Tag Archives: marilyn vos savant

From Coins to Magic

Here’s a great problem or trick for a class exploration … or magic for parties.


Grab a small handful of coins (it doesn’t matter how many), randomly flip them onto a flat surface, and count the number of tails.

Randomly pull off from the group into a separate pile the number of coins equal to the number of tails you just counted.  Turn over every coin in this new pile.

Count the number of tails in each pile.

You got the same number both times!



Marilyn Vos Savant posed a similar problem:

Say that a hundred pennies are on a table. Ninety of them are heads. With your eyes closed, can you separate all the coins into two groups so that each group has the same number of tails?

Savant’s solution is to pull any random 10 coins from the 100 and make a second pile.  Turn all the coins in the new pile over, et voila!  Both piles have an equal number of tails.

While Savant’s approach is much more prescriptive than mine, both solutions work.  Every time.  WHY?


You have no idea the state (heads or tails) of any of the coins you pull into the second pile.  It’s counterintuitive that the two piles could ever contain the same number of tails.

Also, flipping the coins in the new pile seems completely arbitrary, and yet after any random pull & flip, the two resulting piles always hold the same number of tails.

Enter the power (and for young people, the mystery) of algebra to generalize a problem, supporting an argument that holds for all possibilities simultaneously.


The first clue to this is the misdirection in Savant’s question.  Told that there are 90 heads, you are asked to make the number of tails equivalent.  In both versions, the number of TAILS in the original pile is the number of coins pulled into the second pile.  This isn’t a coincidence; it’s the key to the solution.

In any pile of randomly flipped coins (they needn’t be all or even part pennies), let N be the number tails.  Create your second pile by pulling a random coins from the initial pile.  Because the coins are randomly selected, you don’t know how many tails are in the new pile, so let that unknown number of coins be X .  That means 0 \le X \le N, leaving N-X tails in the first pile, and N-X heads in the new pile.  (Make sure you understand that last bit!)  That means if you flip all the coins in the second pile, those heads will become tails, and you are guaranteed exactly N-X tails in both piles.

Cool facts:

  • You can’t say with certainty how many tails will be in both piles, but you know they will be the same.
  • The total number of coins you start with is completely irrelevant.
  • While the given two versions of the problem make piles with equal numbers of heads, this “trick” can balance heads or tails.  To balance heads instead, pull from the initial coins into a second pile the number of heads.  When you flip all the coins in the second pile, both piles will now contain the same number of heads.


If you work on your showmanship, you can baffle others with this.  For my middle school daughter, I counted off the “leave alone” pile and then flipped the second pile.  I also let her flip the initial set of coins and tell me each time whether she wanted me to get equal numbers of heads or tails.  I looked away as she shuffled the coins and pulled off the requisite number of coins without looking.

She’s figured out HOW I do it, but as she is just starting algebra, she doesn’t have the abstractness yet to fully generalize the big solution.  She’ll get there.

I could see this becoming a fun data-gathering project for an algebra class.  It would be cool to see how someone approaches this with a group of students.


Low Floor and High Ceiling Math

Whether you like to solve problems yourself, or are looking for some tidbits for your children or students, I hope this post is informative.

I’ve been reading Jo Boaler‘s brilliant new book, Mathematical Mindsets.  While there’s tons of great information and research there, I’ve been thinking lots lately about her charge to develop more “low floor, high ceiling” tasks into math lessons–problems that are “challenging, but accessible” to a much broader spectrum of students than typical exercises.  In particular, Boaler encourages teachers to use problems that are easily understood, relatively simple to begin, and yet hold deep potential for advanced exploration.  Boaler notes that these problems tend to be very difficult to find.

Here I offer an adaptation of an Ask Marilyn post toward this goal.  While the problem was initially posed in terms of singles at a party; I rephrased it for younger students.  Solving it helped me see variations that I hope address Boaler’s low floor, high ceiling call.


Paraphrasing the original:

Say 100 students stop by the lunchroom for a snack.  Of these, 90 like apples, 80 like pears, 70 like bananas, and 60 like peaches.  At the very least, how many students like all four fruits?


The Ask Marilyn post offered only the answer–zero–but not a solution.  To prove that, I made picture.  Since the question was the least number of commonly liked fruits, I needed to spread out the likes as much as possible.  Ninety liked apples, so when I added the pears, I made sure to include the 10 non-apples among the 80 who did like pears giving


That made 30 (at the bottom) who liked only one of apples or pears, so when I added the 70 bananas, I first added them to those 30, leaving


That made 40 who liked all three, so the 60 peaches could match up to the other 60 who liked only two of the first three, confirming vos Savant’s claim that it was possible in this setup to have no one liking all four.



FIRST:  As a minor extension, one of my students last year would have said the problem could be “complexified” slightly by changing the numbers to percentages.  (I loved my conversations with that student about complexifying vs. simplifying problems to find deeper connections and extensions.)  With enough number sense, students should eventually be able to work with absolute numbers and relative percentages with equal ease.  Mathematically, it doesn’t change anything about the problem.

SECOND:  The problem doesn’t have to be about a single minimum number of students to like all four fruits.  While there is a unique minimum, there are many other non-optimal arrangements.  I wonder how students with developing problem solving skills would approach this.

THIRD:  In my initial attempts, I had used many different variations on my tabular solution above.  Only in the writing of this post did I actually use the above arrangement, and that happened only because I was trying to come up with a visually simple representation.  In doing so, I realized that the critical information here was not what was told, but what was not said.  Where 90, 80, 70, & 60 liked the given fruits, that meant a respective 10, 20, 30, & 40 did not.  And those added up to 100, so I knew that any variation of “not-likes” that also added to 100 could be distributed so that the minimum number who liked all four would also be zero.  So there is an infinite number (if I use percentages) of variations of this problem that have the same answer.  I also realized that any combination of 2 or more fruits whose “not likes” added to 100 could produce the same results.  My ceiling just rose!

FOURTH:  To make the problem more accessible, I could rephrase this in terms of setting out fruit and exploring many different possible arrangements.  i could also encourage learners to support their developing problem solving by translating the problem into pictures.

I’m ready to pose a new variation.  I’d love to hear your thoughts, insights, and variations for raising the ceiling in this problem.


Say 100 students stop by the lunchroom for a snack.  Of these, 90 like apples, 80 like pears, 70 like bananas, and 60 like peaches.  The lunchroom staff knows these numbers, but doesn’t know how much of each fruit to put out.  But putting out too much fruit would be wasteful.

  • What advice can you give them?  Show how you know your solutions are correct.
  • Draw some pictures of the possible numbers of students who like the different fruits.
  • Is there more than one possible answer?
  • It is possible that some students might not like any fruits offered.  How many students might this describe?
  • Some students might like all four fruits.  For how many students might this be true?  How many answers are there to this?

For students who manage all of these, you can challenge them to

  • How can you change the initial numbers in this problem without changing most of the answers?
  • Can you create the same scenarios with more or fewer types of fruit?
  • If the numbers are too big for very young students, you could drop all of the initial numbers by a factor of 10.  How many will see this scaling down simplification (or its scaling up complexification)?

Marilyn vos Savant Conditional Probability Follow Up

In the Marilyn vos Savant problem I posted yesterday, I focused on the subtle shift from simple to conditional probability the writer of the question appeared to miss.  Two of my students took a different approach.

The majority of my students, typical of AP Statistics students’ tendencies very early in the course, tried to use a “wall of words” to explain away the discrepancy rather than providing quantitative evidence.  But two fully embraced the probabilities and developed the following probability tree to incorporate all of the given probabilities.  Each branch shows the probability of a short or long straw given the present state of the system.  Notice that it includes both of the apparently confounding 1/3 and 1/2 probabilities.


The uncontested probability of the first person is 1/4.

The probability of the second person is then (3/4)(1/3) = 1/4, exactly as expected.  The probabilities of the 3rd and 4th people can be similarly computed to arrive at the same 1/4 final result.

My students argued essentially that the writer was correct in saying the probability of the second person having the short straw was 1/3 in the instant after it was revealed that the first person didn’t have the straw, but that they had forgotten to incorporate the probability of arriving in that state.  When you use all of the information, the probability of each person receiving the short straw remains at 1/4, just as expected.

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.