If I were actually in situation A, B, or C, I would expect a 99% chance of a blue door, and in D, E, or F, a 50%, and I would actually bet with this expectation.
There is really no practical way to implement this, however, because of the assumption that random events turn out in a certain way, e.g. it is assumed that there is only a 50% chance that I will survive, yet I always do, in order for the case to be the one under consideration.
Omega runs 10,000 trials of scenario F, and puts you in touch with 100 random people still in their room who believe there is a %50 chance they have red doors, and will happily take 10 to 1 bets that they are.
You take these bets, collect $1 each from 98 of them, and pay out $10 each to 2.
You assume that the 100 people have been chosen randomly from all the people in the 10,000 trials. This is not valid. The appropriate way for these bets to take place is to choose one random person from one trial, then another random person from another trial, and so on. In this way about 50 of the hundred persons will be behind red doors.
The reason for this is that if I know that this setup has taken place 10,000 times, my estimate of the probability that I am behind a blue door will not be the same as if the setup has happened only once. The probability will slowly drift toward 99% as the number of trials increases. In order to prevent this drift, you have to select the persons as stated above.
The reason is that “I” could be anyone out of the full set of two trials. So: there is a 25% chance there both trials ended with red-doored survivors; a 25% chance that both trials ended with blue-doored survivors; and a 50% chance that one ended with a red door, one with a blue.
If both were red, I have a red door (100% chance). If both were blue, I have a blue door (100% chance). But if there was one red and one blue, then there are a total of 100 people, 99 blue and one red, and I could be any of them. So in this case there is a 99% chance I am behind a blue door.
Putting these things together, if I calculate correctly, the total probability here (in the case of two trials) is that I have a 25.5% chance of being behind a red door, and a 74.5% chance of being behind a blue door. In a similar way you can show that as you add more trials, your probability will get ever closer to 99% of being behind a blue door.
But if there was one red and one blue, then there are a total of 100 people, 99 blue and one red, and I could be any of them. So in this case there is a 99% chance I am behind a blue door.
You could only be in one trial or the other.
What if Omega says you’re in the second trial, not the first?
“I could be any of them” in the sense that all the factors that influence my estimate of the probability, will influence the estimate of the probability made by all the others. Omega may tell me I am in the second trial, but he could equally tell someone else (or me) that he is in the first trial. There are still 100 persons, 99 behind blue doors and 1 behind red, and in every way which is relevant, I could be any of them. Thinking that the number of my trial makes a difference would be like thinking that if Omega tells me I have brown eyes and someone else has blue, that should change my estimate.
Likewise with trial 3854 out of 10,000. Naturally each person is in one of the trials, but the persons trial number does not make a significant contribution to his estimate. So I stand by the previous comments.
“I could be any of them” in the sense that all the factors that influence my estimate of the probability, will influence the estimate of the probability made by all the others
These factors should not influence your estimation of the probability, because you could not be any of the people in the other trials, red or blue, because you are only in your trial. (and all of those people should know they can’t be you)
The only reason you would take the trials together as an aggregate is if you were betting on it from the outside, and the person you’re betting against could be in any of the trials.
Omega could tell you the result of the other trials, (1 other or 9999 others,) you’d know exactly how many reds and blues there are, except for your trial. You must asses your trial in the same way you would if it were stand alone.
What if Omega says you are in the most recent trial of 40, because Omega has been running trials every hundred years for 4000 years? You can’t be any of those people. (to say nothing of other trials that other omegas might have run.)
But you could be any of 99 people if the coin came up heads.
If Omega does not tell me the result of the other trials, I stand by my point. In effect he has given me no information, and I could be anyone.
If Omega does tell me the results of all the other trials, it is not therefore the case that I “must assess my trial in the same way as if it stood alone.” That depends on how Omega selected me as the one to estimate the probability. If in fact Omega selected me as a random person from the 40 trials, then I should estimate the probability by estimating the number of persons behind blue door and red doors, and assuming that I could with equal probability have been any of them. This will imply a very high probability of being behind a blue door, but not quite 99%.
If he selected me in some other way, and I know it, I will give a different estimate.
If I do not know how he selected me, I will give a subjective estimate depending on my estimate of ways that he might have selected me; for example I might assign some probability to his having deliberately selected me as one of the red-doored persons, in order to win if I bet. There is therefore no “right” probability in this situation.
In effect he has given me no information, and I could be anyone.
How is it the case that you could be in the year 1509 trial, when it is in fact 2009? (omega says so)
Is it also possible that you are someone from the quite likely 2109 trial? (and so on into the future)
That depends on how Omega selected me as the one to estimate the probability.
I was thinking he could tell every created person the results of all the other trials. I agree that if your are selected for something (information revelatiion, betting, whatever), then information about how you were selected could hint at the color of your door.
Information about the results of any other trials tells you nothing about your door.
If he tells every person the results of all the other trials, I am in effect a random person from all the persons in all the trials, because everyone is treated equally. Let’s suppose there were just 2 trials, in order to simplify the math. Starting with the prior probabilities based on the coin toss, there is a 25% chance of a total of just 2 observers behind red doors, in which case I would have a 100% chance of being behind a red door. There is a 50% chance of 1 observer behind a red door and 99 observers behind blue doors, which would give me a 99% chance of being behind a blue door. There is a 25% chance of 198 observers behind blue doors, which would give me a 100% chance of being behind a blue door. So my total prior probabilities are 25.5% of being behind a red door, and 74.5% of being behind a blue door.
Let’s suppose I am told that the other trial resulted in just one observer behind a red door. First we need the prior probability of being told this. If there were two red doors (25% chance), there would be a 100% chance of this. If there were two blue doors (25% chance), there would be a 0% chance of this. If there was a red door and a blue door (50% chance), there would be a 99% chance of this. So the total prior probability of being told that the other trial resulted in a red door is again 74.5%, and the probability of being told that the other trial resulted in a blue door is 25.5%.
One more probability: given that I am behind a red door, what is the probability that I will be told that the other trial resulted in an observer behind a red door? There was originally a 25% chance of two red trials, and a 50% chance of 1 red and 1 blue trial. This implies that given that I am behind a red door, there is a 1⁄3 chance that I will be told that the other trial resulted in red, and a 2⁄3 that I will be told that the other trial resulted in blue. (Once again things will change if we run more trials, for similar reasons, because in the 1⁄3 case, there are 2 observers behind red doors.)
Applying Bayes’ theorem, then, the probability that I am behind a red door given that I am told that the other trial resulted in an observer behind a red door, is (.255 / .745) x (1/3) = approximately 11.4%. So the probability that I am behind a blue door is approximately 88.6%. Since it was originally only 74.5% with two trials, information about the other trial did contribute to knowledge of my door. The same will happen as you add more trials and more information.
I admit I was blind sided by the possibility that information about the other trials could yield information about your door. I’ll have to review the monty hall problem.
Using your methods, I got:
Being blue given told red=(.745 being blue prior/.745 told red prior) x (2/3 told red given blue)=.666...
Which doesn’t match your 11.4%, so something is missing.
In scenario F, if you’re not told, why assume that your trial was the only one in the set? You should have some probability that the omegas would do this more than once.
Also, I agree that in theory you would have some subjective probability that there were other trials. But this prevents assigning any exact value to the probability because we can’t make any definitively correct answer. So I was assuming that you either know that the event is isolated, or you know that it is not, so that you could assign a definite value.
I’m not sure what it would mean for the event to be isolated. (Not to contradict my previous statement that you have to treat it as a stand alone event. My position is that it is .99 for any number of trials, though I still need to digest your corrected math.)
I’m not sure how different an event could be before you don’t need to consider it part of the set you could have found yourself in.
If you’re in a set of two red-blue trials, and omegas says there is another set of orange-green trials run the same and likewise told about the red-blues, then it seems you would need to treat that as a set of 4.
If you know you’re in a trial with the (99 blue or 1 red) protocol, but there is also a trial with a (2 blue or 1 red) protocol, then those 1 or 2 people will skew your probabilities slightly.
If Omega tells you there is an intelligent species of alien in which male conceptions yield 99 identical twins and female conceptions only 1, with a .50 probability of conceiving female, and in which the young do not know their gender until maturity… then is that also part of the set you could have been in? If not, I’m honestly not sure where to draw the line. If so, then there I’d expect we could find so many such situations that apply to how individual humans come to exist now, so there may be billions of trials.
You’re correct, I made a serious error in the above calculations. Here are the corrected results:
Prior probability for situation A, namely both trials result in red doors: .25;
Prior probability for situation B, namely one red and one blue: .50;
Prior probability for situation C, namely both trials result in blue doors: .25;
Prior probability for me getting a blue door: .745;
Prior probability for me getting a red door: .255;
Prior probability of the other trial getting red: .745;
Prior probability of the other trial getting blue: .255;
Then probability of situation A, given I have a red door = (Pr(A)/Pr(red)) x P(red given A). Pr(red given A)=1, so the result is pr(A given red) = .25/.255 = .9803921...
So the probability that I will be told red, given I have red, is not 1⁄3, but over 98% (namely the same value above)! And so the probability that I will be told blue, given I have red, is of course .01960784, namely the probability of situation B given that I have a red door.
So using Bayes’ theorem with the corrected values, the probability me having a red door, given that I am told the other resulted in red = (pr being red/ pr other red) x pr (told red given red) = (.255/.745) x .9803921… = .33557… or approximately 1⁄3.
You can work out the corresponding calculation (probability of being blue given told red) by starting with the probability of situation C given I have a blue door, and then deriving the probability of B given I have a blue, and you will see that it matches this one (i.e. it will be approximately 2⁄3.)
Thanks! I think this comment is the best so far for demonstrating the confusion (well, I was confused :-) about the different possible meanings of the phrase “you are an observer chosen from such and such set”. Perhaps a more precise and unambiguous phrasing could be used.
This reinforces my feeling that something is deeply wrong with the statement of the problem, or with my understanding of it. It’s true that some random survivor is p=.99 likely to be behind a blue door. It does not seem true for me, given that I survive.
If I were actually in situation A, B, or C, I would expect a 99% chance of a blue door, and in D, E, or F, a 50%, and I would actually bet with this expectation.
There is really no practical way to implement this, however, because of the assumption that random events turn out in a certain way, e.g. it is assumed that there is only a 50% chance that I will survive, yet I always do, in order for the case to be the one under consideration.
Omega runs 10,000 trials of scenario F, and puts you in touch with 100 random people still in their room who believe there is a %50 chance they have red doors, and will happily take 10 to 1 bets that they are.
You take these bets, collect $1 each from 98 of them, and pay out $10 each to 2.
Were their bets rational?
You assume that the 100 people have been chosen randomly from all the people in the 10,000 trials. This is not valid. The appropriate way for these bets to take place is to choose one random person from one trial, then another random person from another trial, and so on. In this way about 50 of the hundred persons will be behind red doors.
The reason for this is that if I know that this setup has taken place 10,000 times, my estimate of the probability that I am behind a blue door will not be the same as if the setup has happened only once. The probability will slowly drift toward 99% as the number of trials increases. In order to prevent this drift, you have to select the persons as stated above.
If you find yourself in such a room, why does your blue door estimate go up with the number of trials you know about? Your coin was still 50-50.
How much does it go up for each additional trial? ie what are your odds if omega tells you you’re in one of two trials of F?
The reason is that “I” could be anyone out of the full set of two trials. So: there is a 25% chance there both trials ended with red-doored survivors; a 25% chance that both trials ended with blue-doored survivors; and a 50% chance that one ended with a red door, one with a blue.
If both were red, I have a red door (100% chance). If both were blue, I have a blue door (100% chance). But if there was one red and one blue, then there are a total of 100 people, 99 blue and one red, and I could be any of them. So in this case there is a 99% chance I am behind a blue door.
Putting these things together, if I calculate correctly, the total probability here (in the case of two trials) is that I have a 25.5% chance of being behind a red door, and a 74.5% chance of being behind a blue door. In a similar way you can show that as you add more trials, your probability will get ever closer to 99% of being behind a blue door.
You could only be in one trial or the other.
What if Omega says you’re in the second trial, not the first?
Or trial 3854 of 10,000?
“I could be any of them” in the sense that all the factors that influence my estimate of the probability, will influence the estimate of the probability made by all the others. Omega may tell me I am in the second trial, but he could equally tell someone else (or me) that he is in the first trial. There are still 100 persons, 99 behind blue doors and 1 behind red, and in every way which is relevant, I could be any of them. Thinking that the number of my trial makes a difference would be like thinking that if Omega tells me I have brown eyes and someone else has blue, that should change my estimate.
Likewise with trial 3854 out of 10,000. Naturally each person is in one of the trials, but the persons trial number does not make a significant contribution to his estimate. So I stand by the previous comments.
These factors should not influence your estimation of the probability, because you could not be any of the people in the other trials, red or blue, because you are only in your trial. (and all of those people should know they can’t be you)
The only reason you would take the trials together as an aggregate is if you were betting on it from the outside, and the person you’re betting against could be in any of the trials.
Omega could tell you the result of the other trials, (1 other or 9999 others,) you’d know exactly how many reds and blues there are, except for your trial. You must asses your trial in the same way you would if it were stand alone.
What if Omega says you are in the most recent trial of 40, because Omega has been running trials every hundred years for 4000 years? You can’t be any of those people. (to say nothing of other trials that other omegas might have run.)
But you could be any of 99 people if the coin came up heads.
If Omega does not tell me the result of the other trials, I stand by my point. In effect he has given me no information, and I could be anyone.
If Omega does tell me the results of all the other trials, it is not therefore the case that I “must assess my trial in the same way as if it stood alone.” That depends on how Omega selected me as the one to estimate the probability. If in fact Omega selected me as a random person from the 40 trials, then I should estimate the probability by estimating the number of persons behind blue door and red doors, and assuming that I could with equal probability have been any of them. This will imply a very high probability of being behind a blue door, but not quite 99%.
If he selected me in some other way, and I know it, I will give a different estimate.
If I do not know how he selected me, I will give a subjective estimate depending on my estimate of ways that he might have selected me; for example I might assign some probability to his having deliberately selected me as one of the red-doored persons, in order to win if I bet. There is therefore no “right” probability in this situation.
How is it the case that you could be in the year 1509 trial, when it is in fact 2009? (omega says so)
Is it also possible that you are someone from the quite likely 2109 trial? (and so on into the future)
I was thinking he could tell every created person the results of all the other trials. I agree that if your are selected for something (information revelatiion, betting, whatever), then information about how you were selected could hint at the color of your door.
Information about the results of any other trials tells you nothing about your door.
If he tells every person the results of all the other trials, I am in effect a random person from all the persons in all the trials, because everyone is treated equally. Let’s suppose there were just 2 trials, in order to simplify the math. Starting with the prior probabilities based on the coin toss, there is a 25% chance of a total of just 2 observers behind red doors, in which case I would have a 100% chance of being behind a red door. There is a 50% chance of 1 observer behind a red door and 99 observers behind blue doors, which would give me a 99% chance of being behind a blue door. There is a 25% chance of 198 observers behind blue doors, which would give me a 100% chance of being behind a blue door. So my total prior probabilities are 25.5% of being behind a red door, and 74.5% of being behind a blue door.
Let’s suppose I am told that the other trial resulted in just one observer behind a red door. First we need the prior probability of being told this. If there were two red doors (25% chance), there would be a 100% chance of this. If there were two blue doors (25% chance), there would be a 0% chance of this. If there was a red door and a blue door (50% chance), there would be a 99% chance of this. So the total prior probability of being told that the other trial resulted in a red door is again 74.5%, and the probability of being told that the other trial resulted in a blue door is 25.5%.
One more probability: given that I am behind a red door, what is the probability that I will be told that the other trial resulted in an observer behind a red door? There was originally a 25% chance of two red trials, and a 50% chance of 1 red and 1 blue trial. This implies that given that I am behind a red door, there is a 1⁄3 chance that I will be told that the other trial resulted in red, and a 2⁄3 that I will be told that the other trial resulted in blue. (Once again things will change if we run more trials, for similar reasons, because in the 1⁄3 case, there are 2 observers behind red doors.)
Applying Bayes’ theorem, then, the probability that I am behind a red door given that I am told that the other trial resulted in an observer behind a red door, is (.255 / .745) x (1/3) = approximately 11.4%. So the probability that I am behind a blue door is approximately 88.6%. Since it was originally only 74.5% with two trials, information about the other trial did contribute to knowledge of my door. The same will happen as you add more trials and more information.
Well you very nearly ruined my weekend. :-)
I admit I was blind sided by the possibility that information about the other trials could yield information about your door. I’ll have to review the monty hall problem.
Using your methods, I got:
Being blue given told red=(.745 being blue prior/.745 told red prior) x (2/3 told red given blue)=.666...
Which doesn’t match your 11.4%, so something is missing.
In scenario F, if you’re not told, why assume that your trial was the only one in the set? You should have some probability that the omegas would do this more than once.
Also, I agree that in theory you would have some subjective probability that there were other trials. But this prevents assigning any exact value to the probability because we can’t make any definitively correct answer. So I was assuming that you either know that the event is isolated, or you know that it is not, so that you could assign a definite value.
I’m not sure what it would mean for the event to be isolated. (Not to contradict my previous statement that you have to treat it as a stand alone event. My position is that it is .99 for any number of trials, though I still need to digest your corrected math.)
I’m not sure how different an event could be before you don’t need to consider it part of the set you could have found yourself in.
If you’re in a set of two red-blue trials, and omegas says there is another set of orange-green trials run the same and likewise told about the red-blues, then it seems you would need to treat that as a set of 4.
If you know you’re in a trial with the (99 blue or 1 red) protocol, but there is also a trial with a (2 blue or 1 red) protocol, then those 1 or 2 people will skew your probabilities slightly.
If Omega tells you there is an intelligent species of alien in which male conceptions yield 99 identical twins and female conceptions only 1, with a .50 probability of conceiving female, and in which the young do not know their gender until maturity… then is that also part of the set you could have been in? If not, I’m honestly not sure where to draw the line. If so, then there I’d expect we could find so many such situations that apply to how individual humans come to exist now, so there may be billions of trials.
You’re correct, I made a serious error in the above calculations. Here are the corrected results:
Prior probability for situation A, namely both trials result in red doors: .25; Prior probability for situation B, namely one red and one blue: .50; Prior probability for situation C, namely both trials result in blue doors: .25; Prior probability for me getting a blue door: .745; Prior probability for me getting a red door: .255; Prior probability of the other trial getting red: .745; Prior probability of the other trial getting blue: .255;
Then probability of situation A, given I have a red door = (Pr(A)/Pr(red)) x P(red given A). Pr(red given A)=1, so the result is pr(A given red) = .25/.255 = .9803921...
So the probability that I will be told red, given I have red, is not 1⁄3, but over 98% (namely the same value above)! And so the probability that I will be told blue, given I have red, is of course .01960784, namely the probability of situation B given that I have a red door.
So using Bayes’ theorem with the corrected values, the probability me having a red door, given that I am told the other resulted in red = (pr being red/ pr other red) x pr (told red given red) = (.255/.745) x .9803921… = .33557… or approximately 1⁄3.
You can work out the corresponding calculation (probability of being blue given told red) by starting with the probability of situation C given I have a blue door, and then deriving the probability of B given I have a blue, and you will see that it matches this one (i.e. it will be approximately 2⁄3.)
Thanks! I think this comment is the best so far for demonstrating the confusion (well, I was confused :-) about the different possible meanings of the phrase “you are an observer chosen from such and such set”. Perhaps a more precise and unambiguous phrasing could be used.
Clearly the bets would not be rational.
This reinforces my feeling that something is deeply wrong with the statement of the problem, or with my understanding of it. It’s true that some random survivor is p=.99 likely to be behind a blue door. It does not seem true for me, given that I survive.