To flip, or not to flip

Marilyn is Wrong Copyright © 1997-1998 Herb Weiner. All rights reserved.

Ask Marilyn ® by Marilyn vos Savant is a column in Parade Magazine, published by PARADE, 711 Third Avenue, New York, NY 10017, USA. According to Parade, Marilyn vos Savant is listed in the "Guinness Book of World Records Hall of Fame" for "Highest IQ."


Fred Moolten <> remembers the following Marilyn column. If you have this column, please send it to me.

You get to keep the monetary value of a coin (or card or any other two sided object that you can write numbers on) that you pick at random from a box. Each coin has a number on one side and a number twice as large on the other, and you can either choose to keep the first number or flip to the other side, in which case you must accept the number on that side. The question is: wouldn't it always be desirable to flip to the other side? The chances are 50/50 that you'll see a larger number, but if you do, your gain (100% of the original value) will be twice what you stand to lose if the flip side is smaller (50% of the original value).


Marilyn said that it wouldn't make any difference if you stay or switch.


Marilyn failed to explain why switching would confer no advantage. Fred suggests that the fallacy lies in the assumption that the probability of a gain is always 50%. In fact, that is only an average, but the higher the initial value (and thus the greater the prospective gain or loss), the more probable it is that you're already looking at the higher side. And if you're already looking at the higher side, your probability of losing is 100%. This fact offsets the cumulative advantages that come from flipping coins with smaller numbers on them. In fact, if you actually knew how the numbers were distributed, you could use this information to compute the probability that you were already looking at the higher side, and optimize your switching strategy.

Herb Responds

This sound like an interesting problem for a computer simulation. Assume that you do not know the distribution of the numbers on the coins (since this makes the problem more interesting), and compute your winnings if you switch, compared to your winnings if you don't switch. My prediction is that Marilyn is correct. And Fred's explanation sounds plausible.

A Computer Simulation

Within two days, Ron Whittle <> responded with a computer program, written in C. The program runs one million trials, each time choosing a coin with a random number between 1 and 100 on the front, and a number twice as high on the back. The program picks either the front side or the back side at random. It keeps track of the winnings accumated by never flipping sides, the winnings accumlated by always flipping sides, and the winnings accumlated by randomly flipping sides half the time. Upon completing all trials, the program prints out the results. Here's the output:

Winnings from never switching:          757119759
Winnings from always switching:         757162968
Winnings from switching half the time:  757144873

Although the results will vary slightly each time the program is run, the results show that, within statistical accuracy, there is no benefit of switching.

A Similar Situation

Brian Scearce <> wrote to describe a similar situation:
This problem is isomorphic to a puzzle I heard years ago: suppose we are presented with the opportunity to open our wallets. Whoever has more money has to give it to the other guy.

A simple analysis suggests that you have a 50/50 chance of winning, and if you do, you'll gain more money than when you lose. So you should take the bet. But the same analysis suggests that I too should take the bet, and it's a zero-sum game, so it can't be advantageous to both of us!

I'm sure I've seen this puzzle in a logic book (in decreasing order of probability: Smullyan, Gardner, Quine), initially stating the puzzle in the form of "two mathemeticians have an agreement that when they meet at a conference, the one with the nicer tie must give it to the one with the uglier tie". He turned it into a wallet puzzle to eliminate the irrelevant factors of personal taste, etc.

A Strategy

Ron Whittle <> wrote to suggest a winning strategy:

Assuming that neither side of any coin has a fractional value (such as half a penny) you should switch whenever you receive an odd number (such as five). Since one side is twice the other, and since two and one half is not a possible value, if one side is five, the other side must be ten. In other words, if one side is odd, the other side must be twice the value.

I originally wrote that if you receive an even number, there is no benefit (nor is there any loss) in switching. However, Fred Moolten <> pointed out that this is incorrect. If odd numbers always dictate a switch, then even numbers must necessarily dictate a stay. The reason is that the total number of coin faces representing the "low" side of coins must equal the total number representing the "high" side. If all the odd numbers are on the low side, what's left (the even numbers) must consist of more high sides than low sides.

The original question does not state how the values of the coins are distributed. However, if the values of the low side of the coin are uniformly distributed, and if neither side of any coin has a fractional value, then half of the low sides would be odd, and half would be even, and all of the high sides would be even. With such a distribution, three quarters of the sides would be even, and two thirds of the even sides would be the high side.

An Alternate Strategy

Stein Kulseth <> wrote to suggest an alternate strategy:

Even when the distribution is not known, and when the amounts on the coins are not restricted to integers there exists a strategy that is better than "always stay" or "always switch".

Pick a random monetary value. If the first number is less than this you switch; if it is greater or equal you stay.

Whenever you pick a coin where both sides are greater than or both are less than your random number, you don't gain anything (you still have a 50% chance of getting the larger number). However there will be some chance that you pick a coin where your random number falls between the two sides' numbers, and in these cases you will always get the larger number using this strategy.

Assumptions, Please

Peter Benson <> reports that
This is a well known problem. For example, I believe John Paulos discusses it in his book Mathematical Illiteracy and Innumeracy.

He does it with a pair of sealed envelopes.

The fundamental assumptions are:

  1. No information is available on the distribution of the values on the low side of the coin (other than being positive). In particular, fractional values are not ruled out, and there is no minimum or maximum value.

  2. You only get one chance (i.e. you are not allowed to gather information about the distribution).
Given these assumptions, there is no benefit to flipping. No harm, either. last updated June 30, 1998 by