The Ten Chests

Reasoning about probability can be very difficult. The puzzle herein is a follow-up of the last one I wrote about, which reminded me a bit of the famous Monty Hall problem. I managed to get the right answer to Smullyan's earlier puzzle without any assistance but it could have been because I already knew what to look for. Smullyan himself admitted that the solution is very counterintuitive and offered several separate justifications to combat intuition. Speaking of counterintuitive reasoning and probability, no less a person than the late Paul Erdős was unconvinced of the optimal decision in the Monty Hall problem until he witnessed a numerical simulation! In the same spirit, I will be applying a Monte Carlo simulation to the following more elaborate version of the previous puzzle:

It took Scheherazade quite a bit of time to get the king to accept the correct answer to the last problem, but she finally succeeded.

"I have thought of a related problem," said Scheherazade. "Suppose we have now ten chests instead of three, and each chest has three drawers. Each of the thirty drawers contains either a diamond, an emerald, or a ruby. They are dispersed in the following manner:

1. D D D
2. D D E
3. D D R
4. D E E
5. D E R
6. D R R
7. E E R
8. E R R
9. E E E
10. R R R

[Of course, D stands for diamond; E for emerald; and R for ruby. So for example, Chest 4 contains one diamond and two emeralds; Chest 7 contains two emeralds and one ruby. There are ten jewels of each of the three types, and they are distributed in all ten possible ways.]
"You open one of the thirty drawers at random and find a diamond. Then you open another drawer of the same chest. What is the probability that it also contains a diamond?"

I will use the power of simulation to attain a nearly perfect answer, like so:

	import copy
	import random
	chests = [
	['D', 'D', 'D'],
	['D', 'D', 'E'],
	['D', 'D', 'R'],
	['D', 'E', 'E'],
	['D', 'E', 'R'],
	['D', 'R', 'R'],
	['E', 'E', 'R'],
	['E', 'R', 'R'],
	['E', 'E', 'E'],
	['R', 'R', 'R']
	]
	iterations = 100000
	found_at_least_one_diamond = 0
	found_two_diamonds = 0
	for _ in range(iterations):
	# Not really necessary, but I abhor the idea of that jeweler's chests being disturbed!
	chest = copy.copy(random.choice(chests))
	random.shuffle(chest)
	if chest[0] == 'D':
	found_at_least_one_diamond += 1
	if chest[0:2] == ['D', 'D']:
	found_two_diamonds += 1
	print(found_two_diamonds / found_at_least_one_diamond)

view raw ten-chests.py hosted with ❤ by GitHub

The most recent run outputting 0.5020185586888407 was extremely close to the analytical solution given by the author of ½. Monte Carlo simulations are really something else.