The Best Gardner Puzzle

Andrew Meier

Peter Winkler 10A 4/17/13 Paper 2

What makes a mathematical puzzle interesting? Some people enjoy odd scenarios with prisoners, while others prefer more common situations. Some people enjoy puzzles that are very simple; some prefer puzzles that are more elaborate. There are many key elements that work together to make up a great puzzle. Many people enjoy puzzles that they can visualize, or relate to. When the puzzle is realistic, the reader can imagine him/herself in the situation. Puzzles that are paradoxical can also be very interesting. If the puzzle or the solution is not clear at first, the reader is likely to continue working on it to fully understand.People enjoy puzzles that they can visualize, are paradoxical, and have a clear, memorable statement. Martin

Gardner mastered the art of creating magnificent mathematical puzzles through the twenty-five years he spent writing them for Scientific American. Gardner has written thousands of puzzles, making it very difficult for me to pick out the one that I think is the best.

The same criteria don’t make a good puzzle in my mind that they do in othersMost of my favorite puzzles all fall into the same category: probability. It isn’t important to me if the puzzle seems too easy, or draws on common experience. I enjoy puzzles that twist my mind even after hearing the solution. When a mathematician tries to write a puzzle, his or her goal is to challenge the reader’s brain to think about something in a different way, or to exploit a weakness within a user’s mathematical understanding. Probability puzzles, such 2 as the Monty Hall puzzle, take advantage of the reader’s misunderstanding of probability.

Of Gardner’s many probability problems, one in particular, Penny Bet, stands out due to its very odd, and maybe unbelievable, solution.

Bill, a student in mathematics, and his friend John, an English major,

usually spun a coin on the bar to see who would pay for each round of beer.

One evening Bill said: “Since I’ve won the last three spins, let me give you a

break on the next one. You spin two pennies and I’ll spin one. If you have more

heads than I have, you win. If you don’t, I win.”

“Gee, thanks,” said John.

On previous rounds, when one coin was spun, John’s probability of

winning was, of course, ½. What are his chances under the new arrangement?

As a reader, this doesn’t sound like a very difficult problem to solve. All that needs to be done is a careful examination of each outcome. Between the three coins, there are eight outcomes (two different results multiplied by the three coins). So, the outcome there could be H

HH, H HT, H TH, H TT, T HH, T HT, T TH, or T TT. Of these different outcomes, there are four in which times where John will win (indicated in red). As it turns out, there are also four in which times where Bill will win. Despite having an extra coin, John’s chances of winning are the same 50% that they were beforehand. Gardner points out that “[John’s] probability of winning remains the same whenever he has one more coin than Bill.” How can this be possible?! With two coins, John has twice the number of chances to flip a heads.

Gardner points out that “[John’s] probability of winning remains the same whenever he 3 has one more coin than Bill.” If we go back to the original scenario, we can shift the way the problem is looked at to fit this condition. Instead of flipping a coin and assigning a side to John and a side to Bill, let’s say that John will flip one coin and Bill will flip zero coins. If

John flips more heads than Bill, he wins and doesn’t have to buy the beer. If he flips tails, then they both had the same number of heads and therefore Bill wins. a coin; and if it is heads he wins and does not have to pay for the drinks, butand if it is tails he loses and has to buy the beer.

LLooking at it this way, we can see that John has one coin and Bill has zero, so John has one more coin than Bill. and the probability of John winning is fifty percent.

But what if John has three coins and Bill only has two? Now there are thirty-two possible results. Surly John will have the upper hand in this case, right? Here are the different possibilities:

1 H HH Wi H H n 2 H HH H T 3 H HT H H 4 H TH H H 5 H HTT H 6 H TTH H 7 H THT H 8 H TTT H 9 HT HH Wi H n 10 HT HH Wi T n 11 HT HT Wi H n 12 HT TH Wi H n 13 HT HTT 14 HT TTH 15 HT THT 16 HT TTT 4

17 TH HH Wi H n 18 TH HH Wi T n 19 TH HT Wi H n 20 TH TH Wi H n 21 TH HTT 22 TH TTH 23 TH THT 24 TH TTT 25 TT HH Wi H n 26 TT HH Wi T n 27 TT HT Wi H n 28 TT TH Wi H n 29 TT TTH Wi n 30 TT HTT Wi n 31 TT THT Wi n 32 TT TTT

Out of the 32 possible outcomes, John will still only win 16 if he flips three coins instead of two. This result can be generalized to show that there is still only a 50% chance of John winning if he has n coins and Bill only has n-1 coins. Why is this the case? If we think of both Bill and John flipping n-1 coins, they have the same chance of flipping more heads than the other. The only time John’s coin will matter is in the event that Bill and John tie during the first n-1 coins. Then the last coin that John flips acts as a tiebreaker, and we can clearly see that the probability of the coin being heads for John is fifty percent.

Maybe this idea would be even easier to understand if we think of the coin as having uneven probabilities. Again John will flip n coins and Bill will flip n-1, but they will flip a special coin where the probability of flipping heads is seventy-five percent instead of fifty like a regular coin. The probability of John flipping k number of heads in n-1 coins is the same as Bill 5 flipping k heads in n-1 coins. Then, only if they tie, does John need to flip the nth coin. There he has a seventy-five percent chance of flipping heads and winning the bet. The probability of John winning comes only from the probability of flipping heads on the last coin.

The scalability of this problem is what makes it great. From it we see the unlikely occurrence that adding additional coins does not alter the probability in any way. This rids the user of any believe he or she may have about probabilities always growing with increased numbers. In a very simple puzzle, this complex and slightly paradoxical probability puzzle can be easily understood.

This puzzle is great for two main reasons. Its solution plays with the reader’s mind through strange probabilities. Also, the puzzle can be generalized to any number of coins.

Martin Gardner wrote many puzzles during his career, but Penny Bet is my favorite.

Andrew,

You do a good job of capturing why the solution to this puzzle is counterintuitive, but nevertheless true. I would challenge you, however, to be more analytical and less descriptive.

Don’t just catalogue the probabilities; explain why they must be that way and why the reader should care. I’ve noticed definite stylistic improvements in this paper over the last, but you could still do better with respect to intensifiers.