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Let's Make A Deal - Page 2

post #31 of 49
Originally Posted by wpg_ski_nut
Your odds of picking the right door originally were 1 in 3. Just like the odds of flipping heads twice is 1 in 4.
flipping a coin repeatedly is a series of independent events, this is a different animal altogether. your argument is not a proof of any kind, it's an explanation and an inconclusive one at best (the word wrong comes to mind . have you been following this thread? if you don't believe the solution read trouble's fist post (he was the first to give it) and find a mistake (there is none).

post #32 of 49
Monty gets to choose which door to open based on knowing where the goats are. You don't. My random grouping to prove my point differs from Monty's grouping because of that fact.
It is also not enough to say that the events are not independant. Here let me make it easy for you. Consider time.

If monty offered you the choice of the door you originally chose, or the other two before opening any doors, telling you that if you opened one of them and it didn't win you could still have the other, which would you choose?

Because Monty got to look, and pick a goat to throw out of the equation, you are not switching for an equivalent door; you are switching for Monty's best door. It's like the odds are 50-50 on the remaining two doors, but half the other doors bad chances are thrown out. The door your switching to had a 50-50 chance of being picked to be opened if it was bad.

Edit: talk about arguing both sides! I guess the worst fundamentalist is a convert!
post #33 of 49
I agree You will always be presented with the 50 50 odds because Monty will eliminate a goat. At this point you have to say are the odds of winning any different if you got to choose again. There are now 2 doors one goat one car, you know nothing else. You are as likely to change a right guess to a wrong as you are a wrong to a right. Your odds of winning are 1 in 2, although your original odds of winning haven't changed at 1 in 3. Even 1 in 2 odds are staggering and have a great deal to do with how large Las Vegas, and they are not much better than 1 in 3.

The original question was do you take the new door the proper answer is to check what the frequency of the new door being a winner has been in the past before making the choice. When you had a choice of 1 in 3 for a car you new that 1 of 3 had a car. The new door is a completely new variable and not connected to the 1 in 3 at all. If monty puts a grand prize behind the new door 80% of the time then your odds of winning with the new door are 4 in 5 choose the new door. If he puts a winner behind the new door only 1 in 5 times stick with your original choice.

The new door doesn't effect your original cahnce of winning at 1 in 3. It doesn't affect your present chance of winning at 1 in 3 or 1 in 2 depending on your view. It offers you a different chance of winning based on the rules of the new door.

Now I'm really confused.
post #34 of 49
Your original choice will be wrong 2 times out of three. Showing you the goat behind another door won't change that. But selecting the OTHER door will be right 1 time out of two.
post #35 of 49
Originally Posted by wpg_ski_nut
Now I'm really confused.
Just look at it this way. When he is giving you a chance to switch, just pretend he gave you a chance to go back in time and pick two doors instead of one.

If your not into time, remember that he eliminated a goat. The door your choosing between is one that has had a chance to be eliminated because it had a goat, but was not, and your own. He eliminated the goat from a choice of two doors. So the likelyhood of a goat being in the door your switching to has been reduced by 1/2.
post #36 of 49
It appears to me that the probablility has been 50/50 since the beginning. The 1 in 3 probability in the original choice has NOTHING to do with the final outcome.

In 100% of the instances one of the two incorrect answers is eliminated. At this point you are either correct or not and a change will be only to correct or not, there is no third option.

I'm sure there is an elegant mathematical proof here, but I'm not an elegant mathematician.

SOMEBODY CALL CHUCK NORRIS. If he can divide by zero he has the answer to this.
post #37 of 49
Get a partner to hide the ace of hearts amongst two other cards and try it! Your chances of guessing right the first time are 1/3. When you switch you are switching to two cards. It's like you are switching to two doors. He had to choose which door to reveal; he didn't just throw away any door, leaving a door just like yours; he had two chances to get a car and threw away the door with the donkey.

The chioce really is between one door and two doors. The fact that he has already opened one of the doors (a bad one) doesn't change that.

What if your had to choose between your door and the two other doors BEFORE he opened the door? I bet you can see your chances are better with two doors now. Why should the choice be any different after he opens the door? Opening the door doesn't move any donkeys or cars!
post #38 of 49
Thread Starter 
You Pick Door 1, Car in door 1, you stick -> Win
You Pick Door 1, Car in door 1, you switch -> Lose
You Pick Door 1, Car in door 2, you stick -> Lose
You Pick Door 1, Car in door 2, you switch -> Win
You Pick Door 1, Car in door 3, you stick -> Lose
You Pick Door 1, Car in door 3, you switch -> Win

You Pick Door 2, Car in door 2, you stick -> Win
You Pick Door 2, Car in door 2, you switch -> Lose
You Pick Door 2, Car in door 1, you stick -> Lose
You Pick Door 2, Car in door 1, you switch -> Win
You Pick Door 2, Car in door 3, you stick -> Lose
You Pick Door 2, Car in door 3, you switch -> Win

You Pick Door 3, Car in door 3, you stick -> Win
You Pick Door 3, Car in door 3, you switch -> Lose
You Pick Door 3, Car in door 1, you stick -> Lose
You Pick Door 3, Car in door 1, you switch -> Win
You Pick Door 3, Car in door 2, you stick -> Lose
You Pick Door 3, Car in door 2, you switch -> Win

Using the method that was proposed a few posts ago, but getting every scenerio properly weighted, we see 18 possible scenerios. In 12 of them switch wins the car and 6 of them switch loses the car. In 12 of them stick loses the car and 6 stick wins the car.

Switching changes the odds of winning from 1/3 to 2/3. For some of you, I know accepting this hurts your brain but believe it's the right answer. Google Monte Hall Problem all you want. Any thorough treatment of the problem will reach this conclusion.

If you still don't believe me, run it out with the Monte Carlo method (the mathematical term for computing propbability by actually running samples). You could write a program (or pull the Perl one off Wikipedia) if you understand programming. If not, find a partner and run through a bunch of samples using a deck of cards.

Trust me on this one. Switch is 2/3. Trust me.

post #39 of 49

Not 18 but 24

Something kept bothering me about this. Woke up in the middle of the night and there it was. There are not 18 possible scenarios, but in fact 24. That's where the trick has been lying.

This is predicated on the premise that Monty knows where the goats are.

If you pick a goat, Monty has to open the other goat. This forms twelve possible combinations, and they are 50/50.

However, if you pick the car Monty has the choice of doors to pick. There are twelve possible combinations here, not six. They are also 50/50!!!


I have it charted and will put it up this afternoon or evening.

I can sleep now.
post #40 of 49
Thread Starter 
Sorry wpg_ski_nut but you almost got it right. You say there are 24 scenerios. The thing is, you are splitting six of the scenerios (in which you pick the correct door on the first guess) in two to do that. That means that each those scenerios are 1/2 as the others. So, nothing has really change.


If you use the switch method, you win every time you do not pick the correct door on the initial guess. That prob of not picking the correct door on the first guess is 2/3. Nothing changes.

Read the wiki write up again. Google "Monty Hall Problem" and see what other references there are. They'll all have the same answer, switching gets you 2/3.
post #41 of 49
There are a million doors. You pick door #1. Monty opens all the others -- except door #777,777. Do you switch?
post #42 of 49
Yup, and you will win more than you loose. Remember Monty only opens doors that don't matter.

This is getting more discussion than I thought was possible about math on this site.
post #43 of 49
Originally Posted by BigE
There are a million doors. You pick door #1. Monty opens all the others -- except door #777,777. Do you switch?
Would you switch if given the chance to switch to the 999 999 doors before he opened the doors? If so, then why not switch just because the decision is made after the other doors are opened?
post #44 of 49
Now if this was a true "Let's make a deal" problem, you would pick door #1, Monty would show you the goat behind door #2, then ask "Do you want to switch to door #3? Or would you take $500 and whatever is behind curtain #1?

Change the problem to a deck of cards and you have to pick the ace of spades from the deck, but the game will work like this: you pick any card from the deck, I will eliminate all the remaining cards except one, which will either be the ace of spades or some other card, meaning you already drew the ace. There are four possible outcomes: you keep your card and win, you keep your card and lose, you switch and win, or you switch and lose. With no knowledge of the preceding events, you have an even 50/50 split to find the ace. BUT with the knowledge of how the game started, you are forced back to the original odds of the game, and in this case, it's fairly obvious that the ace would much more frequently reside with me and you would be better off to switch.
post #45 of 49
Hmmm, you can buy a lot of rice-a-roni with $500.....
post #46 of 49
Or do a lot of shopping in the Speigel catalog!
post #47 of 49
Any chance Carol Merrill is behind door #3?
post #48 of 49
After thinking about this and reading the posts, I think the odds are 2/3.

Look at the million door example. When you pick door #1, your odds were 1:1,000,000. Then Monte opens every other door except #777,777. But Monte knows where the car (a Monte Carlo?) is. So unless you hit the 1:1,000,000, your can be pretty sure the monte carlo is behind door #777,777.

This plays down to 3 doors, but just isn't as obvious. Your odds are 2:3 because you know Monte will pick a goat. Therefore 2 doors of the three will be picked. But it makes me think your odds are 2:3 only if you switch doors, because you started with odds of 1:3 before Monte showed you the goat. Just like the 1:1,000,000 example.
post #49 of 49
I agree, it's not 1/2. It's 2/3....
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