Monty Hall Problem – Solved

thecaligarmo — Thu, 14 Jan 2010 02:18:25 +0000

Before the holidays I posted a question to you all to see what you thought. The question was the famous Monty Hall problem where there are 3 doors and a contestant is told 1 has a fabulous prize while the other 2 have goats. The contestant selects a random door and the host reveals one of the doors that have a goat and gives you a chance to switch doors if you want or not. The question was: is it better to keep the original door, switch doors, or are the odds the same? Here’s the link if you want a refresher.

So most people will look at this problem and say the chances are the same whether you switch or keep the same door. The logic usually follows like this: There are 2 doors left. One has a present, the other a goat, so there is a 50% chance that either of them will have a goat. Thus each door has the same probability of having the prize.

Although this is sound logic, there is only one big mistake, the assumption that the 3rd door doesn’t matter. Let’s start from the beginning and work our way up in order to see what the answer is. Since we first choose a random door let’s pretend it’s door A. We now have 3 options: the present is behind door A, the present is behind door B, the present is behind door C.

If the present is behind door A: We chose door A, so the host then chooses either door B or C (both have goats so it doesn’t matter) and opens it. If we switch to either B or C (whichever the host does not choose) then we lose, if we stay on A then we win.
If the present is behind door B: We chose door A, so the host must select door C to reveal the goat. So if we switch to B we win, and if we stay on A then we lose.
If the present is behind door C: We chose door A, so the host must select door B to reveal the goat. So if we switch to C we win, and if we stay on A we lose.

Since those are the only doors available those are only ways of playing. Now let’s look at the winning ratio. If we stayed the same we would have won 1 game. If we switched then we would have won 2 games. So by switching doors you have increased your chances of winning to 66%! So remember to always switch when given a choice!

Monty Hall Problem

thecaligarmo — Wed, 30 Dec 2009 22:31:48 +0000

Time for a pop quiz! Say you are on a game show. You are presented with 3 different doors and are told that 2 of the doors contain goats and the 3rd door contains 6 quadrillion dollars! You are told to choose which door has the money, so you choose door number 1. At this point the game show host turns to you and says, “What a choice! Let’s look at what was behind door number 2. It was a goat! I’ll give you another chance since you know that the second door was a goat. Do you want to stay with door number 1 or switch?” What is the better choice: to stay with door number 1, to switch to door number 3, or do both doors have the same probability of having the goat/money?

Key points:

you choose a random door out of 3 doors (2 of which are ‘bad’ and 1 of which are ‘good’)
the game show host reveals that one of the other 2 doors is one of the ‘bad’ doors
you are asked whether you would like to keep the original door or switch
Which is better to stay, to switch, or are both the same chance?

This is called the Monty Hall problem because there used to be a game show back in the day that was put on by a guy named Monty Hall. I’m sure it was a super fun show!!!

[Solution after the holidays!]

The Cali Garmo Does Math » Probability

Monty Hall Problem – Solved

Monty Hall Problem