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Ziyuan Liu
Math 07
04/18/2013
Penny Bet
My favorite Gardner puzzle so far is the coin toss puzzle from the probability
section. The puzzle is derived from the following story.
“Two friends spin a coin on the bar to see who would pay for each
round of beer. However, one day, after winning majority of the spins,
the math major 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.’
Now the rule is changed so that if the person with two pennies gets
more heads than the person with one, the person with the two
pennies wins. The person with two pennies will lose if he spins less or
equal number of heads than the person with one penny.”1
(Gardner 43)
The puzzle asks the reader to evaluate the probability of winning for the
person with two pennies. The purpose of this puzzle is to fool readers into thinking
that more chances at something equate to a higher success rate. However, the
wording of the new rule should quickly quash that notion. We know that initially
each person has ½ chance of winning the game. However, the new rule dictates that
the person with two pennies must obtain more heads than the person with one
1
Gardner, Martin. The Colossal Book of Short Puzzles and Problems. New York: Norton,
2006. Print.
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penny, or else he loses. Under this new rule, the player with two pennies must
obtain two heads when his opponent obtains one, or one head when his opponent
obtains none. It is important to note that when they spin equal numbers of head(s),
the player with one penny wins.
Player one
Player two
H
HH
H
HT
H
TH
H
TT
T
HH
T
HT
T
TH
T
TT
Figure 1. All the possible outcomes between the two players. Green indicates a win.
As shown in Figure 1, the player with the additional coin does not receive an
advantage over the player with no additional coin. It turns out that this math major
knew exactly how to appease his friend. What if the math major became too prideful
and allowed his friend to use a third coin under the same rules?
Further Consideration and Generalization.
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Under the same guidelines with four coins total, the player with three coins
gains an upper hand in probability. With four coins, the total number of outcomes is
16. 𝑂𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 2𝑛 , where n is the number of coins involved.
Statistically, the person with three coins has an 11/16 chance of winning.
What would his chance be if he, the player with three coins, gains another one? The
answer is 26/32. The probability will increase dramatically as the number of coins
is increased. The player with the ever-increasing number of coins would never
achieve complete dominance in this game, as there always will be a single case
where all the spins are tails.
What if the math major is reluctant about giving his friend three coins, but he
is unwilling to take back his word? Is there a way that he can make the game fair
again without subtracting coins from his friend? In the case with four coins, the
player with a single coin can rebalance the game by spinning an additional coin. In
this fair case with five coins, there are a total of 32 possibilities, of which both
players will have 16/32 or ½ chance of winning the game. Another approach to
understand this puzzle is to view the necessity of flipping that last coin. If the player
with n coins has already spun n-1 coins and currently has more or less head(s) than
his adversary – the additional coin spin wouldn’t help him, as final results are
already set.
Generally, in order to play a fair game, the math major can allow his friend to
have n coins (where n is larger than or equal to 2), while he must have n-1 coin(s).
The additional coin only matters when two players are at a standstill. In the case of a
standstill, all previous results can be disregarded; the last coin spin will determine
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the winner. It is interesting to point out that with one coin, under the current rule,
the player who will not spin at all will still have the same probability of winning as
the player who has possession of the coin. In explaining the above, we can conclude
that there are two cases with this puzzle: one where the players tie and resort to the
last coin and one where the players don’t tie and don’t need to use the last coin (in
which they spin the same number of coins). The two situations are mutually
exclusive and give both players equal chances. To conclude this puzzle, we have
found that gaining one additional coin for this game is irrelevant, defeating the
notion that more chances represent a higher possibility of winning in this game.