The Monty Hall Problem Explained

5 Responses to “The Monty Hall Problem Explained”

1. carotids says:

   October 4, 2009 at 12:25 am

   Brilliant. Thanks!
2. Philip says:

   October 5, 2009 at 10:50 am

   I would like to add another very simple way of looking at this problem from Edward on this thread: http://groups.google.com/group.....4212?pli=1

   If you don’t switch, the only way you can *win* is if the car is behind your door. If you switch, the only way you can *lose* is if the car is behind your door. There is a 1/3 probability that the car is behind any particular door, so switching *doubles* your odds of winning. This is a plain fact.

   Monty knows where the car is. His showing you an empty door in no way alters the odds.

   The analogy with picking a card is exact. If you switch, your odds of winning are 51/52. That is, the only way you can lose is if you *correctly picked* the ace of spades originally. The person showing you 50 cards **cannot change this fact**. If you don’t switch, your odds of winning are 1/52. If yo switch, the probability *must* be 51/52.

   Similarly, Monty can not change the fact that if you switch the only way you can lose is if you initially picked the right door. The probability of that is 1/3, so the probability of winning is 2/3 if you switch.

   Edward
3. Roy Malone says:

   November 18, 2009 at 8:45 pm

   Hi Philip, that’s a really good explanation you’ve given. Since I am currently trying to learn more about probability, I came across the monty hall problem and I was a bit puzzled with it. I saw saw your video on youtube and how you solve the problem also by listing all the events in the sample space. If we use the frequency approach(successes vs. total trials), it is true that the player only has a 1/3 rd chance if he stays with his initial selection and 2/3 rds if he switches…but the important condition is that the game must be repeated under the similar conditions a large number of times. Do you think these probabilities still hold even if the game is to be played only once ever. And what if in this one and only game setting, the prize is behind the door that the player initially selected and he’s presented with the choice of switching or staying. Do you think that he should still switch. I know that you are only saying that by switching, the player has 2/3rds chance, not 100%. Or, in this case, is it true that probability doesn’t exist just like De Finetti said…I know this might come across as an argument being advanced by an amateur who’s mixing up stuff…I am really interested in developing my probabilistic thinking and I was impressed by the way how you clearly explained the monty hall problem which was a bit confusing to me…I thank you for your effort first and would be grateful if you let me know what you think about my question…cheers
4. Philip says:

   November 19, 2009 at 1:51 am

   Hi, Roy. Yes, the probability is always the same regardless of whether the game is played once or many times. Imagine rolling a die: the chance of getting a six is one out of six. This is true even if you only roll one die in your entire life. Also remember that even if YOU only play the game once in your entire life, many other people are playing it, so the idea that it only happens once is misleading anyway.

   Even if switching gives you the best chance of winning the prize, if you know the prize is behind your door obviously you shouldn’t switch. Probability is only important in unknown situations. This is why people cheat at poker and blackjack etc. If you can somehow mark the cards and know what’s in other people’s hands, you no longer have to rely on probability.

   I think the real question that you are skirting is, “Is probability always important?” Suppose I put a box in front of you and tell you that there is a 90% chance that it contains $1 million, and a 10% chance that it contains a bomb and will explode if you open it. Should you open it? Statistically, on average, over the long run you will be better off opening the box because you will almost always end up rich. Still, you can’t take me up on this offer because you obviously can’t do it a second time if oops the first time you explode. In that sense, it “matters” if you can do it only once or many times, but the probabilities themselves are always the same.

   Another good example is insurance. The chance of me getting into a terrible accident or dying of a horrific disease is very low, and for most people, in hindsight, insurance is a very bad investment. I mean, that’s the whole principle of insurance and why it works in the first place! Still, the odds don’t “matter”: the fact is that I simply must have insurance regardless of what the odds are, because I can’t afford to be injured or get sick.

   Anyway, the point is, the odds are what they are, but sometimes you can’t base your decisions on them.
5. chuck says:

   February 26, 2010 at 7:10 am

   Look at it another, extreme way. Imagine 100 doors, 99 have goats behind them, one has $1000000. Pick a door. There are 99 other doors. Monty then opens 98 of those other doors, leaving only the one you picked (1/100 chance of having the million $ ) and one other. Do you switch?

   Heck yes! Monty knew which door to leave shut.

   If anyone thinks the odds are 50:50 after that, well…