Calculate the odds

[Erasmus00]

Nope no problem with Bayes theorum. All it says is that if you add information, that you increase the accuracy of your prediction by recalculating with the new information. I still fail to see what new information you get by having one shell removed.

 

You are told that one shell is not the correct shell. That is new information. If you need to convince yourself that this is true, write down all the possible situations. You'll see that 2/3 of the time, if you switch, you'll be right. Try googling Monty Hall problem.

-Will

[Turtle]

___To clarify, the coin landing on edge is a possibility & you can not establish a numerical probability for it by theoretical methods, but only by observational methods. As I have seen it once, the probability of it occuring for me again is 1 over the total number of coin flips I have ever witnessed plus all the flips I will witness & until or unless I witness it again.

___I do agree that if you discount that possibility - do overs - that it doesn't figure into the later part of the problem involving the nut cases...er nut shells.:evil:

[Edge]

So, the odds for heads and/or tails is not 1/2... ?

 

Damn, we have been fooled... :)

[Turtle]

So, the odds for heads and/or tails is not 1/2... ?

 

Damn, we have been fooled... :wave:

 

Correct on both statements. May I suggest this smiley::)

[cwes99_03]

___To clarify, the coin landing on edge is a possibility & you can not establish a numerical probability for it by theoretical methods, but only by observational methods. As I have seen it once, the probability of it occuring for me again is 1 over the total number of coin flips I have ever witnessed plus all the flips I will witness & until or unless I witness it again.

___I do agree that if you discount that possibility - do overs - that it doesn't figure into the later part of the problem involving the nut cases...er nut shells.:)

 

Actually you are pretty close but slightly off. The odds for it landing on edge would be fairly approximated by you counting all the coin flips you have ever witnessed between the first coin ever flipped and the second coing ever flipped and to have landed on it's edge divided by two (an average).

 

To clarify you can establish a numerical probability of a coin landing on it's edge, as was established in post number 6 by erasmus, though how accurate that probability is would be hard to calculate. The reason for this is that the coin could land flat on one side and bounce up and land on edge (thus chaos theory takes over, and you have to go back to just flipping a coin as many times as you can to establish an empirical probability.

 

Will look up Monty Hall problem. Be back in a day or so i have to get some sleep.

[Turtle]

Let those who have eyes see.:)

[Edge]

Will look up Monty Hall problem. Be back in a day or so i have to get some sleep.

Look it up, it's very interesting stuff. It is indeed 1/3 the chances you have for winning if you remain with the same choice and 2/3 the chance you have for winning if you switch it.

 

This is because the host will always unveil a door with a bogus prize.

 

There is another example. You are on a test with three options: A, B, and C.

 

You choose A. And then the teacher tells the group that C is not the answer. What are the chances if getting the answer right if you remain with A as an answer?

 

1/2

 

Why? Because the teacher does not know which option you selected. He just gave an option away. You may have as well chosen C and the teacher would have told you that C is not the answer.

[cwes99_03]

Interesting page. Let's have someone take a look at the source code for the page, maybe it is built to be one sided, but look at the statistics.

 

 

 

# of Players Winners Percent Winners

Switched 155317 9187 5.9

Didn't Switch -25621911 -12907113 50.4

http://math.ucsd.edu/~crypto/Monty/monty.html

 

The winning door switches regularly. Just picking a door and playing the game over and over again will not stack the numbers.

The first column is how many people chose to stick with their original choice or switch, the second column is how many of those people won when they made that choice, and the third column is the winning percentage for either sticking with or changing. Seems like sticking with (empirically anyway) has the odds stacked in it's favor.

http://math.ucsd.edu/~crypto/Monty/montybg.html

Anyway, the site goes on to explain why switching would be in your best interest. Let me see if I can debunk this myth.

http://math.ucsd.edu/~crypto/Monty/images/wheel.jpg

If you look at the above image you will understand the following.

 

The inner wheel represents the number of the door that the car is behind, the middle wheel represents the door that is selected by the contestant, and the outer wheel represents the door Monty Hall can show. Spinning this roulette wheel once is equivalent to playing the game once. The outer wheel also tells you what your strategy should be to win. The red means that in order to win the contestant needs to switch doors, and the blue means that the contestant should not switch. Notice that there are twice as many red sections as blue. In other words, you are twice as likely to win if you switch than if you don't switch! What this wheel makes evident is that with probability 1/3 the contestant selects the correct door in which case it would be better not to switch. In the other 2/3 of the cases, Monty Hall is telling the contestant where the car is!

 

No when you initially chose a door, you have a 1 in 3 chance that you are right and a 2 in 3 chance that you are wrong. Monty opens a door that he knows is wrong (of course if you chose the right door then he has two doors to chose from). Once he removes that door, you now have two doors to chose from. You now have a 50/50 chance to win the game. Do you switch? Why would you? It is 50/50 either way. Those who say it isn't 50/50 are assuming that monty is giving something away. Well is he? One cannot know. If you chose the wrong door the first time, then yes he is giving something away. If you chose the right door then he isn't giving anything away. Therefore you have no new information. You don't know whether or not he is giving something away. Instead with no new information except that one possibility has been removed, which means that you now have two choices, you now have a 50/50 chance.

 

Let me demonstrate. Now say you chose a door and monty doesn't know which door you chose. He reveals a door (and it happens to be the door you chose) so now you have a 50/50 chance of getting it right. Or maybe he opens a door you haven't chosen. You still have a 50/50 chance. He won't open the door with the car because that would be giving you the car.

 

Let's say there are two prizes. You chose a door. Monty now opens one of the other two doors. Which will he open? If he opens the one with the goat, then you know you have won at least one of the prizes (assuming you don't win whatever he opens up on the first door). If he opens up the door with the car, you don't win the car, but now you have a 50/50 chance of winning. If he opens up the door with the lesser prize you still have a 50/50 chance of winning the car (though why didn't he open the door with the car behind it? because you must have already chosen it, unless Monty is risking the networks money on the instinct that you won't chance doors.)

 

Let's say there are three prizes. Well I guess you win no matter what, so now it is a big mind game between you and monty. Which one is he gonna show you and why?

 

Well, anyway, there is a 50/50 chance once one door has been removed. It doesn't matter whether you switch or stay. You have a 50/50 chance either way. Monty is not telling you anything when he choses the door, except that you didn't chose one of the wrong doors. Since there were two wrong doors and you can't chose both, you already knew this, so there is no new information.

[Erasmus00]

Well, anyway, there is a 50/50 chance once one door has been removed. It doesn't matter whether you switch or stay. You have a 50/50 chance either way. Monty is not telling you anything when he choses the door, except that you didn't chose one of the wrong doors. Since there were two wrong doors and you can't chose both, you already knew this, so there is no new information.

 

Look, I know its a tough thing to wrap your head around, but this is not actually the case. Its not some myth you are debunking, its a mathematical fact.

 

Think about it like this, if you are wrong the first guess, and you switch, you'll get the prize. You have a 2/3 shot of being wrong the first time, right? So 2/3 of the time, if you switch you should win.

 

Another way to see this: write out all the possible combinations of what could happen with three doors and the prize in a fixed location. You'll notice many more microstates of the switch and win variety.

-Will

[cwes99_03]

Ah but 2/3 only matters before one of the doors is removed. Once a door has been removed you now have a 50/50 chance.

 

Why can't you seem to bend your head around that? Are you so enveloped in the idea that someone told you so with a bunch of pretty words and numbers that you can't see where they went wrong?

 

Have you gone to the site that i showed above? Apparently this was posted in a news column and hundreds or thousands of math and physics professors said that it was wrong.

Have you also looked at the numbers above? This is an actual site with an actual set of three doors to chose from. You can test it yourself. Then when you are done testing it you can look at the data. I bet if you were to go to the site and change your answer every time for a hundred answers you would come out with a 50% win rate. If more people were to test the switch method, the set would be bigger and that awfully small correct percentage would come up to 50%.

Numbers don't lie. People can be wrong, even incredibly smart people.

[cwes99_03]

Let me throw this at you. You have a 2/3 chance that you are wrong with the first choice and a 1/3 chance you are right. One is removed, meaning that your 2/3 chance of being wrong just went to 1/2 because Monty just told you that at least one of the answers you didn't choose was a wrong answer, and your 1/3 chance of being right just went to 1/2 because Monty didn't tell you that you chose wrong.

[Erasmus00]

You seem to be ignoring the initial choice entirely, which is a fallacy of logic.

 

Now: you have a 2/3 choice of being wrong the first time right? You agree with this?

 

Hence, after Monty has opened the new door, you have two choices in front of you. One of the choices you already know will be wrong 2/3 of the time.

-Will

[cwes99_03]

Precisely, the other choice as well would be wrong 2/3 of the time. The second choice was no different than the first choice before one door was removed, nor after.

[Christopher]

Precisely, the other choice as well would be wrong 2/3 of the time. The second choice was no different than the first choice before one door was removed, nor after.

 

 

 

 

The first pick is a 1 in 3 because their are 2 heads and 1 tail. If you stop there this is not going to change no matter what is shown. This a mathematical fact, it will remain 1 in 3.

 

Now the second choice is between a head and a tail 1 in 2 chance this is a new fact. Why should you switch, because your not choosing between 3 coins but only 2 and you know one is a head and one is a tail, correct.

 

Don’t feel too bad, it took me all day to get my head around it.

[wizzkid67]

so your saying that the first pick u made has a 1/3 chance of being right then he takes away a wrong answer now the one u chose is still 1/3 right but if u change your mind and still pick the same one from before it drops to a 1/2 sounds a bit bananas to me but makes a bit of sence

[Erasmus00]

Precisely, the other choice as well would be wrong 2/3 of the time. The second choice was no different than the first choice before one door was removed, nor after.

 

Look, lets say you were presented with two doors, and you were told that Door A has a 2/3 chance of being wrong, wouldn't you pick Door B? Because thats exactly the situation we are in. Removing the door does not change the odds that your initial choice was 2/3 wrong. You are ignoring the original established odds. Removing a door doesn't destroy those probabilities.

-Will

[cwes99_03]

You are missing that the other door also had a 2/3 chance of being wrong. They had equal chances of being wrong before one door was taken away. They now have equal chances once one door has been taken away.