Of Deals and No Deals

We have seen an explanation of the Monty Hall problem of “Let’s make a deal”, using a 100 door-approach. Let’s imagine a 26-door version. You select one, and the host opens 24 doors that do not have the car. Will you switch your choice to the last door standing? The answer is an overwhelming yes.

Now, switch to another game, namely the deal or no deal. The show contains 26 briefcases containing cash values from 0.01 to 1,000,000 dollars. The player selects one box and keeps it aside. Cases are randomly selected and opened to show the cash inside. Periodically, the dealer offers some money to the player to take and quit the game. If the player refuses all offers and reaches the end, she will have to take whatever is in the originally-chosen case.

Monty’s deal or no deal!

Imagine there are just two boxes left – the one you have selected and the one remaining. The remaining cash values are one and one million. Enter Mr Monty Hall and offers a switch. Will you do it? After all, the original recommendation for the case described in the first paragraph was to switch! But here, there is no need to swap, as you have a 50:50 chance of winning a million from your box. Why is that?

Bayes to the rescue

Before I explain the difference, let’s work out the two probabilities using Bayes’ theorem.
First, the original one (Let’s make a deal): Let A be the situation wherein your chosen door has the car behind and B the one where 24 gates did not have it.

P(A|B) = \frac{P(B|A)*P(A)}{P(B|A)*P(A) + P(B|A')*P(A')}

Substitute the values, P(A) = (1/26), P(A’) = 25/26 and P(B|A) = 1 (the chance of 24 doors have no car given the car is behind your chosen door. Now think carefully, what is P(B|A’), the probability of having no cars behind those 24 doors, given yours does not have a car? The answer is: it remains one because it was the host who opened the door with full knowledge; it was not a random choice!

P(A|B) = \frac{1*(1/26)}{1*(1/26)+ 1*(25/26)} = \frac{1}{26}

So, switching increases your chances to 1 – (1/26) = 25/26.

Second, the new one (Deal or no deal version)

P(A|B) = \frac{P(B|A)*P(A)}{P(B|A)*P(A) + P(B|A')*P(A')}

As before, P(A) = (1/26), P(A’) = 25/26 and P(B|A) = 1. Here is the twist, P(B|A’) is not 1 because the situation of 24 cases did not produce a million came at random and was not due to your host. The probability of that happening, given your case doesn’t contain the prize, is 1 in 25. So P(B|A’) = (1/25).

P(A|B) = \frac{1*(1/26)}{1*(1/26)+ (1/25)*(25/26)} = \frac{1/26}{(2/26)} = \frac{1}{2}

Amazing, isn’t it?

Reference

The Monty Hall Problem: The Remarkable Story of Math’s Most Contentious Brain Teaser: Jason Rosenhouse

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