Dan, the whole premise is that you could have chosen the correct one the first time. That isn't revealed. The question is, do you want to stick with the first choice (that could be correct) or switch your choice.

If you chose correctly the first time, that means that Monty Hall can reveal either of the other doors. If you didn't choose correctly the first time, he reveals the one unchosen door that contains the goat. Either way, the player chooses, and Monty reveals a goat.

here is another look at it. if a man has 20 kids and they are all girls, what are the odds that his next kid is a boy. the answer is 50%. because thst conception(50/50) is separate from all the other instances that occured before. that is where howdy falls. he isnt factoring what happens before into his final equation. His ultimate choice is made when only two doors remain, hence the 50/50. and that is true, if you choose between two doors, your chance is 50/50. its the manner in which you get down to two doors that causes the disconnect. so in a manner of speaking he is right and wrong

1/3 of the time the first one will be right

1/3 of the time the one you pick will be right

1/3 of the time the other one will be right

How can this be flawed?

as long as i get some box i will be happy.

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When you always switch, you only lose when you had the car picked with your first choice, in which case switching gets you off the car.
This happens 1/3 of the time, the other 2/3 of the time you win.

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See, this man gets it!

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why switch ATM? because your door has beaten the odds 998 times you don't think it can beat the odds again? opening each door is a separate, unrelated event.

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You are confusing yourself. The probability of flipping heads 10 times in a row does not equal the chances of flipping a head on the 10th flip. Do you agree?

Go watch the re-run of Numb3rs from the other night. He explains this perfectly.

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If you can read this, thank a teacher.
If you can read this in English, thank a Soldier.

i must say that explanation is by far the simplest i have seen.

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See, this man gets it!

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Thanks Racer. I just knew that PhD in Statistics would come in handy.

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Thanks Racer. I just knew that PhD in Statistics would come in handy.

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It is serving you well, my friend!

Did yall give up on me?

aTm - agree 100% because each event is unique and has nothing to do with the previous flip(s)

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Did yall give up on me?

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Pretty much, yeah.

[This message has been edited by RacerX93 (edited 10/18/2005 4:13p).]

Oh well...... SCREW THIS TRICK RIDDLE THING!!

Go Astros!!

What is the probability of them winning 1 of the last 2?

Zero.

Alright, Dan, here were go:

You decide a priori that you will always switch.
Now, everything hinges on your first choice. The circumstances of the game are such that
- If you choose the car first, you lose.
- If you choose a goat first, you win.
Once you convince yourself of that, the 1/3-2/3 thing is easy.

...and GO ASTROS!

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aTm - agree 100% because each event is unique and has nothing to do with the previous flip(s)

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Your not agreeing with me since the probability of flipping 10 heads in a row is much smaller than the probability of getting a heads on the last flip.

but isn't there a 1/3 chance of picking the goat or the car to begin with?

edit: for pwn3d

[This message has been edited by first down down (edited 10/18/2005 4:22p).]

your original question:

The probability of flipping heads 10 times in a row does not equal the chances of flipping a head on the 10th flip. Do you agree?


I agree 100% with that statement. Odds of flipping heads 10 times in a row = 1/2^10. Odds of flipping a head on the 10th flip = 50%.

What am I missing? How am I not agreeing?

You got the math right, I’m just confused with your explanation:

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aTm - agree 100% because each event is unique and has nothing to do with the previous flip(s)

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If the question asked was “you just flipped 9 heads in a row, what is the chances of flipping another head?” then the previous 9 flips have no bearing on the answer. However, when you ask “what is the probability that you will get 10 heads in a row?” Then all 10 flips must be considered together.

Regarding the original question, here are the probability trees. (Assume the prize is behind door A... just copy the tree’s for B and C and you still get the same result)

Always switching:

• You pick A (1/3):
  • Host opens B (1/6). You switch you LOSE
  • Host opens C (1/6). You switch you LOSE
• You pick B (1/3):
  • Host opens C (1/3). You switch you WIN
• You pick C (1/3):
  • Host opens B (1/3). You switch you WIN

Probability of winning: 1/3 + 1/3 = 2/3

Never switching:

• You pick A (1/3):
  • Host opens B (1/6). You don’t switch you WIN
  • Host opens C (1/6). You don’t switch you WIN
• You pick B (1/3):
  • Host opens C (1/3). You don’t switch you LOSE
• You pick C (1/3):
  • Host opens B (1/3). You don’t switch you LOSE

Probability of winning: 1/6 + 1/6 = 1/3

In that tree it completely doesnt accountfor the fact that the host will open "A"(the correct one) 1/3 of the time.

How can this be completely removed from the equation?

and yes im back to argue more, i went to get somethin to eat.

keep up the fight Dan, I'm heading home....

No, he will never open A, since he knows that that one has the prize behind it. He’s not randomly picking a door, he is purposefully opening an empty door in order to create more suspense for the audience.

[This message has been edited by aTmAg (edited 10/18/2005 5:04p).]

atmag, i think it has to be ramdom to work. i believe thats the premise

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In that tree it completely doesnt accountfor the fact that the host will open "A"(the correct one) 1/3 of the time.

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Nope. No matter what, the host will always show you a goat, in order to keep the game going and give you the opportunity to switch.
This is why everything hinges on your first choice. The only way to lose is to pick the car first.

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atmag, i think it has to be ramdom to work. i believe thats the premise

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Which door the prize is behind is random. Which door the host will chose to open after you make your pick is *not* random. He knows which door it is behind. That is the premise.

Well if the host is always going to pick a door with a goat in it. Then That that makes no difference in the game at all. There's only 2 valid doors to choose from....

I’m not sure what this goat thing is. I haven’t read the link, but I have always heard it this way:

There are 3 doors, one has a prize behind it. You pick one. The host says, “now what if I open door X for you?”. He opens a door that you didn’t pick and it is empty. He then gives you the option to switch. Should you switch?

The answer is yes you should (assuming you want to win).

I played the card game several times...

I switched 32 times and stayed the same 32 times...
in 32 switches : 20 wins
in 32 not switching : 11 wins

BQ, how many times did you pick the ace on the first round?

Because if you never picked the ace(cuz you knew where it was) you were only choosing from 2 cards.

What if you are a goat farmer and want a goat instead of a car?

This same argument popped up on here a while back and I was the one arguing against the premise that it is better to switch.

It's been stated here already, but what they are saying is that you have a 1/3 chance of being right when you pick the first door (therefore a 2/3 of being wrong). This means that there is a 1/3 chance the prize is behind the door you picked (one set of doors) and a 2/3 chance it is behind one of the other 2 doors (the other set of doors). This is the important part of this, when the host purposely picks the "other" door it is not a random event. He is purposely picking a door that you didn't already pick that also does not have the prize. By eliminating this "other" door, the odds of the 2 sets have not changed, but the number of options within the 2nd set have been reduced to 1 instead of 2. This means you have a 2/3 chance of getting it right by selecting the only option left in the 2nd set (before the "other" door is eliminated, you have to split those odds in 2).

so, NY Ag, its a trick question basically.