The Uncertain Game Host

Q: You are in a game where the game host has a gold coin covered under one of three hats. However, the game host can choose to not put the gold coin under any hat with a probability of 10%. You open the first two hats and find no coin under it. What is the probability that there is a gold under the 3rd hat?

The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (And Everone Else!) A: The problem clearly requires a Bayesian approach. However, it needs to be framed correctly, or else it may not be easy to grasp. Let us start with the hypothesis we want to test for $$H$$. Clearly if the host has put in a gold coin, then the probability of winning is 1 as you have already opened up two hats. The new evidence here is that two hats have been opened and no coins have shown up. Also the probability of seeing such evidence is 1 if the host has not put coin under any hat. So the way to frame the sought probability is
$$P(H|E) = \frac{P(E|H)\times P(H)}{P(E|H)\times P(H) + P(E|\neg H)\times (1 - P(H))}$$
From the above: $$P(E|\neg H) = 1$$ and $$P(E|H) = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$$. It is important to explain how we got to $$P(E|H) = \frac{1}{3}$$. Given that the host has put in a coin somewhere there is a $$\frac{2}{3}$$ chance that the first hat flip would reveal nothing and given this empty flip there is a $$\frac{1}{2}$$ chance that the next flip would reveal nothing. Putting it all together gives
$$P(H|E) = \frac{\frac{1}{3}\times \frac{9}{10}}{\frac{1}{3} \times \frac{9}{10} + 1 \times (1 - \frac{9}{10})}\\ = \frac{3}{4}$$
Now that you have the answer, it is also intuitive to see why it remains high despite two empty flips. The probability that the host would insert is a coin is high to begin with.

Some good books to learn the art of probability

Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics) This book is a great compilation that covers quite a bit of puzzles. What I like about these puzzles are that they are all tractable and don't require too much advanced mathematics to solve.

Introduction to Algorithms This is a book on algorithms, some of them are probabilistic. But the book is a must have for students, job candidates even full time engineers & data scientists

An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (and Everyone Else!) Introduction to Probability, 2nd Edition The Mathematics of Poker
Good read. Overall Poker/Blackjack type card games are a good way to get introduced to probability theory

Bundle of Algorithms in Java, Third Edition, Parts 1-5: Fundamentals, Data Structures, Sorting, Searching, and Graph Algorithms (3rd Edition) (Pts. 1-5) An excellent resource (students/engineers/entrepreneurs) if you are looking for some code that you can take and implement directly on the job.

Understanding Probability: Chance Rules in Everyday Life A bit pricy when compared to the first one, but I like the look and feel of the text used. It is simple to read and understand which is vital especially if you are trying to get into the subject Data Mining: Practical Machine Learning Tools and Techniques, Third Edition (The Morgan Kaufmann Series in Data Management Systems) This one is a must have if you want to learn machine learning. The book is beautifully written and ideal for the engineer/student who doesn't want to get too much into the details of a machine learned approach but wants a working knowledge of it. There are some great examples and test data in the text book too. Discovering Statistics Using R This is a good book if you are new to statistics & probability while simultaneously getting started with a programming language. The book supports R and is written in a casual humorous way making it an easy read. Great for beginners. Some of the data on the companion website could be missing.

1. You said a probability of 20% of not putting the coin in any hat, but in the equation you used 9/10

Cheers,

2. You are right. Let me correct that. Thanks for pointing out.

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