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Three Curtains

Before any curtain is opened, you have 3 possibilities.

  1. The car is behind curtain #1
  2. The car is behind curtain #2
  3. The car is behind curtain #3
The probabilities for each are the same - 1/3 for and 2/3 against. Combining curtains doesn't change the probability for each individual curtain. However, having the possibility of picking 2 curtains increases the possibility of being correct to 2/3. If one curtain is opened (say, #3) and shown not to be the correct choice, then that changes the probability for the others. There is no longer a 1/3 or 2/3 chance, but a 1/2 chance for each curtain that is left, since there are not 3 but 2 choices left.

  1. The car is behind curtain #1
  2. The car is behind curtain #2
The probability of each is the same - 1/2. Once it has been opened and you have seen that the car is not there, curtain #3 becomes irrelevant to the math.

The TOG​
 
TOG,

Using the 1 curtain in area "A" and the 2 curtains in area "B" example, and before any curtain is opened, do you agree or disagree that there is a 2/3rds chance that the car is in area "B"?
 
There is before the curtain is opened and you still have 3 choices, but once one curtain is opened, you only have 2 left and the probability changes. There can no longer be a 2/3 probability of anything, since there aren't 3 choices.

The TOG​
 
I wasn't good at probability in the 8th grade and it still doesn't make sense to me, especially this Rule being used in this problem. By the end of this in 8th, I couldn't look at black, white, etc marbles without stress. It just does not compute logically in my brain.
 
Deborah13 said:
I wasn't good at probability in the 8th grade and it still doesn't make sense to me, especially this Rule being used in this problem. By the end of this in 8th, I couldn't look at black, white, etc marbles without stress. It just does not compute logically in my brain.
Click to expand...

It's really just fractions, where the number of correct (or incorrect, depending on what you're looking at) choices is the numerator, and the total number of possibilities is the denominator. If you get to choose just one curtain of three (let's suppose you chose curtain #1), then the possibilities are:

You choose #1 and the car is behind curtain #1 - Correct
You choose #1 and the car is behind curtain #2
You choose #1 and the car is behind curtain #3

There are 3 possibilities and one of them is correct. You therefore have a 1/3 chance of getting it right if you choose curtain #1. The same applies if you choose one of the other curtains. Supposing that you have still chosen #1, then if the host opens one curtain (let's suppose it's #3), revealing that the car isn't there, you would have 2 possibilities left.

You choose #1 and the car is behind curtain #1 - Correct
You choose #1 and the car is behind curtain #2

There are now two possibilities and one of them is correct. You therefore have a 1/2 chance of being correct. That probability doesn't change by changing your choice. But it does change if you are allowed to pick both #2 and #3.

You choose #2 and #3 and the car is behind curtain #1
You choose #2 and #3 and the car is behind curtain #2 -Correct
You choose #2 and #3 and the car is behind curtain # - Correct

In this case, 2 out of 3 possibilities are correct, so your chances of being right are 2/3 if you choose both #2 and #3, but just 1/3 if you choose #1. If you make that choice and then it is revealed that the car isn't behind curtain #3, you're back to 2 possibilities and a 1/2 chance of being correct.

The TOG​
 
TOG said:
It's really just fractions, where the number of correct (or incorrect, depending on what you're looking at) choices is the numerator, and the total number of possibilities is the denominator. If you get to choose just one curtain of three (let's suppose you chose curtain #1), then the possibilities are:

You choose #1 and the car is behind curtain #1 - Correct
You choose #1 and the car is behind curtain #2
You choose #1 and the car is behind curtain #3

There are 3 possibilities and one of them is correct. You therefore have a 1/3 chance of getting it right if you choose curtain #1. The same applies if you choose one of the other curtains. Supposing that you have still chosen #1, then if the host opens one curtain (let's suppose it's #3), revealing that the car isn't there, you would have 2 possibilities left.

You choose #1 and the car is behind curtain #1 - Correct
You choose #1 and the car is behind curtain #2

There are now two possibilities and one of them is correct. You therefore have a 1/2 chance of being correct. That probability doesn't change by changing your choice. But it does change if you are allowed to pick both #2 and #3.

You choose #2 and #3 and the car is behind curtain #1
You choose #2 and #3 and the car is behind curtain #2 -Correct
You choose #2 and #3 and the car is behind curtain # - Correct

In this case, 2 out of 3 possibilities are correct, so your chances of being right are 2/3 if you choose both #2 and #3, but just 1/3 if you choose #1. If you make that choice and then it is revealed that the car isn't behind curtain #3, you're back to 2 possibilities and a 1/2 chance of being correct.

The TOG​
Click to expand...

And that makes sense even to my puny math brain.
 
TOG,
re: "But it does change if you are allowed to pick both #2 and #3...If you make that choice and then it is revealed that the car isn't behind curtain #3, you're back to 2 possibilities and a 1/2 chance of being correct."


That is incorrect. Before any curtain is opened, you know that at least one of curtains 2 and 3 doesn't have the car behind it. Yet there is still a 2/3rds chance that it is behind one of them. So the host knowingly opening a curtain without the car doesn't change the 2/3rds odds.
 
Last edited:
We could imagine different scenarios with different numbers of curtains or different numbers of choices. You have already presented a scenario with two choices, for example. Would you agree that the following statement would apply regardless of the number of curtains or choices?

The probability of choosing correctly depends on the number of curtains and the number of choices, but has nothing to do with who makes the choice.​

Is that a correct statement? If not, please explain what's wrong with it.

The TOG​
 
TOG,
re: "The probability of choosing correctly depends on the number of curtains and the number of choices, but has nothing to do with who makes the choice. Is that a correct statement?"

I think it is. What is your point?
 
rstrats said:
TOG,
re: "The probability of choosing correctly depends on the number of curtains and the number of choices, but has nothing to do with who makes the choice. Is that a correct statement?"

I think it is. What is your point?
Click to expand...

I'll get to that. Just one more question first. This one is pretty obvious, but I want to ask it anyway. Then I'll get to my point.

If there were only two curtains, rather than three, wouldn't you agree that there was a 50% chance of being right, regardless of which curtain you chose?

I promise. This will all make sense in my next post.

The TOG​
 
TOG,

re: "If there were only two curtains, rather than three, wouldn't you agree that there was a 50% chance of being right,"

Yes.
 
rstrats said:
TOG,

re: "If there were only two curtains, rather than three, wouldn't you agree that there was a 50% chance of being right,"

Yes.
Click to expand...

So, we agree that it doesn't matter who guesses which curtain the car is behind and that there is a 50% (1/2) chance of being right if there are only 2 curtains. We've previously established that there is a 33% (1/3) chance of being right if there are three curtains. Let's change the scenario slightly and see what happens.

To start with, all three curtains are closed and you are asked to pick one. You have a 1/3 chance of being correct, whichever curtain you choose. Assume you pick curtain #1. The host opens curtain #3, revealing no prize. You sit down, and I take your place. I was not present when you made your choice, and don't know what choice you made. Since I can see that the car is not behind curtain #3, I have only 2 to choose from, which means that I have a 1/2 chance of being correct. Since it doesn't matter who picks a curtain, you would have had the exact same chance if you had been asked to choose again, this time between curtain #1 and curtain #2.

The TOG​
 
(bump) ? .. retired math major here (i accept no credit towards that any more).
after one of the curtains was revealed. that changed the odds. it becames as if that one that was opened wasn't there to start with anymore, as far as the new odds go.
(Eugene was right)
 
TOG,
re: "Since I can see that the car is not behind curtain #3, I have only 2 to choose from, which means that I have a 1/2 chance of being correct."

That is incorrect. There is never a 50/50 chance. I choose curtain #1 which has a 1/3rd chance of having the car. That means that there is a 2/3rds chance that the car is behind one of the other two curtains even though you know that the car is not behind at least one of those curtains. The host knows where the car is so he opens one of those curtains that doesn't have the car. At this point there is still a 2/3rds chance that the car is behind curtain #2. Remember, you knew that at least one of those curtains didn't have the car so the 2/3rds chance still remains.

I sit down and you take over. Let's say you pick curtain #2 which remember has a 2/3rds chance of having the car. So in this instance if you were to switch to curtain #1, you would be going from a 2/3rds chance to a 1/3rd chance.
 
His_nee (Jeff),

re: "after one of the curtains was revealed. that changed the odds."


That is incorrect. What would you do if after your initial pick of curtain #1, and before any curtain is opened, the host tells you that you can switch to both curtains 2 and 3? And note that the host is not trying to trick or influence you in anyway. Even If the car is behind curtain 2 or 3, he will still give you the offer to switch.
 
rstrats said:
TOG,
re: "Since I can see that the car is not behind curtain #3, I have only 2 to choose from, which means that I have a 1/2 chance of being correct."

That is incorrect. There is never a 50/50 chance. I choose curtain #1 which has a 1/3rd chance of having the car. That means that there is a 2/3rds chance that the car is behind one of the other two curtains even though you know that the car is not behind at least one of those curtains. The host knows where the car is so he opens one of those curtains that doesn't have the car. At this point there is still a 2/3rds chance that the car is behind curtain #2. Remember, you knew that at least one of those curtains didn't have the car so the 2/3rds chance still remains.

I sit down and you take over. Let's say you pick curtain #2 which remember has a 2/3rds chance of having the car. So in this instance if you were to switch to curtain #1, you would be going from a 2/3rds chance to a 1/3rd chance.
Click to expand...

You already agreed that there was a 50/50 chance if there were only two curtains. You also agreed that the odds don't change if a different person is choosing a curtain. In the example I gave above, I only had two curtains to choose from, thus a 50/50 chance. The odds would have been the same if you had been choosing. Just look at all the possibilities and count for yourself.

C = Car
N = Nothing
X = Excluded curtain
Leftmost character = Curtain #1, Rightmost character = Curtain #3

With all three curtains closed
  1. C - N - N
  2. N - C - N
  3. N - N - C
There are three possibilities. One of them has the car behind curtain #1, which means you have a 1/3 chance of being right if you choose curtain #1. If you get to choose both curtains #2 and #3, you have a 2/3 chance of being right, since two of the possibilities have the car behind one of those curtains. Now open the third curtain, as you suggested, revealing that there is no car there. Then the possibilities look like this.
  1. C - N - X
  2. N - C - X
There are only 2 possibilities. Either the car is behind curtain #1 or curtain #2. Since there are two possibilities and you have to choose one of them, you have a 1 out of 2 (or 1/2 or 50% or 50/50, if you'd rather use those terms) chance of being right. There is no possible third option, so you can't have a 1 out of 3 or a 2 out of 3 chance. The third option has been removed from the equation and has become irrelevant. You have the same chance as if curtain #3 had never been there.

The TOG​
 
TOG,
re: "You already agreed that there was a 50/50 chance if there were only two curtains."

That is correct, if there were only 2 curtains at the start of the game.



re: "You also agreed that the odds don't change if a different person is choosing a curtain.

Correct. It doesn't matter if the host opens curtain 3 or you personally open it.



re: "There are only 2 possibilities. Either the car is behind curtain #1 or curtain #2. Since there are two possibilities and you have to choose one of them, you have a 1 out of 2 (or 1/2 or 50% or 50/50..."

That is incorrect. Please read post #15.
 
rstrats said:
TOG,
re: "You already agreed that there was a 50/50 chance if there were only two curtains."

That is correct, if there were only 2 curtains at the start of the game.



re: "You also agreed that the odds don't change if a different person is choosing a curtain.

Correct. It doesn't matter if the host opens curtain 3 or you personally open it.



re: "There are only 2 possibilities. Either the car is behind curtain #1 or curtain #2. Since there are two possibilities and you have to choose one of them, you have a 1 out of 2 (or 1/2 or 50% or 50/50..."

That is incorrect. Please read post #15.
Click to expand...

I give up. I have shown you many proofs that you are wrong, but when you start purposely misinterpreting me (and that is what it looks like you're doing), then I won't bother any more. It was obvious from the context that I was referring to the person choosing which curtain to pick as the one with the car, not the one choosing which to open. Believe what you will. It won't change the fact that you're wrong.

The TOG​
 
rstrats said:
You are a contestant on a game show. There are three curtains. Behind one of the curtains is a new car. You are asked to choose one of the curtains. Lets say that you choose curtain #1. The host of the show - who knows where the car is so as not to end the game prematurely - opens curtain #3 and of course there is no car behind it. The host now gives you a choice. You can stay with curtain #1 or you can change your choice to curtain #2. The question now is: would it be to your advantage to stay with curtain #1, or would it be to your advantage to change to curtain #2 or would there be no advantage either way?
Click to expand...

There would be no advantage either way. The car is behind 1 of 2 curtains. The odds are 50/50. Changing from curtain #1 to curtain #2 doesn't change those odds. You still have a 50/50 chance of being right or wrong.
.
 
TOG,

You keep ignoring post #15. Tell me what is wrong with it.
 
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