Problem 7 (Bonnie and Clyde)

Two bandits, Bonnie and Clyde, divide a chest of gold coins between them as follows: One for Bonnie, two for Clyde, three for Bonnie, four for Clyde, and so on until they run out of coins. If they get to 20 for Clyde but there are less than 20 coins, the game stops and Clyde gets the remaining coins. Likewise for Bonnie, if she is the last player, she gets the remaining coins. Who is better off, Bonnie or Clyde ?

Solution

It is clear that the number of coins in the chest determines which person ends up with the most coins. We can approach this problem by simply tabulating the number of coins both Bonnie and Clyde on each of their turns. We have

	TURNS
Bonnie gets 1	Bonnie's turn 1
	Clyde's turn 1	Clyde gets 2
Bonnie gets 3	Bonnie's turn 2
	Clyde's turn 2	Clyde gets 4
Bonnie gets 5	Bonnie's turn 3
	Clyde's turn 3	Clyde gets 6
...	...	...
Bonnie gets 2n - 1	Bonnie's turn n	...
...	Clyde's turn n	Clyde gets 2n

On her nth turn Bonnie says, "I'll take my 2n - 1 coins", giving her a total take of

Bonnie's take = 1 + 3 + 5 + ... + (2n - 1) = n²

coins. Now let's assume that on her nth turn Bonnie takes her 2n - 1 coins, but leaves less than 2n coins for Clyde, which means Clyde takes the remainder r of coins, where r is some number 1, 2, ... , 2n. This means Clyde's total take of coins is

Clyde's take = 2 + 4 + 6 + ... + (2n - 2) + r = n² - n + r

Comparing Bonnie and Clyde's take, we conclude that

• Bonnie ends up with more coins than Clyde if n is greater than r
• Bonnie and Clyde end up with the same amount if n = r
• Clyde ends up with more coins if n is less than r

where n is the number of turns each player has taken coins from the chest.
Now, let's consider the other situation when on his n-th turn, Clyde takes 2n coins from the chest, but leaves only a remainder of r coins for Bonnie, where r is some number 1, 2, ... , 2n + 1, inclusive. In this case, Clyde's total take is

Clyde's take = 2 + 4 + 6 + ... + 2n = n² + n

whereas Bonnie's take is

Bonnie's take = 1 + 3 + 5 + ... + (2n - 1) + r = n² + r

Comparing Bonnie and Clyde's take, we conclude that

• Bonnie ends up with more coins than Clyde if r is greater than n
• Bonnie and Clyde end up with the same amount if r = n
• Clyde ends up with more coins if r is less than n

In conclusion, we see that the person who picks the last remainder of coins ends up with the most coins if and only if the remainder of coins is greater than the number of times the other person has taken coins from the chest. When the remainder of coins is the same as the number of turns of each player, the two players end up with the same number of coins.
For example, suppose Bonnie takes 7 coins on her n = 4th turn, but leaves r = 3 coins for Clyde. We know immediately that Bonnie ends up with more coins since the number of her turns, n = 4, is greater than Clyde's remainder, r = 3. To check this, we know that Bonnie's total take is

• Bonnie's take = 1 + 3 + 5 + 7 = 16
• Clyde's take = 2 + 4 + 6 + 3 = 15

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Last modified on Tuesday, January 12, 1999