁄₃ of the time will the opened door #3 mislead you into changing from the winning door to a losing door. These probabilities assume you change your choice each time door #3 is opened, and that the host always opens a door with a goat. This response provoked letters from thousands of readers, nearly all arguing doors #1 and #2 each have an equal chance of success. A follow-up column reaffirming her position served only to intensify the debate and soon became a feature article on the front page of The New York Times. Parade received around 10,000 letters from readers who thought that her workings were incorrect.[19]

Under the "standard" version of the problem, the host always opens a losing door and offers a switch. In the standard version, Savant's answer is correct. However, the statement of the problem as posed in her column is ambiguous.[20] The answer depends on what strategy the host is following. If the host operates under a strategy of only offering a switch if the initial guess is correct, it would clearly be disadvantageous to accept the offer. If the host merely selects a door at random, the question is likewise very different from the standard version. Savant addressed these issues by writing the following in Parade magazine, "the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. Anything else is a different question."[21]

She expounded on her reasoning in a second follow-up and called on school teachers to show the problem to classes. In her final column on the problem, she gave the results of more than 1,000 school experiments. Most respondents now agree with her original solution, with half of the published letters declaring their authors had changed their minds.[22]

Like the Monty Hall problem, the "two boys" or "second-sibling" problem predates Ask Marilyn, but generated controversy in the column,[23] first appearing there in 1991–1992 in the context of baby beagles:

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

When Savant replied "one out of three", readers[24] wrote the odds were 50–50. In a follow-up, she defended her answer, saying that "If we could shake a pair of puppies out of a cup the way we do dice, there are four ways they could land", in three of which at least one is male, but in only one of which none are male.

The confusion arises here because the bather is not asked if the puppy he is holding is a male, but rather if either is a male. If the puppies are labeled (A and B), each has a 50% chance of being male independently. This independence is restricted when at least A or B is male. Now, if A is not male, B must be male, and if B is not male, A must be male. This restriction is introduced by the way the question is structured and is easily overlooked – misleading people to the erroneous answer of 50%. See Boy or Girl paradox for solution details.

The problem re-emerged in 1996–97 with two cases juxtaposed:

Say that a woman and a man (who are unrelated) each have two children. We know that at least one of the woman's children is a boy and that the man's oldest child is a boy. Can you explain why the chances that the woman has two boys do not equal the chances that the man has two boys? My algebra teacher insists that the probability is greater that the man has two boys, but I think the chances may be the same. What do you think?

Savant agreed with the teacher, saying the chances were only 1 out of 3 that the woman had two boys, but 1 out of 2 the man had two boys. Readers argued for 1 out of 2 in both cases, prompting follow-ups. Finally she began a survey, asking female readers with exactly two children, at least one of them male, to give the sex of both children. Of the 17,946 women who responded, 35.9%, about 1 in 3, had two boys.[25]

On January 22, 2012, Savant admitted a mistake in her column. In the original column, published on December 25, 2011, a reader asked:

I manage a drug-testing program for an organization with 400 employees. Every three months, a random-number generator selects 100 names for testing. Afterward, these names go back into the selection pool. Obviously, the probability of an employee being chosen in one quarter is 25 percent. But what is the likelihood of being chosen over the course of a year?

Her response was:

The probability remains 25 percent, despite the repeated testing. One might think that as the number of tests grows, the likelihood of being chosen increases, but as long as the size of the pool remains the same, so does the probability. Goes against your intuition, doesn't it?

The correctness of the answer depends on how the question is asked. The probability of being chosen each time is 25% but probability of being chosen at least once across the 4 events is higher. In this case, the correct answer is around 68%, calculated as the complement of the probability of not being chosen in any of the four quarters: 1 – (0.75⁴).[26]

On June 22, 2014, Savant made an error in a word problem. The question was: "If two people could complete a project in six hours, how long would it take each of them to do identical projects on their own, given that one took four hours longer than the other?" Her answer was 10 hours and 14 hours, reasoning that if together it took them 6 hours to complete a project, then the total effort was 12 "man hours". If they then each do a separate full project, the total effort needed would be 24 hours, so the answer (10+14) needed to add up to 24 with a difference of 4.[27] Savant later issued a correction, as the answer ignored the fact that the two people get different amounts of work done per hour: if they are working jointly on a project, they can maximize their combined productivity, but if they split the work in half, one person will finish sooner and cannot fully contribute. This subtlety causes the problem to require solving a quadratic equation and thus to not have a rational solution. Instead, the answer is ${\displaystyle 4+{\sqrt {40}}}$ (approximately 10.32) and ${\displaystyle 8+{\sqrt {40}}}$ (approximately 14.32) hours.[28]