## Monty Hall Problem

24 Oct

I came across this puzzle yesterday which was quite puzzling. It is termed the famous Monty Hall problem. It goes like this.

There are 3 doors one of which contains a “car” and the other two have “goat”. The contestant has the option to choose a door. The host, before opening the opted door, opens another door which always has a goat. Remember, the host knows what lies behind all the 3 doors. Now, the host asks the contestant if he will be willing to switch his earlier decision about the opted door. The question is this. What is the probability that the contestant will win by switching his earlier decision versus going with his original decision?

I will give you the solution. But, the more interesting part lies in understanding the solution. It looks like, by switching the earlier decision, the contestant has a 2/3 probability of winning when compared to going with his original decision which only has a 1/3 probability of winning…. Our intuition might suggest that, after the host revealed 1 door, we have 2 doors and the probability of winning, irrespective of switching should actually be 1/2. But, that is not true… Something to think about….or not….;)

Posted by on October 24, 2007 in From AM-KICKING blog, Informative, Trivia

### 5 responses to “Monty Hall Problem”

1. October 25, 2007 at 7:46 am

@mindframes: I don’t know about a mathematic proof, but since the possiblities are only 3 its easy to conclude the odds as 2/3.

image the three choices are G1, G2, C

1. contestant chooses G1. Host has to open up G2. Contestant switches choice (as per definition) and gets C. Win!
2. Contestant chooses G2. Host has to open up G1. Contestant switches choice to C and wins!
3. contestant chooses C. Host opens up G1 or G2 (doesn’t matter), and contestant switches to G2 or G1 and fails

So the wins are 2 out of 3.

2. October 25, 2007 at 3:28 pm

@mano: that’s a very good way to think about it… Basically, the host’s move gives out information which doesnt seem evident… Infact, if a new guy is introduced into the room and if he hasnt been told about the initial selection, his probability of getting it right will be 50% as we would normally expect…

Now, you could think on this next problem called the cereal box problem… Let us assume that there is a prize in each cereal box and there are a total of 5 prizes uniformly distributed in the cereal boxes. What is the minimum number of cereal box you should buy in order to get all 5 prizes?

3. October 25, 2007 at 7:02 pm

Very interesting counter intutive puzzle.
Good explanation Mano.

4. October 28, 2007 at 5:44 am

Super interesting problem…in fact mathematicians are at odds as to what is the correct solution…will give my 2 cents worth tomomrrow 🙂

5. October 29, 2007 at 12:04 am

@ Mano: very good intuition…this is a classic case of bayesian updating.. think about it this way…you have made a choice of a particular door and the host reveals another door..now the probability that the gift is behind the third door that was not chosen is the mirror problem of the assumption that the initial choice made by the contestant is wrong…which is 2/3, hence we should switch…however note that the interpretation of the problem is critical for the answer to be 2/3 and switching is optimal (Marilyn Savant outlines the conditions under which this solution is correct).