Skip to content


Three doors…and probability

You’ve likely seen the “three doors of choice” in cartoons, TV shows, and movies. In theory, the general view is (and I saw this recently in the movie 21) that if you have options of door A, B, and C, you have a 33% chance of getting it “right”. Let’s say you select door A. If one door (let’s say door C) is then eliminated, the question becomes: should you change your choice? On the surface, it would appear as if you are still dealing with the same percentages…but in reality, you have a better chance of getting the right door if you change your choice to door B, as it now has a greater possibility of being correct (door A still has 33% possibility (not 50% as would be assumed with the existence of only two doors), but the potential of door B being correct is now 66% as the probability of door C is subsumed into door B). What’s the point of this? Well, according to an article in NY Times, many prominent experiments on cognitive dissonance don’t account for the potentiality shift to the remaining unselected option…and as such, these well known experiments may not be as authoritative as is often thought.

Posted in Uncategorized.

3 Responses

Stay in touch with the conversation, subscribe to the RSS feed for comments on this post.

  1. if, by removing Door C, you’re saying it’s not the right on, then Door A does have a 50/50 chance of being right. If you remove Door C without indicating if it was “right” or not, then Door A still has 33% chance, as does Door B. Nothing would have changed in that case…

  2. I love this problem and will continue to follow the experimental stats on this site: http://math.ucsd.edu/~crypto/Monty/monty.html

  3. gsiemens said

    Hi D’Arcy - just to clarify - if by removing door C and saying “it’s not it”, then, in terms of probability, door B (mathematically/statistically) is more likely to be correct…i.e. you are ahead if you change your guess from door A to door B. When you chose A, B, C, all three doors had a 33% chance. When one door is eliminated, then door A retains its possibility of being correct (33)…and door B is now statistically more likely to be correct. The new information of what is NOT there changes the mathematical odds of the remaining doors being the correct choice.

    The second question you raise - i.e. about the possibility of doors being correct if you haven’t told anyone what was behind the one door you eliminated, then I would suspect odds would remain the same. It is the introduction of new information (”it’s not here”) that changes the possibilities.

Some HTML is OK

(required)

(required, but never shared)

or, reply to this post via trackback.