# Python for (Cartalk) Puzzlers (Shallow Thoughts)

Akkana's Musings on Open Source Computing and Technology, Science, and Nature.

## Python for (Cartalk) Puzzlers

Last week's Car Talk had a fun puzzler called "Three Pieces of Paper":

Three different numbers are chosen at random, and one is written on each of three slips of paper. The slips are then placed face down on the table. The objective is to choose the slip upon which is written the largest number.

Here are the rules: You can turn over any slip of paper and look at the amount written on it. If for any reason you think this is the largest, you're done; you keep it. Otherwise you discard it and turn over a second slip. Again, if you think this is the one with the biggest number, you keep that one and the game is over. If you don't, you discard that one too.

What are the odds of winning? The obvious answer is one in three, but you can do better than that. After thinking about it a little I figured out the strategy pretty quickly (I won't spoil it here; follow the link above to see the answer). But the question was: how often does the correct strategy give you the answer?

It made for a good "things to think about when trying to fall asleep" insomnia game. And I mostly convinced myself that the answer was 50%. But probability problems are tricky beasts (witness the Monty Hall Problem, which even professional mathematicians got wrong) and I wasn't confident about it. Even after hearing Click and Clack describe the answer on this week's show, asserting that the answer was 50%, I still wanted to prove it to myself.

Why not write a simple program? That way I could run lots of trials and see if the strategy wins 50% of the time.

So here's my silly Python program:

```#! /usr/bin/env python

# Cartalk puzzler Feb 2011

import random, time

random.seed()

tot = 0
wins = 0

while True:
# pick 3 numbers:
n1 = random.randint(0, 100)
n2 = random.randint(0, 100)
n3 = random.randint(0, 100)

# Always look at but discard the first number.
# If the second number is greater than the first, stick with it;
# otherwise choose the third number.
if n2 > n1 :
final = n2
else :
final = n3

biggest = max(n1, n2, n3)
win = (final == biggest)
tot += 1
if win :
wins += 1
print "%4d %4d %4d %10d %10s %6d/%-6d = %10d%%" % (n1, n2, n3, final,
str(win),
wins, tot,
int(wins*100/tot))
if tot % 1000 == 0:
print "(%d ...)" % tot
time.sleep(1)
```

It chooses numbers between 0 and 100, for no particular reason; I could randomize that, but it wouldn't matter to the result. I made it print out all the outcomes, but pause for a second after every thousand trials ... otherwise the text scrolls too fast to read.

And indeed, the answer converges very rapidly to 50%. Hurray!

After I wrote the script, I checked Car Talk's website. They have a good breakdown of all the possible outcomes and how they map to a probability. Of course, I could have checked that first, before writing the program. But I was thinking about this in the car while driving home, with no access to the web ... and besides, isn't it always more fun to prove something to yourself than to take someone else's word for it?

[ 21:17 Feb 22, 2011    More programming | permalink to this entry | comments ]
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