One approach that really helped me out was to extend the problem from three doors to a million doors.<p>You're allowed to pick one door, out of a million. Then, the host opens 999,998 of the other doors which he <i>knows</i> have a goat. This leaves the closed door you picked and one remaining closed door.<p>So, there's now two possibilities. Out of <i>one million</i> doors you somehow just happened to have guessed right and picked the right one. Or, you're wrong. If you're wrong then the correct door would be the other closed one.
Nice explanation. Small critique: I think this page would be better if the animations could be played manually rather than on a continuous loop which is quite distracting. Also, maybe add in a bit where the user can play the game themselves to see the logic work out.<p>Something like this was useful when I was first learning about the Monty Hall problem.<p><a href="http://www.grand-illusions.com/simulator/montysim.htm" rel="nofollow">http://www.grand-illusions.com/simulator/montysim.htm</a>
The best explanation I've seen of the Monty Hall problem is still Oscar Bonilla's: <a href="http://oscarbonilla.com/2009/05/the-monty-hall-problem/" rel="nofollow">http://oscarbonilla.com/2009/05/the-monty-hall-problem/</a>
I liked the simulation, and it reminds me of a tangentially related probability issue from early in my career.<p>While still in college, I had a part-time job testing a spreadsheet product (back in the days when there were other commercial choices). One of the things we testers did to keep ourselves entertained was that we each picked a casino game, and implemented that game in spreadsheet macros. My game was blackjack, but another guy did craps.<p>Just for kicks, the guy who built the craps game re-wired it as an autoplay simulation, and left it playing against itself overnight. When we came back in in the morning, we found that the virtual player was rich, which seemed odd. So the next night we left it running again, and again the following morning, the player had made a bundle.<p>It turns out that my coworker had accidentally discovered a bug in the random number generator: the distribution of the results was a bit non-uniform.
Very cool way of illustrating it. It looks like you just used iframes for the animations so they were in separate pages with distinct stylesheets, JS resources, etc., why did you do that rather than having the Jquery animations happen in the main document? Was it just to keep things nicely modular?
I've usually found the problem poorly worded (probably intentionally). An assumption that Monty knows which door is which is required but often not stated. If Monty doesn't know which door is which, the odds are 50-50 like you'd imagine.
I suspect that much of the contention around this famous problem is based on the original boundaries / rules of the problem being unclear.<p>The problem as stated sometimes leaves out crucial information that affects the outcome.<p>Example:<p>Game show, 3 doors. Contestant chooses one door. The host opens another door, which has a goat. Do you switch?<p>In this situation, it is not explained whether the host randomly picked one of the two remaining doors to open, without knowledge that one of them would contain a goat. This is the key missing information. If the host ALWAYS reveals the goat, then it is beneficial to switch as 2 out of 3 times the contestant wins by switching.<p>However, if the host is simply opening another door at random, and the door happens to contain a goat, then there is no advantage to switching, as the odds are 50/50.<p>edit: it appears that from the original version of the problem stated in Parade magazine, the implication is that the host knowingly reveals a goat.<p>"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
I've tried to think of the best way to make this obvious. This is what I came up with, and I'd like to hear if it's accurate:<p>Contestant picks a door out of three. 1/3rd chance he's got the car. Then he is given the choice to either stick with his door, or do a coin-toss where you have to pick heads.<p>In this case, it seems obvious that option for the coin-toss is the better option, because it gives you a 50% chance of winning.<p>The confusing part of the problem is that you feel like you're doing one thing, where really it's two independent games where one clearly offers a bigger chance of winning.<p>That said, some people here have mentioned that it's not actually a 50% chance if you switch doors. So maybe my explanation is incorrect? And if it's incorrect, is it still correct enough to show that switching doors (i.e. opting for the coin-toss) is the better option?
Looking this up on Wikipedia reveals it's relationship with the Three Prisoners Problem[1] as well. The solution presented in that case helped me in reasoning about it.<p>[1]: <a href="http://en.wikipedia.org/wiki/Three_Prisoners_problem" rel="nofollow">http://en.wikipedia.org/wiki/Three_Prisoners_problem</a>
Here's how it helps me to look at it. As others have said, you can play with the numbers, but the mechanics are basically this:<p>First step: you are choosing one option from all available choices.
Second step: you are now betting whether your original choice was the correct one.<p>By consciously narrowing down all the remaining options to one, the host is essentially asking you: "Do you think you guessed right?" As long as there are more than two available options at the start, the odds are greater than you guessed wrong.<p>As others have said, the important dynamic here is that the host _knows_ where the prize is and proceeds to open all the other doors, leaving only yours and one remaining closed. That remaining door then symbolizes an aggregate of all the doors you didn't select.
Like everyone who encountered this in their intro stats class, I love this problem. And it bothers everyone the first time.<p>Are there other problems like the Monty Hall problem, perhaps in other disciplines?
Given two choices (equal probability; redundant?) - which is the case once you have two doors - switching cannot change the odds. Have someone flip a coin. Call it. But before the flipper looks at the coin - (whether he KNOWS the result or not) he asks you to consider switching. Explain how the option of switching changes the probability of a correct or incorrect call. Inference?
The best explanation I've seen was in series 3, episode 3 of Jame May's Man Lab, where they repeatedly iterated the problem and showed that when repeated enough times, the frequency of winning/losing tends to a different value with a change of door than without (by having the presenters open booby-trapped beer cans and either get splashed with beer or not)
I can't see how your generalization of the problem to more doors is correct.<p>In the classical problem with 3 doors, if you pick a door with a car, the host will only open 1 other door.<p>In your demo, if I use 40 doors and pick one, the host opens all other doors but one. Why?<p>If the one of those 40 I pick has a car, will the host open all remaining doors? That doesn't seem to be the case for the 3-door variant.
I have a much easier explanation than this: By offering you the switch, the host has given you two doors to open - it's irrelevant that he opened one for you - so, it comes down to picking one door (the one you're on) or two doors (of which he has opened one for you, again, irrelevant - the effect is the same) ergo, it's 1/3 or 2/3.
That's a good looking demo, quite smooth. Implementing this was actually a recent DailyProgrammer challenge (<a href="http://www.reddit.com/r/dailyprogrammer/comments/1qdw40/111113_challenge_141_easy_monty_hall_simulation/" rel="nofollow">http://www.reddit.com/r/dailyprogrammer/comments/1qdw40/1111...</a>)
I always explain this problem with stars: I'm thinking of a star in the night's sky... guess which one.<p>Alright, good guess, but would you consider changing? You can either keep your original guess, or switch to this one other star I'm pointing to now.<p>Everyone switches.
Heh, Monte Carlo simulation of Monty Hall problem. Mildly amusing... if only we could work a Monty Python reference in there somewhere!<p>Watching this play out really kills any remaining gut response that would say "it's the same either way!"
some background:
<a href="http://en.wikipedia.org/wiki/Monty_Hall_problem" rel="nofollow">http://en.wikipedia.org/wiki/Monty_Hall_problem</a>
Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong<p>(per above post At least this made it to stats intro class)
I like the simulation at the bottom. Sometimes it's best to just say, "The math gets confusing, let's run this a couple times and see what happens."
In another 50 years, no one is going to know who "Monty Hall" is.<p>Except for this "problem" which people will still be arguing about.
got into a huge fight with a family member over this. i even programmed a simulation and showed it run over 1000 tries and he still refused to budge from the 50% answer.