## Puzzle #142: Bills and Two Hats

This one is a simple problem, is a variation of a probability puzzle posted about 3 years back – try it with your teenage children! This one is again taken from The Colossal Book of Short Puzzles and Problems by Martin Gardner. Here it goes:

“No,”said the mathematician to his 14-year old son, “I do not feel inclined to increase your allowances this week by 10 dollars. But if you will take a risk, I will make you a sporting proposition.”

The boy groaned. “What is it this time, Dad?”

“I happen to have,” said the father, “10 crisp new 10-dollar bills and 10 crisp new one-dollar bills. You may divide them any way you please into two sets. We’ll put one set into hat A, the other set into hat B. Then I’ll blindfold you. I’ll mix the contents of each hat and put one hat on the right and one on the left side of the mantel. You pick either hat at random, then reach into that hat and take out one bill. If it’s a 10, you may keep it.”

“And if it isn’t?”

“You’ll mow the lawn for a month, with no complaints.”

The boy agreed. How should he divide the 20 bills between the two hats to maximise the probability of his drawing a 10-dollar bill, and what will that probability be?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy lawn mowing!

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### 3 Responses to Puzzle #142: Bills and Two Hats

1. Bhanu says:

If one set is one 10 dollar bill and the second set is the rest of the nineteen bills (ten 1s and nine 10s) then the probability of picking a 10 dollar bill is 1/2*(1+9/19) or 14/19 or .737.

2. Pratik Poddar (@pratikpoddar) says:

Lets say right hat has x \$10 bills and y \$1 bills.
The probability you get \$10 in right hat is then x/(x+y)
The probability you get \$10 in left hat is then (10-x)/(20-x-y)

Expected probability of getting something is:
f(x,y) = 1/2*( x/(x+y) + (10-x)/(20-x-y) ) where x and y are independent and integer from 0 to 10 both inclusive

We need to maximise f(x,y)
f(0,0) = 1/4
f(0,y) = 5/(20-y) for y!=0
f(x,0) = 1/2(1+(10-x)/(20-x)) = (15-x)/(20-x) = 1+5/(x-20) for x!=0
f(10,y) = 5/(10+y) for y!=10
f(x,10) = 1/2 + 1/2*x/(x+10) for x!=10
f(10,10) = 1/4

f(x,y) is maximised at either y=0, x=1 or y=10,x=9 which is effectively the same configuration

So, f(x,y) is maximised at x=1, y= 0

Put one \$10 bill on one hat that ensures you are getting the \$10 bill if you choose that hat. The other hat would have half probability as well almost. Maximising the probability to get \$10 almost to 0.7.

3. Karan says:

This puzzle took me back to 9th grade, when you asked me the “balls in boxes puzzle”, when you were visiting us in Jaipur. 🙂

Put 1 ten dollar note in one hat and remaining 9 ten dollar notes and 10 one dollar notes in another.
The probability will be 1/2(1+9/19)=0.73