Math problem

[millring]

Jun 16, 2009 15:56:04 GMT -5 patrick said:

But in that new game, you always have a 50/50 chance of picking the winner.

And beating the dead horse in midstream...

why, if given a 50/50 choice, is changing the initial one going to INCREASE my odds of getting the car?

[Fingerplucked]

Russell, it would seem that you have a 50/50 chance, if you consider only the two remaining doors.

But you have three doors when you make your initial choice. And Monte always opens one of them. He knows what's behind the doors, so he always opens a "goat" door.

When you make your initial choice, you have a 1/3 chance of getting it right. By always changing your guess when offered the chance, your odds of getting it right change to 2/3.

[Deleted]

Zogg no give flying fungus about numbers--there only three anyway. Zogg want goat. Zogg want both goats. Good eating, better than ground sloth or gamy old possum.

Just one thing. What is "door"?

[omaha]

Jun 16, 2009 15:56:04 GMT -5 patrick said:

Jun 16, 2009 15:27:36 GMT -5 @omaha said:

It doesn't, and just because there is a "best" strategy, it does not follow that you can't lose. If you pick the prize with your first guess, you'll lose, even if you follow this "optimum" strategy.

Here's a quick scenario: Assume Door #1 has the prize.

You pick Door #1 : You lose (Monte reveals door #2, you switch to #3, you're hosed).
You pick Door #2 : You win (Monte reveals door #3, so you switch to #1)
You pick Door #3 : You win (Monte reveals door #2, so you switch to #1)

Bottom line: You can improve your odds from a raw 1/3 to a respectable 2/3 with this strategy.

I think the way the odds are being calculated in this game are misleading. Because of the way the game is played, your odds were always 1/2.

Nope. With the optimum strategy, your odds are 2/3.

When you pick a door, any door, you don't get to find out if that door was the winner before the game changes. It doesn't matter which door you choose first, because they will always discard one door and then have you play again. But in that new game, you always have a 50/50 chance of picking the winner.

Imagine this scenario: I'm playing the game with Monty. He tells me to pick a door out of three. Instead, I begin doing a medley of Neil Young hits in my most obnoxious quavering falsetto voice. The producer, afraid that the audience will just run away in nausea, tells Monte to just continue and they reveal one of the losers. My game is now 1/2, without having picked a door or having to switch.

Yes, you lowered your odds by giving Monte three options instead off two.

The important thing about the scenario is not what Monte tells you, its what you tell him: By picking a door, you force his hand. He has to reveal one of the goats to you.

But all you need to do is run through the analysis from my last post. There is no scenario where switching doesn't create a 2/3 chance of winning in the end.

[Greg B]

Here's the way I visualize it.

You have three doors: One with a prize and two with a "goat".

Funny thing about that, look at this snippet of the code I wrote for the test program of the Monty Hall problem.

            int winnerDoor = rndOne.nextInt(3);
            int playerChoice = rndOne.nextInt(3);
            int goat = 0;
            switch (winnerDoor)
            {
                case 0 : if(playerChoice == 1){goat=2;}else{goat=1;}
                if(playerChoice == 0) {
                    int newRnd = rndOne.nextInt(2); 
                    if(newRnd== 1) {
                        goat = 1;
                    } else {
                        goat = 2;   
                    }
                }
                    break;

There are goats everywhere.

[loopysanchez]

The Wiki page put it in terms everyone can understand, and I paraphrase:

Let's say that instead of 3 doors, there are a million. And once you make your pick, Monty opens 999,998 doors with goats behind them, leaving only one other unopened door besides yours. Now, are you confident enough that you picked the one door out of a million with a car behind it that you're not going to change your door, or are you going to switch to the one other door out of a million that Monty, with full knowledge of which door contains the car, chose not to open?

There's a 1 in a million chance your first pick was right; There's a 999,999 in a million chance the other door that Monty willfully chose not to open has the car behind it.

Now, just reduce the number of doors from 1,000,000 down to 3, and use the same logic.

[millring]

Jun 16, 2009 20:51:35 GMT -5 loopysanchez said:

The Wiki page put it in terms everyone can understand, and I paraphrase:

Let's say that instead of 3 doors, there are a million. And once you make your pick, Monty opens 999,998 doors with goats behind them, leaving only one other unopened door besides yours. Now, are you confident enough that you picked the one door out of a million with a car behind it that you're not going to change your door, or are you going to switch to the one other door out of a million that Monty, with full knowledge of which door contains the car, chose not to open?

There's a 1 in a million chance your first pick was right; There's a 999,999 in a million chance the other door that Monty willfully chose not to open has the car behind it.

Now, just reduce the number of doors from 1,000,000 down to 3, and use the same logic.

That still doesn't really do it for me. In the case of the million doors, I might get more suspicious when Monte cautiously avoids door #549 in the midst of ALL THE OTHER DOORS that Carol Merrill opens...

But when it's only three doors standing side by each, it doesn't make much difference. Monte and Carol don't seem, in the three door instance, to be so carefully avoiding a door.

Now if it were Brer Rabbit instead of Monte Hall, and he was telling me, "Please, oh please brer John, don't go checkin' behind doa numma 3! Please, oh please, suh!" Then I might change my answer to door number three, having read Harris and stuff.

[Deleted]

Jun 17, 2009 5:17:21 GMT -5 millring said:

Jun 16, 2009 20:51:35 GMT -5 loopysanchez said:

The Wiki page put it in terms everyone can understand, and I paraphrase:

Let's say that instead of 3 doors, there are a million. And once you make your pick, Monty opens 999,998 doors with goats behind them, leaving only one other unopened door besides yours. Now, are you confident enough that you picked the one door out of a million with a car behind it that you're not going to change your door, or are you going to switch to the one other door out of a million that Monty, with full knowledge of which door contains the car, chose not to open?

There's a 1 in a million chance your first pick was right; There's a 999,999 in a million chance the other door that Monty willfully chose not to open has the car behind it.

Now, just reduce the number of doors from 1,000,000 down to 3, and use the same logic.

That still doesn't really do it for me. In the case of the million doors, I might get more suspicious when Monte cautiously avoids door #549 in the midst of ALL THE OTHER DOORS that Carol Merrill opens...

But when it's only three doors standing side by each, it doesn't make much difference. Monte and Carol don't seem, in the three door instance, to be so carefully avoiding a door.

Now if it were Brer Rabbit instead of Monte Hall, and he was telling me, "Please, oh please brer John, don't go checkin' behind doa numma 3! Please, oh please, suh!" Then I might change my answer to door number three, having read Harris and stuff.

Millring: I heard it explained with the hypothetical of a hundred doors, which made it a little easier to grasp. But, I think the mistake you are making is assuming that Monte is randomly opening doors. But, of course, Monte KNOWS where the prize is, and is only opening goat-doors.

But, here's the fundamental thing: Your odds of being correct with your first choice was FIXED at the time you made the choice, and nothing Monte did by opening doors later improved those odds. He's NOT gonna open the prize door, and he has the advantage of knowing where that is. So, your initial choice was made with a 2/3 (or 999k/1m) chance you were WRONG.

Here's why you change your pick, as gamblers explain it: The second you made your first choice, you get stuck with your 1/3 chance of being correct. But, there's a 2/3 chance that the prize is behind the other, unchosen doors, AND NOTHING MONTE DID IN OPENING THE WRONG DOOR(S) CHANGED THAT. He KNEW the unchosen door would have the goat, not the prize.

So, when he offers you a chance to change your pick, you trade your 1/3 chance for a 2/3 chance.

Hope that helps.

[Russell Letson]

Maybe I'm being dense, but I tend to think of this process as a series of "rounds" in which the odds depend on the actual available choices. In Round 1, I have three doors from which to choose, but after the goat has been revealed behind one of them, Round 2 begins--and I eliminate that door from consideration and am choosing between (or, if there are more than three doors originally, among) the remaining unrevealed possibilities. My chances are always 1/n in which n=number of unrevealed doors. Is this whole thing a matter of defining how much of the process "counts"? If n=total doors revealed and unrevealed, then the fancy arithmetic applies. Of course, the chances that Russell = dense are much higher than 1/3.

[millring]

Jun 17, 2009 11:12:55 GMT -5 @dharmabum said:

Jun 17, 2009 5:17:21 GMT -5 millring said:

That still doesn't really do it for me. In the case of the million doors, I might get more suspicious when Monte cautiously avoids door #549 in the midst of ALL THE OTHER DOORS that Carol Merrill opens...

But when it's only three doors standing side by each, it doesn't make much difference. Monte and Carol don't seem, in the three door instance, to be so carefully avoiding a door.

Now if it were Brer Rabbit instead of Monte Hall, and he was telling me, "Please, oh please brer John, don't go checkin' behind doa numma 3! Please, oh please, suh!" Then I might change my answer to door number three, having read Harris and stuff.

Millring: I heard it explained with the hypothetical of a hundred doors, which made it a little easier to grasp. But, I think the mistake you are making is assuming that Monte is randomly opening doors. But, of course, Monte KNOWS where the prize is, and is only opening goat-doors.

But, here's the fundamental thing: Your odds of being correct with your first choice was FIXED at the time you made the choice, and nothing Monte did by opening doors later improved those odds. He's NOT gonna open the prize door, and he has the advantage of knowing where that is. So, your initial choice was made with a 2/3 (or 999k/1m) chance you were WRONG.

Here's why you change your pick, as gamblers explain it: The second you made your first choice, you get stuck with your 1/3 chance of being correct. But, there's a 2/3 chance that the prize is behind the other, unchosen doors, AND NOTHING MONTE DID IN OPENING THE WRONG DOOR(S) CHANGED THAT. He KNEW the unchosen door would have the goat, not the prize.

So, when he offers you a chance to change your pick, you trade your 1/3 chance for a 2/3 chance.

Hope that helps.

It doesn't really help.

I'm really this dense. I understand that Monte knows ahead of time what's behind the doors. What I don't understand is...

"So, when he offers you a chance to change your pick, you trade your 1/3 chance for a 2/3 chance."

I understand that my choice is sort of narrowed to a 50/50. But I cannot conceive of the notion that because Monte revealed one of the goat choices, that that had any bearing on whether or not my initial choice was wrong.

There is NOTHING about Monte's reveal that made my initial choice any more wrong. In a sense, one might say that all Monte's reveal illustrated was that I had chosen one of the two possible right chances.

So, again, Monte's reveal and subsequent option (that I could change my mind) changes my odds from 1 in 3 to 1 in 2. But I don't understand why changing my initial choice increases my odds to two in three. Nothing about Monte's reveal suggested my initial choice was the wrong one.

[Deleted]

Well, Millring, rest assured no one will ever convince me of your density. The fact is, this is damn hard to explain properly.

I'll try this.... Forget everything that happens AFTER you make the initial choice. At that time ALL the doors are divided into two universes:
1. The door you picked (with it's 1/3 chance of being correct); and
2. The doors you didn't pick, with their 2/3 chance of containing the prize.

Now, as you know, Monte did nothing when he opened the other door, because he KNEW it was a goat door. NOTHING CHANGED. Just a clever trick, smoke and mirrors, basically, that does not change the odds of either universe. Your universe STILL has it's 1/3 chance, and the remaining door still has it's 2/3 chance. When he offers you the chance to change your pick, he's essentially offering to let you choose the entire universe of doors you did not pick initially (by eliminating the goat door, that is now out of play, so the entire 2/3 chance of that universe is now held by the remaining door you did not pick, initially).

If any of you know a really competant gambler, try this on them. I'm not talking about the casual poker player, but the type of guy who can make a living at blackjack or dice. They may or may not be educated, but it has amazed me that they tend to GET the right answer on this, while folks who have actually have math degrees do not. For whatever reason, it's NOT counterintuitive to a gambler. Anyway, the way a gambler explained his reasoning to me was: Monte is essentially giving you the chance to pick ALL the doors you originally excluded. In the 100-door scenario, monte has opened 98 of the doors you did NOT pick, leaving one closed among the whole universe of doors you did NOT choose. THAT single closed door you did not choose now has enormous odds of being THE door. And the single door you initially chose still has it's lousy 1% chance of being THE door.

Gotta change my pick, Monte.

[Russell Letson]

Hmph. Imagine this scenario: Posit an array of N numbered doors--say 100 to make the arithmetic easier. Above each is a red light (they have to be red or the problem won't work) and the behind-the-doors environment is populated by one car and N-1 angora goats. (No substitutes. Angora is required for obscure mathematical and metaphysical reasons that may not be revealed to the uninitiated.)

Monty (not Monte; see m&m provisions above) throws a knife switch (no buttons), causing three red lights to illuminate (see: Illuminati) and the player is invited to guess which of the thus indicated doors conceals a car, which a Federal regulatory agency mandates be actually present. (And we know that these guys are never wrong.)

With the choice made, Monty opens one red-lit door, revealing a goat and makes the now-familiar offer to change or stand pat. (Alternate wording is permitted at this point.) The kicker is this: when Monty opened the goat door, subtle machines concealed in the lintels teleported the contents of all of the remaining door cabinets to randomly-selected destinations. The red lights, however, remained lit at their original numbered positions. Are your chances of finding the car better or worse now? What if the car has been replaced by a tiger and shackles have mysteriously been teleported around your ankles? Do ya feel lucky, punk?

[millring]

Jun 17, 2009 11:44:18 GMT -5 @dharmabum said:

Well, Millring, rest assured no one will ever convince me of your density. The fact is, this is damn hard to explain properly.

I'll try this.... Forget everything that happens AFTER you make the initial choice. At that time ALL the doors are divided into two universes:
1. The door you picked (with it's 1/3 chance of being correct); and
2. The doors you didn't pick, with their 2/3 chance of containing the prize.

Now, as you know, Monte did nothing when he opened the other door, because he KNEW it was a goat door. NOTHING CHANGED. Just a clever trick, smoke and mirrors, basically, that does not change the odds of either universe. Your universe STILL has it's 1/3 chance, and the remaining door still has it's 2/3 chance.

I think MAYBE this is one of my problems with your explanation....

1. The door you picked (with it's 1/3 chance of being correct); and
2. The doors you didn't pick, with their 2/3 chance of containing the prize.
Now, as you know, Monte did nothing when he opened the other door, because he KNEW it was a goat door. NOTHING CHANGED. Just a clever trick, smoke and mirrors, basically, that does not change the odds of either universe. Your universe STILL has it's 1/3 chance, and the remaining door still has it's 2/3 chance

Actually, before Monte's reveal, each door has a 1 in three chance of being right. There is nothing "collective" about the two doors I did not choose. Before the reveal, they each still had a 1 in 3 chance of being the car door.

But after the reveal, you realize that the game was, all along, going to be a game of 50/50.

There was NOTHING about the two doors I didn't choose that made them collectively a 2 in 3 chance of being right. Thus, nothing about Monte's reveal conferred upon the other unchosen door a 2 out of three probability of being right. It was, at the beginning AND after the reveal, ALWAYS just one of the 50/50 possibilities that the game had built into it.

[omaha]

Jun 17, 2009 11:35:10 GMT -5 millring said:

It doesn't really help.

I'm really this dense. I understand that Monte knows ahead of time what's behind the doors. What I don't understand is...

"So, when he offers you a chance to change your pick, you trade your 1/3 chance for a 2/3 chance."

How about thinking about it this way...change the rules around just a tad (the math stays the same, but I think its easier to understand).

Step one : Monte says "Pick a door!" And you do.

Step two : Monte says : "I'll give you a choice: You can either stay with your pick, OR you can switch and pick both of the remaining doors. If the prize is behind either one of the two remaining doors, you win!"

[Fingerplucked]

Russell, this is a great time to buy a car. The dealers are starving and so are the manufacturers. Quit playing around with the doors and red lights. Your best chance is to walk away from Monty and buy a severely discounted car through the usual channels.

[millring]

Jun 17, 2009 11:55:58 GMT -5 @omaha said:

Jun 17, 2009 11:35:10 GMT -5 millring said:

It doesn't really help.

I'm really this dense. I understand that Monte knows ahead of time what's behind the doors. What I don't understand is...

"So, when he offers you a chance to change your pick, you trade your 1/3 chance for a 2/3 chance."

How about thinking about it this way...change the rules around just a tad (the math stays the same, but I think its easier to understand).

Step one : Monte says "Pick a door!" And you do.

Step two : Monte says : "I'll give you a choice: You can either stay with your pick, OR you can switch and pick both of the remaining doors. If the prize is behind either one of the two remaining doors, you win!"

See my post above yours.

[Russell Letson]

Actually, I was hoping for a goat. Don't ask. (But remember--angora.)

[millring]

Jun 17, 2009 11:58:48 GMT -5 Russell Letson said:

Actually, I was hoping for a goat. Don't ask. (But remember--angora.)

Dyslexic Santa worshipper.

[omaha]

Jun 17, 2009 11:57:31 GMT -5 millring said:

Jun 17, 2009 11:55:58 GMT -5 @omaha said:

How about thinking about it this way...change the rules around just a tad (the math stays the same, but I think its easier to understand).

Step one : Monte says "Pick a door!" And you do.

Step two : Monte says : "I'll give you a choice: You can either stay with your pick, OR you can switch and pick both of the remaining doors. If the prize is behind either one of the two remaining doors, you win!"

See my post above yours.

You are almost there.

You are exactly right that Monty is being theatrical. But you are slicing things wrong.

It boils down to this: You are offered a choice of one door out of three (1/3 odds). You are then offered the choice of taking two doors out of three (2/3). Ignore all the theatrics taking place between the two choices. You can either pick one door and take your chances, or you can pick two doors and take your chances.

[Russell Letson]

What if Monty has a pointed stick?