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This is a two part question:

1. You toss a fair coin four times in a row and it lands on tails all four times. What is the probability that your next coin toss will be tails?

2. A carnie at the fair has hidden a gold ring under one of three cups, marked A, B, and C. He lets you name one cup as your choice. You pick cup A. He then lifts cup C, showing you that it's empty, and allows you to choose again. What are the odds that the ring is under cup B?

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#1: always 50/50 chance whenever coin is flipped.

#2: Monty Hall paradox. 1/3 chance of Goat. Better switch cups! ;)

This is a two part question:

1. You toss a fair coin four times in a row and it lands on tails all four times. What is the probability that your next coin toss will be tails?

2. A carnie at the fair has hidden a gold ring under one of three cups, marked A, B, and C. He lets you name one cup as your choice. You pick cup A. He then lifts cup C, showing you that it's empty, and allows you to choose again. What are the odds that the ring is under cup B?

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the answer for both questions is 50/50

This is a two part question:

1. You toss a fair coin four times in a row and it lands on tails all four times. What is the probability that your next coin toss will be tails?

2. A carnie at the fair has hidden a gold ring under one of three cups, marked A, B, and C. He lets you name one cup as your choice. You pick cup A. He then lifts cup C, showing you that it's empty, and allows you to choose again. What are the odds that the ring is under cup B?

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1. If you are calculating for the last coin only, its 50/50. if you want the probability to get 5 tosses and having all of them the same, then do a probability calcualtion (dont think that applies here tho)

2. 2/3 chance that its in the B cup. The carnie by ruling out the empty cup has given u 1/3 (edit from 2/3 error) chance for free if u switch cup.

Edited by Ronin
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This is a two part question:

1. You toss a fair coin four times in a row and it lands on tails all four times. What is the probability that your next coin toss will be tails?

2. A carnie at the fair has hidden a gold ring under one of three cups, marked A, B, and C. He lets you name one cup as your choice. You pick cup A. He then lifts cup C, showing you that it's empty, and allows you to choose again. What are the odds that the ring is under cup B?

1. 50%

2. 2/3

Edited by renan
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He asks what is the probability the ring is under cup B, not the probability you'll be subject to find the ring! So I agree with 1/3.

I think they are related. When the person who knows the correct location give away one or more wrong answer possibility, the possibility of yours initial chice also been wrong increses by that amount.

In the beginign, the possibility was 1/3. Now, knowing that the other C cant be the right anwser, the chance of youre initial choice been the correct keeps in the initial possibility (1/3) but, the chances of the B is changed by the present impossibility of C, so that it is now 2/3.

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1. If you are calculating for the last coin only, its 50/50. if you want the probability to get 5 tosses and having all of them the same, then do a probability calcualtion (dont think that applies here tho)

2. 2/3 chance that its in the B cup. The carnie by ruling out the empty cup has given u 1/3 (edit from 2/3 error) chance for free if u switch cup.

Right on both counts. I paired the two problems for a reason. In the first scenario, knowing the events leading up to the toss has no impact on the odds. In the second, knowing the events leading up to the guess can substantially improve your odds compared to someone simply seeing a choice between cup A and cup B.

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I think they are related. When the person who knows the correct location give away one or more wrong answer possibility, the possibility of yours initial chice also been wrong increses by that amount.

In the beginign, the possibility was 1/3. Now, knowing that the other C cant be the right anwser, the chance of youre initial choice been the correct keeps in the initial possibility (1/3) but, the chances of the B is changed by the present impossibility of C, so that it is now 2/3.

Shouldn't the fact that it is a "carnie" be factored into the equation?

Adding the carnie factor tells me the likelihood of the ring being under any cup selected is zero.

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Shouldn't the fact that it is a "carnie" be factored into the equation?

Adding the carnie factor tells me the likelihood of the ring being under any cup selected is zero.

Ok, thats a possibility

:D

But if so, we could consider, in the first question, taht the cois could land and stand by its side, so that it's not head neither tails. So the chance would be 1/3. ;)

Lets say hipoteticaly that we have a ring under one of the cups B))

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Ok, thats a possibility
:D

But if so, we could consider, in the first question, taht the cois could land and stand by its side, so that it's not head neither tails. So the chance would be 1/3. ;)

Lets say hipoteticaly that we have a ring under one of the cups B))

the answer to the first question is 50%. The probability of getting one outcome out of a possible two outcomes is the same, no matter what the previous outcomes were.

I know there is a complicated answer to the second question, and that the probability is not 50%. I have absolutely no idea what the probability is, though. I know it's the classic game show, or "Let's Make a Deal" problem. The contestant's chances improve if he or she chooses the other cup.

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If we agree that the first answer is 50%, then why isn't the answer to the second 50%? if the events before the last last flip don't matter then why does it affect the cup scenario. I know the math, but look at it this way, you choose cup A and cup C is revealed, and we assume that the odd's of it being A are 33% and "not A" is 66%. So the logic is choose B, but if someone were to come in after cup C was removed, wouldn't the odds for them be 50% for both A and B?

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If we agree that the first answer is 50%, then why isn't the answer to the second 50%? if the events before the last last flip don't matter then why does it affect the cup scenario. I know the math, but look at it this way, you choose cup A and cup C is revealed, and we assume that the odd's of it being A are 33% and "not A" is 66%. So the logic is choose B, but if someone were to come in after cup C was removed, wouldn't the odds for them be 50% for both A and B?

i don't really understand how this works, and i never have. no one has really ever been able to effectively explain this to me so that I understand, so I have never remembered the answer.

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i don't really understand how this works, and i never have. no one has really ever been able to effectively explain this to me so that I understand, so I have never remembered the answer.

Its easier to understand if you think with largers proportions. Lets change a little: If it was 100.000 cups instead of only 3. One has a ring. You choose one cup and, right now, you have only 1 possibility in 100.000 to had chossen the right one.

But, the guy that knows were the ring is removes 99.998 cups, showing that the ring was not in any of them.

Now, we have only 2 cups. The one you chose before (that has a chance of 1/100.000 of been the correct) and the only other one left in the game, that had adquired, when the guy shows the others cups impossibilities, all these cups chances of beign the correct one.

Just think whats more possible: The cup you initaly choose has the ring (1/100.00, youre a very luck guy) or the only one left, with all the rest of possibilities (99.999/100.000)?

Now can you understand it?

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Its easier to understand if you think with largers proportions. Lets change a little: If it was 100.000 cups instead of only 3. One has a ring. You choose one cup and, right now, you have only 1 possibility in 100.000 to had chossen the right one.

But, the guy that knows were the ring is removes 99.998 cups, showing that the ring was not in any of them.

Now, we have only 2 cups. The one you chose before (that has a chance of 1/100.000 of been the correct) and the only other one left in the game, that had adquired, when the guy shows the others cups impossibilities, all these cups chances of beign the correct one.

Just think whats more possible: The cup you initaly choose has the ring (1/100.00, youre a very luck guy) or the only one left, with all the rest of possibilities (99.999/100.000)?

Now can you understand it?

Oh,

So, since he chose C when he could have just as easily chosen B, that means that there is a higher chance that B has the ring? And so, if you know the carnie will be lifting a cup, you should try to choose the one you think doesn't have the ring!

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Oh,
So, since he chose C when he could have just as easily chosen B, that means that there is a higher chance that B has the ring? And so, if you know the carnie will be lifting a cup, you should try to choose the one you think doesn't have the ring!

You got it.

The important thing is: the carnie knows where is the ring and it will not show it. So when he lifts the cup that does not has the ring, the possibility of the other one increases.

You dont have to worry about choosing the cup that does not has the ring. Just choose randomly and the statistic will help you to choose the wrong one :D .

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Right on both counts. I paired the two problems for a reason. In the first scenario, knowing the events leading up to the toss has no impact on the odds. In the second, knowing the events leading up to the guess can substantially improve your odds compared to someone simply seeing a choice between cup A and cup B.

WRONG! you should know the answer to your own riddle. yes, the carnie has given you 1/3 of the possibility, but you have to spread that between the two unknowns, so the possibility of B having the ring is 1/3 plus 0.5/3 or 1.5/3. I'm pretty sure that equals 50/50 (50%). Stop thinking so much. There are only two possibilities left and you have no idea which cup the ring is under. 50% chance it's B.

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Its easier to understand if you think with largers proportions. Lets change a little: If it was 100.000 cups instead of only 3. One has a ring. You choose one cup and, right now, you have only 1 possibility in 100.000 to had chossen the right one.

But, the guy that knows were the ring is removes 99.998 cups, showing that the ring was not in any of them.

Now, we have only 2 cups. The one you chose before (that has a chance of 1/100.000 of been the correct) and the only other one left in the game, that had adquired, when the guy shows the others cups impossibilities, all these cups chances of beign the correct one.

Just think whats more possible: The cup you initaly choose has the ring (1/100.00, youre a very luck guy) or the only one left, with all the rest of possibilities (99.999/100.000)?

Now can you understand it?

like a fallacy to me. his removing the other incorrect choices does increase the chances that the other cup (the one you didn't choose) has the ring in it--i will grant you that. but does this not also increase the chances that your cup has the ring under it? Originally, the probability of the ring being underneath any of the three cups was 1/3. Once one of the incorrect options is eliminated, the chances of the ring being under either of the two remaining cups would have to be 1/2. it is under one of the two cups, but it seems completely ridiculous to assume that one cup has a higher probability of having the ring than the other. i'm still not convinced.

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If the carnie knows where the ring is and reveals that it's not under cup C, then I would definately stick with cup A. If the ring was under cup B, then the carnie shouldn't give the option of choosing cup B instead of cup A. He would just reveal that I was wrong and take my money!

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WRONG! you should know the answer to your own riddle. yes, the carnie has given you 1/3 of the possibility, but you have to spread that between the two unknowns, so the possibility of B having the ring is 1/3 plus 0.5/3 or 1.5/3. I'm pretty sure that equals 50/50 (50%). Stop thinking so much. There are only two possibilities left and you have no idea which cup the ring is under. 50% chance it's B.

No, you are incorrect, the probability of the ring being under cup B are 2/3.

This is why:

The initial probability of the ring being under cup A was 1/3. 1 time out of 3, the ring is under A, 2 times out of 3, the ring is not under A.

Now, one of the cups has been removed, and it did not have a ring under it.

So now, 2 out of three times the ring was NOT under A, and is under the other cup.

And 1 out of 3 times the ring WAS INDEED under A.

But the probability of it being under the new cup (not the cup you originally selected) is 2/3.

Sources:

http://en.wikipedia.org/wiki/Monty_Hall_problem

http://mathforum.org/dr.math/faq/faq.monty.hall.html

Two excellent sources that can explain it better than me.

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like a fallacy to me. his removing the other incorrect choices does increase the chances that the other cup (the one you didn't choose) has the ring in it--i will grant you that. but does this not also increase the chances that your cup has the ring under it? Originally, the probability of the ring being underneath any of the three cups was 1/3. Once one of the incorrect options is eliminated, the chances of the ring being under either of the two remaining cups would have to be 1/2. it is under one of the two cups, but it seems completely ridiculous to assume that one cup has a higher probability of having the ring than the other. i'm still not convinced.

The reason this seems like a fallacy is because the person removing the cup knows which cup the ring is under and did not remove it. For example:

Its easier to understand if you think with largers proportions. Lets change a little: If it was 100.000 cups instead of only 3. One has a ring. You choose one cup and, right now, you have only 1 possibility in 100.000 to had chossen the right one.

But, the guy that knows were the ring is removes 99.998 cups, showing that the ring was not in any of them.

Now, we have only 2 cups. The one you chose before (that has a chance of 1/100.000 of been the correct) and the only other one left in the game, that had adquired, when the guy shows the others cups impossibilities, all these cups chances of beign the correct one.

Just think whats more possible: The cup you initaly choose has the ring (1/100.00, youre a very luck guy) or the only one left, with all the rest of possibilities (99.999/100.000)?

Now can you understand it?

This is an excellent example. Suppose there are, as stated 100,000 lottery tickets, and one and only one will win. You have one lottery ticket.

Now a person WHO KNOWS WHICH LOTTERY TICKET IS THE WINNING ONE rips up 99,998 of the tickets, all except yours and one other. One of them has to win and, obviously, the odds of yours winning are only 1/100,000. The odds of the other one winning are 99,999/10000 because it is the only one left besides yours.

Now suppose that 99,998 tickets were RANDOMLY ripped up by someone WHO DID NOT KNOW THE WINNING TICKET.

Your probability of winning is still 1/100,000 and the probability of the other ticket winning is 1/100,000. The probability is 99,998/100,000 that one of the ripped-up tickets was the winning ticket.

The probability can change in the first example because the person KNEW WHICH WAS THE WINNING TICKET and accordingly did not rip it up.

Here is the problem, asked in a slightly different way, already on Brainden:

http://brainden.com/forum/index.php?showto...p;hl=monty+hall

EDIT: added link

Edited by rossbeemer
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