Marilyn vos Savant and Conditional Probability

The following question appeared in the “Ask Marilyn” column in the August 16, 2015 issue of Parade magazine.  The writer seems stuck between two probabilities.


(Click here for a cleaned-up online version if you don’t like the newspaper look.)

I just pitched this question to my statistics class (we start the year with a probability unit).  I thought some of you might like it for your classes, too.

I asked them to do two things.  1) Answer the writer’s question, AND 2) Use precise probability terminology to identify the source of the writer’s conundrum.  Can you answer both before reading further?


Very briefly, the writer is correct in both situations.  If each of the four people draws a random straw, there is absolutely a 1 in 4 chance of each drawing the straw.  Think about shuffling the straws and “dealing” one to each person much like shuffling a deck of cards and dealing out all of the cards.  Any given straw or card is equally likely to land in any player’s hand.

Now let the first person look at his or her straw.  It is either short or not.  The author is then correct at claiming the probability of others holding the straw is now 0 (if the first person found the short straw) or 1/3 (if the first person did not).  And this is precisely the source of the writer’s conundrum.  She’s actually asking two different questions but thinks she’s asking only one.

The 1/4 result is from a pure, simple probability scenario.  There are four possible equally-likely locations for the short straw.

The 0 and 1/3 results happen only after the first (or any other) person looks at his or her straw.  At that point, the problem shifts from simple probability to conditional probability.  After observing a straw, the question shifts to determining the probability that one of the remaining people has the short straw GIVEN that you know the result of one person’s draw.

So, the writer was correct in all of her claims; she just didn’t realize she was asking two fundamentally different questions.  That’s a pretty excusable lapse, in my opinion.  Slips into conditional probability are often missed.

Perhaps the most famous of these misses is the solution to the Monty Hall scenario that vos Savant famously posited years ago.  What I particularly love about this is the number of very-well-educated mathematicians who missed the conditional and wrote flaming retorts to vos Savant brandishing their PhDs and ultimately found themselves publicly supporting errant conclusions.  You can read the original question, errant responses, and vos Savant’s very clear explanation here.


Probability is subtle and catches all of us at some point.  Even so, the careful thinking required to dissect and answer subtle probability questions is arguably one of the best exercises of logical reasoning around.


As a completely different connection, I think this is very much like Heisenberg’s Uncertainty Principle.  Until the first straw is observed, the short straw really could (does?) exist in all hands simultaneously.  Observing the system (looking at one person’s straw) permanently changes the state of the system, bifurcating forever the system into one of two potential future states:  the short straw is found in the first hand or is it not.

CORRECTION (3 hours after posting):

I knew I was likely to overstate or misname something in my final connection.  Thanks to Mike Lawler (@mikeandallie) for a quick correction via Twitter.  I should have called this quantum superposition and not the uncertainty principle.  Thanks so much, Mike.


4 responses to “Marilyn vos Savant and Conditional Probability

  1. One caveat: the original letter-writer didn’t state the problem rigorously and Marilyn didn’t correct the loose ends (though she later makes a comment in her responses that ties up one of the bigger ones). Thus, there have been scholarly articles over the years that point out the ways the problem as stated in the original article was ill-posed. I’m not sure that lets any of the arrogant Ph.Ds off the hook. Having posed the problem myself to some colleagues who unlike me had Ph.Ds in mathematics, I saw exactly how rigid and stubborn even professionals can get when they misunderstand a probability problem. One tack Marilyn didn’t take in the linked responses is to exaggerate the conditions by extending it to more doors: let there be 100 doors, 99 of which have goats and one of which has a car. Make your pick. The host opens 98 doors with goats (key points: he always knows which door has the car, and ALWAYS offers you the choice of staying or switching regardless of whether you picked correctly the first time. Without that caveat, all bets are off). Should you stay or switch? Most people at this point start to see that switching is the better play. But even when I changed the problem that way back in the late ’80s, one math Ph.D insisted that it was then 50-50 and made no difference. I’m still amazed by that.

    I love the sexist comments. Shameless, reflecting a slightly different era, not that there isn’t plenty of sexism out there, particularly in the world of mathematics, even if a woman finally got a Fields Medal.

    By the way, I have no faith in the notion of IQ and find Marilyn to be a bit dense when it comes to higher mathematics. Her book on the proof of Fermat’s Last Theorem is an embarrassment (she argued that it wasn’t proved when in fact it was). She’s not immune to arrogance herself.

  2. Thanks, Chris. I should add that I actually have a pretty good idea why a lot of people react with the “it makes no difference” response almost automatically. People with a little (or even a lot) of knowledge of some basic principles of probability know well the fallacy of thinking, for example, that coins have memories so that the probability of tossing a head with a fair coin on the NEXT toss remains 1/2 regardless of how many heads have previously been tossed in a row (though too many in a row would make most of us reasonably suspect the “fairness” of the coin). Independent events are independent and they have no impact on the probability of other independent events.

    Of course, that’s true. But as you stated, there’s a difference between that and making decisions based on new information. When I first heard this problem, I really didn’t know how to prove mathematically that it was better to switch (I’m a relatively late arrival to math education as a profession), but I knew that the information changed once a non-winning door was revealed and that the increased information changed the likelihood that a previously unchosen door was the correct one. I found that I could convince a lot of non-math people with that argument, but that my one stubborn colleague was unmoved by any argument I made. An empirical demonstration by experiment might have worked, but that didn’t occur to me at the time and I’m not sure I would have known a good way to conduct such an experiment that would have properly modeled the situation and also convinced someone so sure he was right no matter what.

    People do NOT want to fall into the trap of thinking that independent events can be affected by what happens after they’ve occurred, but in this case, choosing again occurs under different conditions than does the original choice. Clearly, the first choice has a probability of being correct = 1/n for n doors. And that probability is true for each door. But after, say, one door is opened, the original choice is still 1/n but any other options have a 1/(n – 1) probability of being correct. And that is better than 1/n. How much better is a function of n and how many losing doors are opened by the host. I can see creating a family of problems stemming from the original that could teach a lot of interesting facts about conditional probability. So thanks for getting me thinking about this classic again.

  3. Pingback: Marilyn vos Savant Conditional Probability Follow Up | CAS Musings

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