Decide for Yourself

[TonyV]

There are two ways I've found to make the answer more intuitive.

First, imagine that instead of three doors, there are 1,000,000 doors, and Monty knows which one the prize is behind.  You pick your door (let's say, door number one), then Monty opens up 999,998 "goat" doors to build suspense.  Okay, so now there are two doors left: door number one (which you picked), and door number 411,037.  It should be obvious that although you may have been extremely lucky and you happened to pick the one out of a million doors that has the prize, it's much more likely--999,999 times more likely, to be precise--that the prize is behind door number 411,037.

The second way is to put it like this.  When you first pick a door, your odds of picking the right one are 1 in 3.  Then Monty opens up a "goat" door.  This isn't random, it's a foregone conclusion.  There's no correlation between this action and your odds at all.  Whether you pick the prize door or a goat door, whether you pick door one, two, or three, it's irrelevant.  There is a 100% chance that the door that Monty opens has a "goat" behind it.  Because of this, it has absolutely zero bearing on your probability--it's still 1 in 3.  Since your probability is still 1 in 3 that the prize is behind your door, the probability that the prize is behind the other door is 2 in 3.

Guy has fallen into what I call the "comet fallacy."  (There might be a more formal name for it, but I don't want to look it up now.)  It goes something like this: Either a comet will hit the earth tonight and kill us all, or a comet won't hit the earth tonight and kill us all.  Therefore, there is a 50/50 chance that we're all going to go up in a fiery blaze tonight.  Sleep tight!"

Of course, it doesn't take into account the possibility that these two choices have unequal weighting.  This is illustrated by his dogged insistence that because there are two doors and one prize, the odds are 50/50.  He's not taking anything into account that has preceded his choice.  What he's doing is claiming now that he misunderstood the question.  Actually, because he doesn't even want to admit that, he's presenting it as if I stated the question wrong in some way.  He's trying to present it is if my initial description of the scenario involved two separate, unrelated events--choosing one of three doors, and choosing one of two doors.

However, if you re-read the transcript, it's plainly evident that that is not what I described, that I made it perfectly clear exactly how we arrived at there being two doors.  Reading the Wikipedia article, my description is as classic a description of the Monty Hall problem as there can be.  Truth is, he just plain got it wrong and now doesn't want to admit it.

[eabrace]

I believe it would be a 2/3 chance of winning, not 9/15.

For the Monty Hall scenario Tony is after, yes.  For the scenario I outlined with the 15 outcomes, no.  Note that my example assumed the rule of "if you pick the correct door on the first try, the game is over and you win."

[GuyPerfect]

What he's doing is claiming now that he misunderstood the question.  Actually, because he doesn't even want to admit that, he's presenting it as if I stated the question wrong in some way.  He's trying to present it is if my initial description of the scenario involved two separate, unrelated events--choosing one of three doors, and choosing one of two doors.

However, if you re-read the transcript, it's plainly evident that that is not what I described, that I made it perfectly clear exactly how we arrived at there being two doors.

I misunderstood the question. I'll even go so far as to state that it happened because of a carelessly hasty interpretation without adequate consideration to make sure I knew what you were talking about. If that's the case, though, then it's less about Guy falling into Tony's ingenious trap and more about Tony wishing he did. So by all means, carry on.

Having said that--and this is the part where we continued to go back and forth with Tony being a block head and me being the only rational mind in the conversation (see what I did thar?)--prior actions or knowledge or Monty's shady past have no bearing on the likelihood that either of the two remaining doors after elimination have the prize behind them. The 2/3 figure comes from a process where you make two distinct selections, the second of which based on the first. Saying "which of the remaining two doors is more likely to contain the prize" explicitly presents a scenario where there is only one decision, which is where the 1/2 figure comes from.

A less ambiguous way to present the mental exercise, Tony--which I know you'll keep in mind though openly refute here in public--is to phrase it like this: "The question is, statistically speaking, if you want to win the nice prize, should you stay with your original choice, or switch to the other one?"

[TonyV]

prior actions or knowledge...have no bearing on the likelihood that either of the two remaining doors after elimination have the prize behind them.

This is exactly what happens.

Before Monty opens the "goat" door, thus eliminating it as a possibility, that door has a 33.33...% chance of having a prize behind it.  After Monty opens the "goat" door, that door has a 0% chance of having the prize behind it.  So that right there is proof that something has happened, "likelihood"-wise.

Now, the probability must add to up 100%.  If Monty opens up door number two, then before he opened it, doors one and three each had a 33.33...% chance of having the prize.  Now, the 33.33...% chance that door number two had must have been distributed somehow between doors one and three.  Therefore, prior actions have had some impact on the probability of at least one of those doors, one or three, that each has the prize behind it.

Thus...

prior actions or knowledge...have no bearing on the likelihood that either of the two remaining doors after elimination have the prize behind them.

has been demonstrated as false.

The knowledge part of it comes in determining how exactly that probability is split.  If Monty is ignorant of which is the prize door and he just happened to open a goat door, then the probability gets split evenly between the doors that are left.  If Monty knows which door the prize is behind and he deliberately opens a goat door, then his actions are not random, and your door's probability hasn't changed.

I can write that program to test it too, if you want.

[TonyV]

By the way, I was a bit amused today.  I asked a guy at work this very problem.  When I got to the, "Which is the better option, staying with door number two, or changing to door number three?" he said without hesitation, "stay with door number two."  When I asked him what he thinks the odds are that the door he originally picked has the prize, he told me, "Around 70%."

 

In all the years I've asked people this, I don't think anyone has ever contended that their original door was more likely.  I asked him why he thought that.  His answer was, to paraphrase, "Because you have to be confident in your choices.  If you're confident, you increase your odds of success."

I have to admit, I just plain don't have an answer for that.  There's a tiny little evil part of me that thinks that I can use that to my advantage, though.

[eabrace]

That's not probability, that's quantum mechanics.

[Sekoia]

For the Monty Hall scenario Tony is after, yes.  For the scenario I outlined with the 15 outcomes, no.  Note that my example assumed the rule of "if you pick the correct door on the first try, the game is over and you win."

I was responding to your scenario of 15 outcomes. Your fifteen outcomes aren't all equally likely. Some are weighted twice what the others are.

Consider a scenario with two doors, X and Y. You choose one, then you see if you picked the right one. There are four scenarios:

If door X is the winning door and you select door X, you win 100% of the time.
If door X is the winning door and you select door Y, you win 0% of the time.
If door Y is the winning door and you select door X, you win 0% of the time.
If door Y is the winning door and you select door Y, you win 100% of the time.

Each of those four scenarios are equally likely so your odds of winning is the average of their respective chances.

(100% + 0% + 0% + 100%) / 4
= 200% / 4
= 50%
= 1/2

Now consider a scenario with three doors, A, B, and C. You choose one, then you see if you picked the right one. If you didn't, you're allowed to choose a second door, then see if you picked the right one. (This is the scenario you outlined.)

Well, if you pick right the first time, you win 100% of the time. If you pick wrong the first time, you end up in the first scenario above -- which means your second choice will win 50% of the time. We thus actually have nine scenarios:

If door A is the winning door and you select door A first, you win 100% of the time.
If door A is the winning door and you select door B first, you win 50% of the time.
If door A is the winning door and you select door C first, you win 50% of the time.
If door B is the winning door and you select door A first, you win 50% of the time.
If door B is the winning door and you select door B first, you win 100% of the time.
If door B is the winning door and you select door C first, you win 50% of the time.
If door C is the winning door and you select door A first, you win 50% of the time.
If door C is the winning door and you select door B first, you win 50% of the time.
If door C is the winning door and you select door C first, you win 100% of the time.

Thus the odds of winning are:

(100% + 50% + 50% + 50% + 100% + 50% + 50% + 50% + 100%) / 9
= 600% / 9
= 66.67%
= 2/3

[GuyPerfect]

If Monty opens up door number two, then before he opened it, doors one and three each had a 33.33...% chance of having the prize.  Now, the 33.33...% chance that door number two had must have been distributed somehow between doors one and three.  Therefore, prior actions have had some impact on the probability of at least one of those doors, one or three, that each has the prize behind it.

Don't you just love this guy? This is what I stared at for over three hours, and somehow, my brain is still intact.

If you have two doors and one prize behind one of them, then the likelihood of each having the prize is 1/2. That's a simple calculation based on exactly that definition: prizes / doors. So you started with three doors and eliminated one? You're left with two doors. Maybe you've eliminated five doors? Still left with two doors. And one prize. Contradict as you may, Tony, that calculation will never change. Two doors with one prize is a 50% chance for both doors having the prize, regardless of how many Alpha Bits you had for breakfast.

In the broader context of the conversation, there are two very different potential scenarios present: 1) If you pick a door, one is eliminated, and you pick a door again, you can consistently average in your favor if you know how to pick. 2) If you pick a door, you have a 1/x chance of picking the one with the prize no matter what color underpants Hitler liked to wear.

What you're doing, Tony, is taking the logic and statistics of scenario #1 and trying to apply them to scenario #2 with the smugly arrogant desire to paint me as an error-prone dunce. No joke: you felt so strongly about it that you literally suggested a coin toss will result in a 33/67 split among other strange and unquestionably false claims. If you want to write that program, feel free: flip a coin when there are 2 doors left and you will find that you pick the correct door 50% of the time (in direct contrast to your assertion, as cited in the linked post).

Naturally, you don't have to do that because it's not the situation you were asking about in the first place. But it was certainly the situation I had in mind when you asked, and it's the situation you continue to attempt to apply the 2/3 statistic to every time I bring it up. And so certain are you that your communicative skills are pristine and infallible--so determined are you to disagree with me when my answer is not the one you've convinced yourself to be the only applicable truth--that you begin to say such nonsensical things as how 1/2 is equal to 67% (but only when Monty Hall knows certain things).

There is misinterpretation present here, and there are also egomania and irrationality. These are distributed between us two persons, and 2/3 of them get distributed to you. (-:


Bonus Material

Since Tony and I have been over this before, I'm gonna save the thread the suspense and reveal in advance how he will reply:

He'll start by disregarding the validity of my first paragraph in its entirety, claiming it's utterly inapplicable for one or both of two reasons: 1) it's not what he had in mind, and/or 2) he's still caught up on there having been a door eliminated and totally misses the point where the context no longer applied to that door. Bonus points if he says something about what Monty or the contestant knows, and additional bonus points if he refers to me with the words "not getting it."

Next, he'll say "No, there never was scenario number 2. It was scenario number 1, exactly as I repeatedly described." Again, he'll totally gloss over the fact that it's possible to interpret his original question in the context of scenario 2.

Since the next paragraph contains a link to something dumbfounding that he said, and since he's not actually a total looney and is ballsy enough to accept when he's said something dumbfounding, he'll creatively deflect the point I was making using one or more of an undefined number of clever comebacks. While there's no way to know for certain what he'll say, the two most likely candidates involve him telling me "you're just wrong," and referencing a program that he did write last night (which enumerates failures and successes in the context of both picks rather than only the pick where there were two doors left). In a streak of consistency, he will be three-for-three oblivious to the possibility that I may have had a different scenario in my head than he did in his and presented my answers accordingly. Bonus points if his first three words are "no I'm not."

The last paragraph is basically troll bait in the form of a big (albeit subdued) flame, and he may or may not respond to it. Should he decide to respond, expected remarks range from the "you shouldn't have misinterpreted me you shameless twit" to "just admit that you've made a mistake." Whatever his response (if there is one), there will be no accommodation made for the notion that I might actually know what I'm talking about.

Finally, after he reads the bonus material, he'll have to re-think what he wants to say. Tony's an odd beast who likes to throw wrenches in the works, so it's difficult to predict accurately what he will do. Typical responses (in the general context of "the internet") when faced with such a situation is for the poster in question to say something along the lines of "you think you can predict me but you're just as wrong as ever." Having said that (literally because I typed it just now), it's possible he may very well not bring it up. Though again, the general trend is "you say that, so you must be wrong a lot" or some other feeble retort that Tony's too civilized to try... I dunno. It'll be exciting to see what happens next!

[Diellan]

Don't you just love this guy? This is what I stared at for over three hours, and somehow, my brain is still intact.

If you have two doors and one prize behind one of them, then the likelihood of each having the prize is 1/2. That's a simple calculation based on exactly that definition: prizes / doors. So you started with three doors and eliminated one? You're left with two doors. Maybe you've eliminated five doors? Still left with two doors. And one prize. Contradict as you may, Tony, that calculation will never change. Two doors with one prize is a 50% chance for both doors having the prize, regardless of how many Alpha Bits you had for breakfast.

In the broader context of the conversation, there are two very different potential scenarios present: 1) If you pick a door, one is eliminated, and you pick a door again, you can consistently average in your favor if you know how to pick. 2) If you pick a door, you have a 1/x chance of picking the one with the prize no matter what color underpants Hitler liked to wear.

Except that option (2) was never in consideration. He was always asking it in terms of option (1). You have tried to turn it into option (2), which is where the whole confusion comes up (because Tony is always thinking in terms of option (1)).

The issue of "having chosen first (correctly or incorrectly, you don't know) and then eliminated doors and given a chance to repick" is absolutely crucial and necessary to the problem. If you try to rule that out in any way, then you're talking about an entirely different problem and by bringing it up, you're only serving to confuse your opponent.

His initial description of the problem was correct and valid for option (1). You answered incorrectly, and when he told you the answer you didn't believe it. From there on out, the conversation got increasingly confused because of the introduction of option 2.

[GuyPerfect]

That's the conversation in a nutshell, yeah, except the "introduction of option 2" happened before I "answered incorrectly."

[Diellan]

That's the conversation in a nutshell, yeah, except the "introduction of option 2" happened before I "answered incorrectly."

I guess that depends upon whether or not something happened before this:

TonyV: Well, here's the set up.  You're on Let's Make a Deal with Monty Hall.  He gives you a choice of three doors to pick.  Behind two of them are "goat" prizes--worthless stuff you don't want.  Behind one door is something really nice worth a lot. So you pick a door.  Go ahead, pick a door.  Which one do you want?  Door number one, door number two, or door number three?

Guy Perfect: Let's go with door number one.

TonyV: Okay, now Monty, like me, knows which door the prize is behind.  But like any good showman, he wants to build up the drama and suspense. So when you pick door number one, he says, "Okay, let's open up one of the other doors first.  How about we see what's behind...  Door number two!" Obviously, he's going to open up one of the goat doors. So now there are two doors left.  Monty says, "Okay Guy, I'll tell you what.  Are you still feeling good about door number one?  Because if you want, I'll let you change your mind and go with door number three.  Of course, if you like door number one, you are more than welcome to stick with that one." The question is, statistically speaking, if you want to win the nice prize, should you stay with door number one, or switch to door number three?  Or does it make a difference?

Guy Perfect: Well, my brain is telling me that since I still know nothing about doors 1 or 3, there may as well be an even 50/50 for it being the prize.

TonyV: The answer is, statistically speaking, if you stay with door number one, you only have a 1/3 chance of winning.  By switching to door number three, you increase your chances to 2/3. So if you want the prize, you should switch to door number three.

Guy Perfect: I don't believe you.

TonyV: No one ever does.  Would you believe Marilyn vos Savant? The reason why basically boils down to one fact: Monty KNOWS which door the prize is behind. By opening up a "goat" door, that door's original probability (1/3) is shifted to any doors left that you didn't pick, but not yours, because you made your choice in ignorance.

Guy Perfect: I'm going to say that Marilyn vos Savant is full of hot air. The situation wouldn't have been any different if I'd picked door number 3.

Because this only talks about option (1), not option (2), and you answered incorrectly.

[GuyPerfect]

In the context of option 1, I would have answered incorrectly, yes. But didn't I already say I misinterpreted it as option 2? I read it as "Which is better: door 1 or door 3?" instead of "Which is better: stay or switch?"

[eabrace]

Now consider a scenario with three doors, A, B, and C. You choose one, then you see if you picked the right one. If you didn't, you're allowed to choose a second door, then see if you picked the right one. (This is the scenario you outlined.)

OK, I can see how you get 2/3 out of that, but the math looks a little different in my head.  Let's see if it falls out the same.

With three doors, any door you pick has a 1/3 chance of winning on your first pick and a 2/3 chance of losing.

These outcomes represent all the possible outcomes for your first pick, and 1/3 + 2/3 = 1, so we're not violating any laws of probability.

If you don't win on your first pick (2/3 chance), then you now have a 1/2 chance of winning and a 1/2 chance of losing on your second pick.  Since that represents all of the possible outcomes of your second pick and 1/2 + 1/2 = 1, we're still not violating any laws of probability.  But, since we only had a 2/3 chance of getting to the second pick, the odds of winning from a second pick are 1/2 * 2/3 = 1/3 and the odds of losing on a second pick are also 1/3.

Odds of winning on first pick + odds of winning off second pick + odds of losing are 1/3 + 1/3 + 1/3 = 1, so again, still not breaking any laws.

And odds of winning off first pick + odds of winning off second pick are 1/3 + 1/3 = 2/3.

So, slightly different approach, but same answer.  I'll buy that.

[TonyV]

It really boils down to one thing: Your odds after your first pick don't change.  On your first pick, you had a 1/3 chance of picking the correct door.  After Monty opens up a goat door, your odds that the door you picked has the prize behind it are still 1/3.  There really is no probability of 1/2 that ever enters the equation.

If there were four doors, you picked one, then Monty opened up two goat doors, the odds that the first door you picked has the prize remains 1/4.  If there were seven doors originally, the odds are 1/7.  For any n doors, if you pick one, then Monty opens n-2 doors, leaving just the door you originally picked and one other door, the odds that the door you originally picked has the prize behind it are 1/n.  By the laws of probability, that necessarily means that the odds that the other door has the prize behind it are (n-1)/n.

The thing that Guy is overlooking is that he fell into the trap that the problem is designed to make people fall into: to disregard circumstances leading up to a decision, then assuming that all probabilities must be equal.  It's really diabolical.  There's an old trick question that exploits the converse.  It goes something like this: If you flip a fair coin 587 times and it comes up heads 587 times straight in a row, what are the odds that it will come up heads on the next toss?  The answer, of course, is 50%.

Most people have that pounded into their heads at some point during their mathematical education, and so when a problem such as the Monty Hall problem is presented to them, they automatically mentally make it equivalent.  There are two doors, one prize, and thus the odds are 50/50 for each door; it doesn't matter which you pick.  Guy actually says this explicitly in his OP:

[10:00:16 PM] Guy Perfect: You said there was one prize.
[10:00:18 PM] TonyV: True.
[10:00:22 PM] Guy Perfect: And two doors.
[10:00:25 PM] TonyV: Yes.
[10:00:32 PM] Guy Perfect: Which means one of two doors has a prize.
[10:00:37 PM] TonyV: Yes.
[10:00:42 PM] Guy Perfect: Ta dah! 50%.
[10:00:46 PM] TonyV: Nope!

*spring!  He fell into the trap.  There was no miscommunication, there was no misunderstanding, there was no, "I thought you meant x and so I answered that way."  I didn't misrepresent the problem, I didn't screw up my presentation of it in any way.  There weren't two scenarios.

He really shouldn't feel bad.  According to the New York Times, "The experts responded in force to Ms. vos Savant's column [in which she presented the Monty Hall problem]. Of the critical letters she received, close to 1,000 carried signatures with Ph.D.'s, and many were on letterheads of mathematics and science departments."

But of course, we're expected to believe that Guy is a mathematical genius who didn't fall into the same trap that people with mere Ph.D.s fall into.  No, he came to the exact same conclusion--even for the exact same stated reason ("Well, my brain is telling me that since I still know nothing about doors 1 or 3, there may as well be an even 50/50 for it being the prize") because he simply misunderstood the problem, which was perfectly stated.

But the thing that he and other people aren't taking into account is another little fact: Just because a coin comes up heads 50% of the time after several tosses does not mean that the coin is, in fact, a fair coin!  If one side, for example, has been weighted to be significantly heavier, it could shift the probability of it coming up heads to, I dunno, let's say, 1/3.  Two possible outcomes, not 50/50.  If you bet on heads, you're probably going to lose.

That's exactly what's happening with the Monty Hall problem.  A sequence of events leads up to a choice between two options that are not 50/50, but it cleverly disguises that fact by presenting a specific sequence that, on the surface, doesn't seem to matter much.

What I find amusing is that at this point, Guy either 1) made a mathematical error that experienced mathematicians, including many Ph.D.s have made, or 2) made a reading comprehension error that a fifth-grader would have nailed.  In his extraordinary effort to try to sound smart, he is using the excuse of doing something ten times dumber.  He thinks that as long as he has technically not admitted that he was wrong, everyone will somehow magically buy it.  

Personally, I don't mind admitting I was wrong.  I've done it before, and I'll do it again.  I mean, I don't relish in it or anything, and I do diligently try to be right most of the time, and in fact, I'm a pretty bright guy who is right most of the time.  But I won't hesitate to say that when I first encountered the Monty Hall problem, I screwed it up in the exact same way that Guy did.  I was talking about it with a friend, and like Guy, I tried six ways to Sunday to convince my friend that he was utterly and completely wrong.  But once I got it, I got it.  I admitted I was wrong, he had a good laugh at my expense, and I added it to my repertoire of really cool and clever brain teasers.

Meanwhile, it's two days later, and Guy still insists that he was right.  Dude, if your goal is to look smarter, it ain't working too good.  Cut your losses already; you'd probably be surprised how much more respect admitting your wrong garners when you've been gotten than making such a fuss like this.

[Diellan]

In the context of option 1, I would have answered incorrectly, yes. But didn't I already say I misinterpreted it as option 2? I read it as "Which is better: door 1 or door 3?" instead of "Which is better: stay or switch?"

Actually, they're exactly the same question (since "stay" = "door 1" and "switch" = "door 3"). "Misinterpreting" is exactly what happens when people get it wrong.

But of course, we're expected to believe that Guy is a mathematical genius who didn't fall into the same trap that people with mere Ph.D.s fall into.  No, he came to the exact same conclusion--even for the exact same stated reason ("Well, my brain is telling me that since I still know nothing about doors 1 or 3, there may as well be an even 50/50 for it being the prize") because he simply misunderstood the problem, which was perfectly stated.

The fact that he didn't recognize the Monty Hall problem as soon as it was given tells me a lot.

[GuyPerfect]

Once more for the record: the two options in question have an equal likelihood. Also for the record, in order to say otherwise, Tony argues that a coin that comes up heads 50% of the time is not necessarily a fair coin. Can you spot how that applies? Me neither.

Say the prize is behind door 1, I pick door 1, and door 2 is eliminated. At this point, I have two options:

Choice 1: Door 1 + Choice 2: Door 1 = Win
Choice 1: Door 1 + Choice 2: Door 3 = Lose

As you, I, anyone and in fact even TonyV himself can see, that's a 50% chance of picking the correct door. Would you, the reader, suggest that it was inappropriate of me to have that situation in mind when TonyV asks the question "[I]f you want to win the nice prize, should you stay with door number one, or switch to door number three?"

The scenario Tony had in mind, on the other hand (using the above example), was this one:

Choice 1: Door 1 + Choice 1: Door 1 = Win
Choice 1: Door 1 + Choice 2: Other Door = Lose
Choice 1: Door 2 + Choice 2: Door 2 = Lose
Choice 1: Door 2 + Choice 2: Other Door = Win
Choice 1: Door 3 + Choice 2: Door 3 = Lose
Choice 1: Door 3 + Choice 2: Other Door = Win

This, as I'm sure anyone who's still awake can point out, is not a situation about making one choice ("Do you choose A or B," which happens to be the format of the question TonyV asked), but a situation about making two. When you make two decisions, you can make it so you'll average a 2/3 success rate.

A 2/3 success rate for two decisions. A question that asks you to make one. Reading comprehension of a fifth-grader, or a delusional narcissist with a bone to pick?

You decide!

[TonyV]

As you, I, anyone and in fact even TonyV himself can see, that's a 50% chance of picking the correct door.

Prove it.  I want you to propose a test to simulate this exact scenario.  Make sure that you account for door 2 being eliminated specifically because it is not a prize door.

Here's my entry, which should make it pretty spectacularly clear.  I want you to put one out there too.  And don't give me this, "pick one of two doors at random" without taking into account the stuff that led up to it, because again, that was never the scenario presented to you.  I'm at work so I don't have the Java program I gave to you, but I do have C# on my work workstation, so here's a C# console app:

Code: [Select]

using System;
using System.Collections.Generic;
using System.Text.RegularExpressions;

namespace MontyHall {
  class Program {
    /// <summary>Random number generator</summary>
    private static Random r = new Random();

    static int win_by_staying = 0;
    static int win_by_switching = 0;
    static int lose_by_staying = 0;
    static int lose_by_switching = 0;

    static void Main(string[] args) {
      // Default to running one million tests.
      int tests = 1000000;

      if (args.Length > 0 && Regex.Match(args[0], @"^\d+$").Success)
        tests = int.Parse(args[0]);

      // Re-run the test as many times as you want!  Hit Ctrl-C to stop.
      while (true) {
        win_by_staying = 0;
        win_by_switching = 0;
        lose_by_staying = 0;
        lose_by_switching = 0;

        for (int i = 0; i < tests; i++)
          Test();
        Console.WriteLine("Win by sticking with original door: {0}",
          win_by_staying);
        Console.WriteLine("Win by switching to other door: {0}",
          win_by_switching);
        Console.ReadLine();
      }
    }

    /// <summary>This is a simlation of one test.</summary>
    static void Test() {
      // First, let's put the prize behind a door, numbered 0 - 2.
      int prize_door = r.Next(3);

      // Now let's assign a door for the contestant.
      int contestant_door = r.Next(3);

      // Monty needs to open a door, so let's give him some doors to choose
      // from.  A list is created to keep track of possible doors.
      List<int> monty_doors = new List<int>(3) { 0, 1, 2 };

      // Monty can't open the contestant's door.
      monty_doors.Remove(contestant_door);
      
      // If the prize door is not the contestant's door, he can't open the
      // prize door, either.
      if (contestant_door != prize_door)
        monty_doors.Remove(prize_door);

      // Okay, which door will Monty open?
      int monty_door = monty_doors[r.Next(monty_doors.Count)];

      // Figure out what door is left.  Notice the following:
      // Contestant's door: 0, Monty's door: 1, Switch Door: 2 (= 3 - 0 - 1)
      // Contestant's door: 0, Monty's door: 2, Switch Door: 1 (= 3 - 0 - 2)
      // Contestant's door: 1, Monty's door: 0, Switch Door: 2 (= 3 - 1 - 0)
      // Contestant's door: 1, Monty's door: 2, Switch Door: 1 (= 3 - 1 - 2)
      // Contestant's door: 2, Monty's door: 0, Switch Door: 1 (= 3 - 2 - 0)
      // Contestant's door: 2, Monty's door: 1, Switch Door: 0 (= 3 - 2 - 1)
      int swtich_door = 3 - contestant_door - monty_door;

      // Now let's tally the results.
      if (prize_door == contestant_door) {
        win_by_staying = win_by_staying + 1;
        lose_by_switching = lose_by_switching + 1;
      }
      else if (prize_door == swtich_door) {
        win_by_switching = win_by_switching + 1;
        lose_by_staying = lose_by_staying + 1;
      }
      else {
        // Should never get here, because the prize door has to be either
        // the door the contestant picked or the remaining door.  Still, just
        // for the sake of being paranoid, I'll put an exception in here.
        throw new Exception("Oops, something weird happened!");
      }
    }
  }
}
Here are ten trials running 1,000,000 tests each:

Code: [Select]

Win by sticking with original door: 332982
Win by switching to other door: 667018

Win by sticking with original door: 333050
Win by switching to other door: 666950

Win by sticking with original door: 332931
Win by switching to other door: 667069

Win by sticking with original door: 333976
Win by switching to other door: 666024

Win by sticking with original door: 333057
Win by switching to other door: 666943

Win by sticking with original door: 333559
Win by switching to other door: 666441

Win by sticking with original door: 333382
Win by switching to other door: 666618

Win by sticking with original door: 333290
Win by switching to other door: 666710

Win by sticking with original door: 332551
Win by switching to other door: 667449

Win by sticking with original door: 333287
Win by switching to other door: 666713
Look pretty obvious to me that the odds of winning are not 50/50 as you keep trying to make it out to be.  So come on, man, show us your experimental trials replicating the scenario I gave you as precisely as you can.

[GuyPerfect]

And don't give me this, "pick one of two doors at random" without taking into account the stuff that led up to it, because again, that was never the scenario presented to you.

Translation:

It's impossible for someone of my irreproachable stature to fail to express something with anything less than utmost unambiguity, which therefore proves that you interpreted my question exactly as I say you did and not as you say you did.

So have you considered yet why I titled the thread "Decide for Yourself" rather than something like "Math Skills of the Wonder Boy"?

[Big King]

So how about it, folks?  Can I get a, "Hell, yeah!  Tony's right!"

Hell, yeah! Tony's right!

Actually, at this point, it sounds like you both agree mathematically/logically, and the discussion has devolved into a semantic argument.

But the real issue at hand is who let Guy out of the coding dungeon while I still don't have Sentinel 1.0 at my disposal. 

[TonyV]

TonyV: Well, here's the set up.  You're on Let's Make a Deal with Monty Hall.  He gives you a choice of three doors to pick.  Behind two of them are "goat" prizes--worthless stuff you don't want.  Behind one door is something really nice worth a lot. So you pick a door.  Go ahead, pick a door.  Which one do you want?  Door number one, door number two, or door number three?

Guy Perfect: Let's go with door number one.

TonyV: Okay, now Monty, like me, knows which door the prize is behind.  But like any good showman, he wants to build up the drama and suspense. So when you pick door number one, he says, "Okay, let's open up one of the other doors first.  How about we see what's behind...  Door number two!" Obviously, he's going to open up one of the goat doors. So now there are two doors left.  Monty says, "Okay Guy, I'll tell you what.  Are you still feeling good about door number one?  Because if you want, I'll let you change your mind and go with door number three.  Of course, if you like door number one, you are more than welcome to stick with that one." The question is, statistically speaking, if you want to win the nice prize, should you stay with door number one, or switch to door number three?  Or does it make a difference?

What's so ambiguous about this?  No one I have presented this to has had any trouble whatsoever understanding it.  I want to know exactly where it is you got confused to the point where you give the wrong answer.  You're the only person contending that it is confusing--not just that you misunderstood, oh no.  That somehow that description misrepresents the problem into some scenario you have made up after missing it--again, with the exact result and the exact stated reason that almost everyone misses this problem.

No matter how you slice it or how you dice it, there is no way that the answer to this problem is 50/50.  The only way that can happen is if you completely neglect everything that lead up to choosing between the two doors and take the second choice as a completely independent event separate from everything that led up to it.  Troll as much as you want, but it's plainly obvious to everyone that I wasn't just telling that lead-up for my health, that it is pertinent to the story.  Otherwise, my question would simply have been, "You have two doors, one of which has a prize behind it..."

As the mathematical genius that you think you are, you should be more apt to spot dependencies like this that affect probability, not less.  I mean, just before this, we were discussing probability as it relates to dice rolls.  We talked about the odds of rolling at least one one if you roll a standard six-sided die six times.  (66.5%, incidentally.)  You never said, "Oh, it's 1/6!" under the mistaken impression that I was only talking about the last die roll.  In fact, you only made your, "of course it's 1/3 / 2/3 if you take into account the preceding events" comments after I had thoroughly explained why it's not 50/50.  And then I find it telling that when you posted here, you misrepresented the conversation completely, not even posting the original presentation of the problem.

So yeah, decide for yourself indeed.  It's pretty clear to me what's going on, you're under the misguided impression that this is "yanking my chain" or something.  I just have to wonder why you're continuing to do it, surely knowing that it's impacting your credibility.