This is a discussion on Probability 101: Monty Hall within the Off Topic & Humor Discussion forums, part of the The Back Porch category; OK, Here is the story. My mother and her friend go out to lunch. After lunch, Sandy drags my Mom into a high end children's ...
OK, Here is the story.
My mother and her friend go out to lunch. After lunch, Sandy drags my Mom into a high end children's store located next door.
Sandy sees a Brass Cradle and my mother looks down at the price and replies who in the right mind is going to spend $1400.00 on this. So they look around some more and Sandy leave without the cradle.
A few weeks later my mom,dad,aunt and uncle go to Cali for a vacation. They go to see the Price is Right.
Everybody registers before the show. It allows you to get a name tag and a tag with a number.
Before you walk into the studio you meet with a producer who asks you a few questions. The producer remembers the number.
The show starts and they cross reference the # to your name. To call you down. So you have no idea that you will be on the show.
My mother gets called to come on down. The item that she has to bid on is the exact same Brass Cradle. She hits the $1400.00 right on the head.
So now Mom has to pick the highest priced item out of the 3
2 rocker recliner
side by side refridge/frezer
Solid oak roll top desk.
She picked the desk and wins all three.
During the commercial breaks there was this lady yelling like a loon. "Pick me Pick me I am a winner, I will show you I am a winner.
Well, the pick her. And she gets on stage.
Now the chance to spin the big wheel. Mom gets something (I do not remember). This woman spin $1.00 on the head.
Paymeister, as has been said, you are correct in that any series of random events is random, and does not depend on preceding events (roulette spins, coin flips, etc). This, however, does NOT hold true here: there is only one event,; you are just given more information about it DURING the event. Imagine you spun a roulette wheel after betting on black, then found out (magically) that the ball would NOT land on 10 of the black squares...would you be smart to change your bet to red if you were given the opportunity? Of course - you know have 10 less winning options... (Note that this is NOT analogous to the Monty Hall problem, it's just to show that the MH problem is one event, with information added during the event. The idea that some “random” events do have a memory is the key to “card counting” in Black Jack and the basis of all odds predictions in poker – the odds change DURING each shoe/deal as more information, i.e. more cards, are shown to the player.)
Ok, I thought the explanation (and example) I gave would show the answer, but apparently not. I'll write it here again, and try to expand on it so it's more clear...
At the outset, you have three choices, and NO information about what is behind any of the doors. Picking one at random, you have a 1/3 chance of picking the BP and a 2/3 chance of picking a G. Assume the BP is behind door number one (it doesn’t matter which for the explanation, you can just change the numbers...)
You have three options; you chose 1, 2, or 3.
1) If you chose 1, Monty could show you 2 or 3 (both being goats). If you switch, you lose. If you stay, you win.
2) If you chose 2, Monty MUST show you 3 (1 being the BP). If you switch, you win. If you stay, you lose. It is important to note that he cannot show you 1, because that would reveal where the BP is. He MUST show you the other losing door.
3)If you chose 3, Monty MUST show you 2 (1 being the BP). If you switch, you win. If you stay, you lose. Same here – he MUST show you the other losing door.
After he has shown the one losing door, you now have another option, with more information. You know that if you picked a goat (which you had a 2/3 chance of doing at the outset), then Monty was just forced to show you where the OTHER goat was. So, if you had guessed wrong (2/3 chance, remember?), you now know where both goats are – behind the door you choice, and behind the one Monty revealed. That means the BP is behind the door neither originally chosen, nor revealed.
If you originally guessed right (1/3 chance), Monty could reveal either door, and you don’t really gain any new information. Remember, though, that you had a 2/3 chance of picking a goat at the outset, and only a 1/3 chance of picking the BP. So, we assume (based on probability) that we originally picked a goat, and now Monty has shown us the other goat, therefore (again based on probability – not certainty) you now know where both goats (and therefore the BP) are.
Check the play by play again – if you stay, you only win if you had originally picked correctly (1/3 chance). If you switch, you win if you had originally picked incorrectly (2/3 chance). Remember there are only three original options – picking 1, 2, or 3….
Clear yet? It’s counterintuitive to some extent, that’s what makes it an interesting problem, but if you simply look at all the possible outcomes you will see that, by switching, you will win in 2 of the 3 possible outcomes and by staying you will win in only 1 of the 3 possible outcomes….
A man fires a rifle for many years, and he goes to war. And afterward he turns the rifle in at the armory, and he believes he's finished with the rifle. But no matter what else he might do with his hands - love a woman, build a house, change his son's diaper - his hands remember the rifle.
I know most people around here are not math geeks, so I just want to add one little thing to OPFOR's excellent explanation.
There is a difference between unconditional probability (where you know nothing about the system prior to the next event) and conditional probability (where you are now estimating odds given some knowledge of previous conditions).
For coin tosses, they happen to be the same, because the next coin toss has nothing to do with the previous one, two, or ten thousand tosses.
In the Monty Hall problem, though, the door that Monty shows you is conditioned on your first choice---if you already chose one of the goats (which is 67% likely), he has to show you the other loser. In essence, opening one losing door gives you extra information about the choice you already made and lets you reassess the odds for the second round.
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OPFOR, thank you! Your latest explanation was crystal clear! Now I understand it, and I'm now sure that the bad guy knew the two bullets were consectutive!
(Just Kidding) Thanks!
Last edited by JonInNY; April 23rd, 2010 at 02:26 PM.
"Democracy is two wolves and a lamb voting on what to have for lunch; Liberty is a well-armed lamb contesting the vote."
-- Benjamin Franklin
Does anyone know any good recipes for goat?
God is love (1 John 4:8)
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Choose a door, and Monty shows that one of the other two was a loser.
You now have a 1/2 chance of choosing the correct door. If you stick with the same door then your decision to do so has a 50% chance of being right, the same chance as switching to the other door. People are skipping over the idea that sticking with your original door is a choice in and of itself, and bears an equal chance.
"The flock sleep peaceably in their pasture at night because Sheepdogs stand ready to do violence on their behalf."
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I may be missing something. A game like this is played only one time, and the odds for that one time are still 50:50. All you know for the one play is that there may or may not be a goat behind the door you chose. For any door you choose, there may or may not be a goat behind it. That doesn't change. It is still 50: 50. It was 50:50 for each door from the start. Each door either did or did not have a goat or a prize.
You know nothing more and got no more info by having been shown where one of the goats actually is located. All you know is that behind a door there is either a goat or a prize. And you know the prize isn't behind the door monty showed, but that is irrelevant because it is now out of play and never really was in play once you made a choice.
Or, I'm too dense to follow your explanation-- certainly a 50:50 possibility.
Very very, simple math.
Suppose there were fifty doors. You pick number one. Monty then opens number three through forty nine revealing 48 goats (pass the curry please).
There are now two unopened doors, number one and number two, and again you have chosen number one.
You stick with number one and your chances are 50/50.
If you switch to number two your chances are...........guess what, 50/50.
It does not matter how many doors you start out with.
CCW permit holder for Idaho, Utah, Pennsylvania, Maine and New Hampshire. I can carry in your country but not my own.
This is what I don't get. We don't know if Money was FORCED to show you that goat, or if he just chose one.After he has shown the one losing door, you now have another option, with more information. You know that if you picked a goat (which you had a 2/3 chance of doing at the outset), then Monty was just forced to show you where the OTHER goat was. So, if you had guessed wrong (2/3 chance, remember?), you now know where both goats are – behind the door you choice, and behind the one Monty revealed. That means the BP is behind the door neither originally chosen, nor revealed.
I'm in the 50/50 camp.
In the beginning you had 3 choices. You have a 2/3 chance of picking a goat.
Once he reveals a goat, you have 2 doors, with 2 outcomes. You have no idea what is behind either door.
50/50.
OK, I am now one of the "converted" by OPFOR, and I agree! And remember, the original question is "Do you want to switch your choice!"
Try this:
On a table, put two pennies, and a quarter. Go ahead, do it, I'll wait.
Now, pretending they are covered (or not), your goal is to try and choose the quarter.
Does everyone agree that your chances right now of making a bad choice (a penny), is 2:3, and a good choice (the quarter), is 1:3? Good! If you don't agree you can stop right here.
Remember, sometimes you LOSE this game... you don't always win. You just want to choose the best odds.
Now, pick a coin. Let's say you picked the quarter, ok? Now one of the other pennies becomes visible. If you switch, you lose. Too bad. Go home.
Let's say you picked penny #1. Now the OTHER penny (#2) becomes visible. You switch... you win!
Let's say you picked penny #2. Now the OTHER penny (#1) becomes visible. You switch... you win!
So, your chances are 2:3 that if you switch, you win and 1:3 that if you switch, you lose! Just remember, that sometimes you lose! That's life.
Thanks again OPFOR, I just re-phrased your original explanation! Did I do ok?
Last edited by JonInNY; April 23rd, 2010 at 05:07 PM.
"Democracy is two wolves and a lamb voting on what to have for lunch; Liberty is a well-armed lamb contesting the vote."
-- Benjamin Franklin
Hopyard, I want to try to answer this, and please understand that I am in no way, shape, or form, trying to pick on you.
The fact of the matter is, most people simply don't have a gut feeling for probabilities. Probabilities are useful as expressions of long-term trends in outcomes. If you roll a 6-sided die a couple hundred times, about one-sixth of your rolls will be 1, about one-sixth will be 2, and so on.
Mathematically, we say that on your very next roll of that die, you have a 1/6th chance (or probability) of getting, say, a 6. That's the definition of probability.
You, on the other hand, like a lot of people, look at it from the very direct point of view that either you get a 6 or you don't. Two outcomes. And because most people really don't get probabilities at a gut level, everything tends to look to them like "certainly true", "certainly false", or "50/50"---that is, probability 100%, 0%, or 50%---even though that's not what the math says about the situation.
Let me give another example---you're playing poker and you have four of a kind. You are about to rake in a huge pot against one guy still betting into you. Maybe he has a royal flush and maybe he doesn't. Those are the possibilities, but it is certainly not a 50% chance that he has a royal flush. In fact, I think the odds of a royal flush are about 650,000:1 against, but if he does pull that royal flush, those low odds aren't going to be much consolation to you...
By the way (and this may be a rather tangential aside), these are not just simple little game-playing consequences. The fact that people in positions of power and authority don't understand odds bites folks in the butt all the time. What were the odds of a plane being downed by the volcanic ash over Europe this past week? Pretty slim; maybe even infinitesimal. But because of the fact that EU functionaries can't tell the difference between "some small risk of disaster" and "near certainty of disaster," the EU bureaucracy made a decision that cost the airline industry a couple billion dollars.
</math_rant>
“What is a moderate interpretation of [the Constitution]? Halfway between what it says and [...] what you want it to say?” —Justice Antonin Scalia
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JOIN IN-- I follow your explanation somewhat better than OPFOR's but:
The chances of winning or losing are still 50:50 (I think), because the options of choices 1 and 2 or 1 and 3 each give you only an overall chance of picking correctly of 50:50.
Take choice number 1; the other penny is revealed, there is still only either a penny or quarter behind the door and you don't know whether switching or not (50:50) will be beneficial.
Start with your statement: "Now, pick a coin. Let's say you picked the quarter, ok? Now one of the other pennies becomes visable. If you switch, you lose. Too bad. Go home." Yes, but IF you don't know that you picked the quarter and choose "don't switch" (an option not given in your example) you win.
Put in ALL 4 of the options and the odds stay at 50:50.
The explanation appears to work only because one of 4 options is missing.
OR, I'm not smart enough to see what you guys are driving at---a possibility I readily admit to. And this possibility is probably better than 50:50. :)