Which Twin has the Tony? [Archive] - Two Plus Two Older Archives

binions

06-29-2005, 02:00 PM

[ QUOTE ]
Two girls are identical twins. One was born first so is technically the older of the two. One of the girls has a Tony haircut while the other does not. We do not know which girl has the Tony and have no reason to think one is more likely than the other to have the Tony. One girl is behind door #1 and the other is behind door #2 with a 50-50 chance for either to be behind either door.

Now, door #1 is opened to show that the twin behind it has the Tony haircut. What is the probabilty that the girl behind door #2 is the older twin?

PairTheBoard

[/ QUOTE ]

50% unless you define "the twin" to mean the younger is "the twin" of the older, ie the older is the original, the younger is the twin.

In which case, when you say "the twin" has the Tony haircut, then the older girl must be behind Door #2.

PairTheBoard

06-29-2005, 05:44 PM

To be clear on the experiment; The twin girls and their haircuts are fixed. One of them is older. One has the Tony haircut, we just don't know which. The repeatable, random part of the experiment involves the doors they stand behind. If the experiment is repeated, half the time the older twin will be behind door #1 and half the time she will be behind door #2.

Before any door is opened the probabilty that the older twin is behind door #2 is 50%. But now suppose door #1 is opened to reveal the twin with the Tony. We still don't know whether she is the older or younger twin. But those who say that the probabilty is still 50% that the older twin is behind door #2 are wrong.

Do you see why?

PairTheBoard

BettyBoopAA

06-29-2005, 06:01 PM

NO
There's only 4 combinations
D1 has tony and is older
D1 has tony and is younger
D1 no tony and is older
D1 no toney and is younger
once you take away 2 options, it's still 50-50.

Stephen H

06-29-2005, 07:30 PM

[ QUOTE ]
To be clear on the experiment; The twin girls and their haircuts are fixed. One of them is older. One has the Tony haircut, we just don't know which. The repeatable, random part of the experiment involves the doors they stand behind. If the experiment is repeated, half the time the older twin will be behind door #1 and half the time she will be behind door #2.

Before any door is opened the probabilty that the older twin is behind door #2 is 50%. But now suppose door #1 is opened to reveal the twin with the Tony. We still don't know whether she is the older or younger twin. But those who say that the probabilty is still 50% that the older twin is behind door #2 are wrong.

Do you see why?

PairTheBoard

[/ QUOTE ]

The clarification, oddly enough, does not spell out the key facts given in the original problem; namely, that the Tony haircut being on the older twin isn't any more likely than the Tony haircut being on the younger twin (and, incidentally, there's no chance that both twins have the Tony haircut).

I'm also not sure what you mean by saying that the haircuts are fixed, but the door they are standing behind is not. Given the information we have, the Tony haircut being on the older or the younger is a 50/50 probability, nothing more, nothing less. If you repeat the experiment without possibility of moving the haircut, then the Tony location becomes known, and you aren't repeating the experiment, are you?

I don't see any reason why putting the sisters behind doors has changed the probability that the sister without the Tony is the older sister.

PairTheBoard

06-29-2005, 08:02 PM

[ QUOTE ]
[ QUOTE ]
To be clear on the experiment; The twin girls and their haircuts are fixed. One of them is older. One has the Tony haircut, we just don't know which. The repeatable, random part of the experiment involves the doors they stand behind. If the experiment is repeated, half the time the older twin will be behind door #1 and half the time she will be behind door #2.

Before any door is opened the probabilty that the older twin is behind door #2 is 50%. But now suppose door #1 is opened to reveal the twin with the Tony. We still don't know whether she is the older or younger twin. But those who say that the probabilty is still 50% that the older twin is behind door #2 are wrong.

Do you see why?

PairTheBoard

[/ QUOTE ]

The clarification, oddly enough, does not spell out the key facts given in the original problem; namely, that the Tony haircut being on the older twin isn't any more likely than the Tony haircut being on the younger twin (and, incidentally, there's no chance that both twins have the Tony haircut).

I'm also not sure what you mean by saying that the haircuts are fixed, but the door they are standing behind is not. Given the information we have, the Tony haircut being on the older or the younger is a 50/50 probability, nothing more, nothing less. If you repeat the experiment without possibility of moving the haircut, then the Tony location becomes known, and you aren't repeating the experiment, are you?

I don't see any reason why putting the sisters behind doors has changed the probability that the sister without the Tony is the older sister.

[/ QUOTE ]

I didn't say that the probabilty that the older twin has the Tony is 50%. Either she does or she doesn't. We just don't know. We do know that one and only one twin has the Tony.

Also, the experiment can be repeated by bringing in new observers each time. The same girls with the same haircuts get randomly shuffled behind the doors, and each time a new observer comes in and is asked for the probability that the older twin is behind door #2 after being shown the twin behind door #1. For those cases where the observer sees the twin with the Tony behind door #1, is it correct for him to say the probabilty is 50% that the older twin is behind door #2?

Or put another way, given the experiment as described, what is the conditional probabilty that the older twin is behind door #2 GIVEN that the twin with the Tony is behind door #1?

PairTheBoard

sully4321

06-29-2005, 08:02 PM

[ QUOTE ]
[ QUOTE ]
To be clear on the experiment; The twin girls and their haircuts are fixed. One of them is older. One has the Tony haircut, we just don't know which. The repeatable, random part of the experiment involves the doors they stand behind. If the experiment is repeated, half the time the older twin will be behind door #1 and half the time she will be behind door #2.

Before any door is opened the probabilty that the older twin is behind door #2 is 50%. But now suppose door #1 is opened to reveal the twin with the Tony. We still don't know whether she is the older or younger twin. But those who say that the probabilty is still 50% that the older twin is behind door #2 are wrong.

Do you see why?

PairTheBoard

[/ QUOTE ]

The clarification, oddly enough, does not spell out the key facts given in the original problem; namely, that the Tony haircut being on the older twin isn't any more likely than the Tony haircut being on the younger twin (and, incidentally, there's no chance that both twins have the Tony haircut).

I'm also not sure what you mean by saying that the haircuts are fixed, but the door they are standing behind is not. Given the information we have, the Tony haircut being on the older or the younger is a 50/50 probability, nothing more, nothing less. If you repeat the experiment without possibility of moving the haircut, then the Tony location becomes known, and you aren't repeating the experiment, are you?

I don't see any reason why putting the sisters behind doors has changed the probability that the sister without the Tony is the older sister.

[/ QUOTE ]

this is the dumbest thing i have ever heard of. it is 50/50 or you haven't given enough information. end of story.

PairTheBoard

06-29-2005, 08:12 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
To be clear on the experiment; The twin girls and their haircuts are fixed. One of them is older. One has the Tony haircut, we just don't know which. The repeatable, random part of the experiment involves the doors they stand behind. If the experiment is repeated, half the time the older twin will be behind door #1 and half the time she will be behind door #2.

Before any door is opened the probabilty that the older twin is behind door #2 is 50%. But now suppose door #1 is opened to reveal the twin with the Tony. We still don't know whether she is the older or younger twin. But those who say that the probabilty is still 50% that the older twin is behind door #2 are wrong.

Do you see why?

PairTheBoard

[/ QUOTE ]

The clarification, oddly enough, does not spell out the key facts given in the original problem; namely, that the Tony haircut being on the older twin isn't any more likely than the Tony haircut being on the younger twin (and, incidentally, there's no chance that both twins have the Tony haircut).

I'm also not sure what you mean by saying that the haircuts are fixed, but the door they are standing behind is not. Given the information we have, the Tony haircut being on the older or the younger is a 50/50 probability, nothing more, nothing less. If you repeat the experiment without possibility of moving the haircut, then the Tony location becomes known, and you aren't repeating the experiment, are you?

I don't see any reason why putting the sisters behind doors has changed the probability that the sister without the Tony is the older sister.

[/ QUOTE ]

this is the dumbest thing i have ever heard of. it is 50/50 or you haven't given enough information. end of story.

[/ QUOTE ]

What's dumb is saying, "it is 50/50" and not knowing what you mean.

PairTheBoard

itsmesteve

06-29-2005, 08:13 PM

Well, the probability is now either 100% or 0%, we just don't know which, since we don't know which twin is older.

PairTheBoard

06-29-2005, 09:27 PM

[ QUOTE ]
Well, the probability is now either 100% or 0%, we just don't know which, since we don't know which twin is older.

[/ QUOTE ]

Correct

PairTheBoard

TomCollins

06-29-2005, 09:57 PM

And therefore God exists and Jeebus will save us all.

ddubois

06-29-2005, 10:22 PM

How is this not like saying: "I've flipped a coin, but I haven't looked down at it yet; the odds of it being heads is not 50:50, instead it's either 100% or 0% heads, I just don't know which". Because such a statement is, at best, a semantic argument, at worst a huge waste of our collective time.

Siegmund

06-29-2005, 10:26 PM

[ QUOTE ]
[ QUOTE ]
Well, the probability is now either 100% or 0%, we just don't know which, since we don't know which twin is older.

[/ QUOTE ]

Correct

PairTheBoard

[/ QUOTE ]

Sorry, no (with one condition, see (*) below.)

The question you asked was what is the probability the older twin is behind door #2.

The probability that the older twin has the Tony was 100% or 0%, but we don't know which, before we did the experiment. And this probability is still either 100% or 0% but we don't know which, after we do the experiment.

But, you placed the two twins randomly behind the two doors - so at the start of the experiment, the older twin had a 50% chance of being behind door #2 - and when you opened the door, you learned ABSOLUTELY NOTHING about the age of the girls. Whether neither, either, or both doors are open is completely irrelevant since you seeing a girl doesn't help you know which one she is. Posterior probabilities change only when additional information is gained.

The chance that the older twin is behind door #2 remains 50% from the time they are randomly positioned until the time you learn something new about how to tell the two of them apart.

(*) - You could, if you wanted, say "the probability was always 100% or 0%, as soon as the girls were placed behind the doors, and will forever remain so, since no amount of door-opening, questioning, or hair-cutting will cause them to physically move."

If you take that approach, however, you've basically chosen to regard probability theory in its entirety as meaningless - regarding your two hole cards as being best with probability 100% or 0% as soon as they are dealt, since the deck has already been shuffled and the identities of the flop turn and river are already fixed but unknown. True, in a vacuous way - but a useless attitude to take if you're interested in deciding whether to bet those cards or not.

PairTheBoard

06-29-2005, 10:44 PM

[ QUOTE ]
And therefore God exists and Jeebus will save us all.

[/ QUOTE ]

I wouldn't say that. But I think it does illustrate how incorrect it is when many people use the term "probability" in a haphazard way. In the repeatable experiment I describe above, if it were correct for the observers to say that the probability is 50% that the older twin is behind door #2 after seeing the twin with the Tony behind door #1, Then half the time the older twin would be shown to the observer behind door #2 and half the time the younger twin would be shown behind door #2. But this clearly does not happen. If the younger twin has the Tony then 100% of the time the repeating experiment will show the older twin behind door #2. If the older twin has the Tony then 0% of the time the repeating experiment will show the older twin behind door #2. Just because we don't know which it is doesn't make it 50-50.

When a nonexpert uses the "probability" term incorrectly it's understandable. But it irks me when David says things like, he estimates the probabilty that God exists to be about 1 in a Trillion - or whatever. As a recognized expert in probabilty his Misuse of the term amounts to an attempt to infuse his opinion with an authority it doesn't deserve. He can say he's 99.9999999999999999999% sure that God doesn't exist if he wants. That's fine. But delivering his opinion like it's an expert analysis of probability on the subject is just nonsense.

PairTheBoard

TomCollins

06-29-2005, 11:09 PM

After reading this thread, it is quite clear that your understanding of the term probability is far worse than the people you describe, especially DS.

PairTheBoard

06-30-2005, 12:47 AM

[ QUOTE ]
How is this not like saying: "I've flipped a coin, but I haven't looked down at it yet; the odds of it being heads is not 50:50, instead it's either 100% or 0% heads, I just don't know which". Because such a statement is, at best, a semantic argument, at worst a huge waste of our collective time.

[/ QUOTE ]

It depends on what the repeated experiment is. If the coin gets flipped one time and is now to be revealed repeatedly to various observers then with respect to that experiment the probabilty that it is heads is either 100% or 0%, we just don't know which.

If the coin is reflipped before each time it's revealed then with respect to that experiment the probability that it is heads is 50%.

This is not a trivial exercise in semantics. In the 2 Envelope problem where the amounts in the envelopes are fixed, having been chosen by some unknown and irrelevant means, this is exactly the point on which the so called paradox twists. It's known that one envelope has twice the amount of money in it than the other. Before opening any envelope the probabilty is 50% that the Second Envelope has twice the amount as the first and 50% that it has half the amount of the first. But after opening the First envelope and seeing it has a certain amount in it, say X dollars, it is no longer true that the probabilty is 50% that the second envelope has twice the known amount X and 50% probabilty that it has half the known amount X in the first envelope. This even though we have no idea whether X is the larger or smaller amount.

If the probabilities were still 50-50 then we could calculate the expected value of the second envelope to be 1.25X which would imply that it's always better to take the second envelope after seeing what's in the first - clearly a mistaken conclusion. If the amounts in the envelopes are A and 2A then the proper probabilty statements upon seeing X in the first envelope are: The conditional probabilty the Second Envelope has 2X given X=A is 100%. The conditional probabilty the Second Envelope has .5X given X=A is 0%. Similarly for the case X=2A. These proper conditional probabilties allow us to make the correct expected value calculation for the second envelope to be equal to X as it should be. Notice these conditional probablities are working exactly like the Twin with the Tony.

PairTheBoard

MikeL05

06-30-2005, 07:40 AM

[ QUOTE ]
I'm going to be really disappointed if the answer is shown to be pointless and a huge waste of my time

[/ QUOTE ]

FMP

inlemur

06-30-2005, 11:49 AM

The reason this issue is cloudy is due to different interpretations of probability. DS takes a Bayesian approach to probability, and clearly PTB does not.
http://en.wikipedia.org/wiki/Bayesian_probability

Stephen H

06-30-2005, 12:43 PM

It seems your point is "probability only applies to future events; events that have already occurred (such as a girl getting a haircut) cannot have probability assigned to them." This is perhaps semantically true, but useless. My contention is that we can equally use probability to apply to events that have happened in the past but that we do not know the outcome of, and that this is just as valid as using probability on future events.

Case in point:
You have the A/images/graemlins/spade.gifK/images/graemlins/spade.gif. The board comes Q/images/graemlins/spade.gifJ/images/graemlins/spade.gif2/images/graemlins/heart.gif. What is the probability your flush will hit?
We use probabilty in poker all the time. But the deck is already shuffled; if you have perfect information about the state of the deck, you *know* that the flush either will come (100%) or won't come (0%). The randomizing event has happened. But we don't know the outcome yet; it's hidden. Therefore we use probability to analyze the problem. I can say "the repeatable part of the experiment is flipping the cards over; I'm not going to reorganize the deck" all I want - the use of probability is still valid.

Clearly you disagree with this, but you aren't convincing me that your point is valid very easily.

icepoker

06-30-2005, 02:16 PM

To PairTheBoard: I read the whole thread, and I've come to the conclusion that you are a moron. Please never again use the word probability, you have no idea what it means.

probman

06-30-2005, 02:27 PM

I have to agree with icepoker. Form your posts in this thread, it is quite clear that you have fundamental flaws in your understanding of probability. What makes it worse is your biting attempts to lecture others on their misuse. While the former is acceptable, when combined with the latter, they make you out to be a complete idiot.

Stephen H

06-30-2005, 03:48 PM

I disagree with the analogy to the 2 envelope problem. In that case, the apparent paradox, as you've said in a past thread, relates to an incorrect assumption on the probability distribution:
[ QUOTE ]

This is the heart of the problem - the distribution from which the envelope amounts are chosen. Even advanced probabilty students miss the underlying bogus assumption that all envelope amounts are equally likely. ie. that the envelope amounts were chosen from a uniform distribution over the positive real numbers. Of course there is no such probabilty distribution.

[/ QUOTE ]
However, in this case, the probability distribution of haircuts onto twins is a lot simpler than money in envelopes. We're talking discretes and Booleans, not analysis across the real number line. I'm still unconvinced that we cannot model it as an equal probability of each twin having the haircut and get useful results.

AaronBrown

06-30-2005, 05:39 PM

I'll stay out of the meaning of probability discussion, but I have to defend the envelope paradox. It is considerably deeper than suggested here. Like all good paradoxes, the point is not to explain it, but to peel away the layers.

You do not need to assume that the amount in the envelope is equally likely to be any real number. It is true that from a Bayesian perspective if seeing the amount of money in one envelope does not change your estimate of the probability of it being the larger amount, then your prior probability distribution must have an infinite expectation. If you expect infinity, any finite number is a disappointment, so you should always switch.

That's not a logical inconsistency, although it's hard to think of an example that doesn't involve God or the Devil.

But in the first place, that argument only makes sense to Bayesians. In the second place, you can make the paradox work with finite expectations as well.

One example is an American bets a European that the US dollar will buy less that 0.80 Euros in one year (or set any rate such that the probability is 50%). If the American wins, he gets 800 Euros. If he loses, he pays US $1,000.

The American will either lose $1,000 or win Euros he can exchange for more than $1,000. The European will either win 800 Euros or dollars he can exchange for more than 800 dollars. So both parties have positive expectations. In fact, the American can convert his bet to a 50/50 proposition for $1,000 and get paid $50, by selling an option in the financial markets. The European can get a 50/50 800 Euro bet and be paid 40 Euros.

PairTheBoard

07-01-2005, 01:34 AM

[ QUOTE ]
The reason this issue is cloudy is due to different interpretations of probability. DS takes a Bayesian approach to probability, and clearly PTB does not.
http://en.wikipedia.org/wiki/Bayesian_probability

[/ QUOTE ]

Thanks for that link inlemur. I did not realize there was such a "Baysian" camp out there. The term "Baysian" is a little misleading I think. I believe all Mathematical Probabilty is being done by "Relative Frequentists" and they all make extensive use of Bayes Theorem. According to the link, "Baysian Probabilty" refers to a Philosophy. The philosophy evidently asserts that it is meaningful to use the term probabilty in ways that don't have the precise mathematical meanings that are used in Mathematical Probabilty - ie. Relative Frequency Probabilty. It's not suprising that as the link points out, Mathematical probablists don't attend the Conferences of Baysian Probablists. They are basically speaking a different language.

But you are correct in pointing out that David has the right to go with the Baysian Crowd in his use of language if he so desires. I don't see the need for creating the confusion though myself. You can use "degree of certainty" language just as easily without constantly making us wonder whether the "probabilty" you refer to carries the precise meaning of mathematical probabilists or the Philisophically Debatable meaning of the "Baysians".

PairTheBoard

PairTheBoard

07-01-2005, 01:46 AM

[ QUOTE ]
I have to agree with icepoker. Form your posts in this thread, it is quite clear that you have fundamental flaws in your understanding of probability. What makes it worse is your biting attempts to lecture others on their misuse. While the former is acceptable, when combined with the latter, they make you out to be a complete idiot.

[/ QUOTE ]

I appologize to all if my tone is unpleasant. However, the point I'm making with the Twin with the Tony is correct, it's not flawed, it's not dumb, it's not meaningless, and I'm not a moron or an idiot.

The idea that you need to think about the repeatable experiment in order to make sense - from the point of view of mathematical probabilty that's being done and taught in Universities - of the probabilities you assign is fundamental to the ability to actually work with probablities.

PairTheBoard

PairTheBoard

07-01-2005, 02:53 AM

"I'll stay out of the meaning of probability discussion, but I have to defend the envelope paradox. It is considerably deeper than suggested here. Like all good paradoxes, the point is not to explain it, but to peel away the layers."

--AaronBrown

I realize the Two Envelope problem as usually stated stirs most discusion around the amounts in the envelopes being "randomly chosen". I am eliminating that part of the discussion by defining the problem more precisely and specifying the repeated experiment. Here is the version of the Two Envelope Problem I am talking about.

There are Two Envelopes, Env1 and Env2. A certain amount of money has been placed in each. We have no idea what the amounts are but we do know that one amount is twice as much as the other. In the repeated experiment the exact same amounts are used over and over. But we have numerous players of the game, none of which know the amounts in the envelopes. We also assume that if an envelope is opened and it's contents are revealed, it gives no information to a player as to whether the amount is the larger or smaller amount.

The envelopes are randomly shuffled and presented to a player. At this point there is a 50% probability that Env2 has twice the amount as Env1 and a 50% probability that Env2 has half the amount as Env1. Now, Env1 is opened and the amount X dollars is revealed to the player. The player is now asked to either choose the X dollars she sees or switch to Env2.

The player says to herself, I still have no idea whether the larger amount is in Env2 or not. But I now know there is X dollars in Env1. So I can calculate the expected value of the amount in Env2 as follows:

50%(.5X) + 50%(2X) = 1.25X

The player figures it's a nobrainer and takes the amount in Env2. In fact, every player makes the same calculation and if the calculation is correct the players on average must come out 25% better by always switching. What's wrong with this picture? Clearly it can't be right so what's wrong with the calculation?

What's wrong with the calculation is exactly the same mistake being made by people who insist after seeing the Twin with the Tony that there is still a 50% probability that the older Twin is behind door #2. Maybe a Philosophical Baysian can use that language but it sure doesn't help when it comes time to actually calculate the expected value in the Two Envelope problem.

Here is the correct way to calculate the expected value of the amount in Env2. Let the amounts in the envelopes be A and 2A and let the amount in Env1 which is revealed to a random player be Z. Then the conditional probabilty that Env2 contains 2Z given Z=A is 100%. The conditional probabilty that Env2 contains .5Z given Z=A is 0%. Just like the Twin with the Tony. It's either 100% or 0%, we don't know which. Similarly when Z=2A. For a random player the amount Z is equally likely to be A or 2A. So the expected value for Env2 is:

50%(100%(2Z) + 0%(.5Z) : when Z=A) +
50%(0%(2Z) + 100%(.5Z) : when Z=2A) =

50%(2A) + 50%(A) = 1.5A

While the expected value of Z for a random player is also

50%(2A) + 50%(A) = 1.5A

For all of you who have been saying mean things to me, I accept your apologies unspoken.

PairTheBoard

mmmmmbrother

07-01-2005, 02:17 PM

[ QUOTE ]
To PairTheBoard: I read the whole thread, and I've come to the conclusion that you are a moron. Please never again use the word probability, you have no idea what it means.

[/ QUOTE ]

owned

PairTheBoard

07-02-2005, 12:47 AM

[ QUOTE ]
It seems your point is "probability only applies to future events; events that have already occurred (such as a girl getting a haircut) cannot have probability assigned to them." This is perhaps semantically true, but useless. My contention is that we can equally use probability to apply to events that have happened in the past but that we do not know the outcome of, and that this is just as valid as using probability on future events.

Case in point:
You have the A/images/graemlins/spade.gifK/images/graemlins/spade.gif. The board comes Q/images/graemlins/spade.gifJ/images/graemlins/spade.gif2/images/graemlins/heart.gif. What is the probability your flush will hit?
We use probabilty in poker all the time. But the deck is already shuffled; if you have perfect information about the state of the deck, you *know* that the flush either will come (100%) or won't come (0%). The randomizing event has happened. But we don't know the outcome yet; it's hidden. Therefore we use probability to analyze the problem. I can say "the repeatable part of the experiment is flipping the cards over; I'm not going to reorganize the deck" all I want - the use of probability is still valid.

Clearly you disagree with this, but you aren't convincing me that your point is valid very easily.

[/ QUOTE ]

One difference here is that the known cards, AsKsQsJs2h do not DETERMINE what the next card to be dealt will be. However they DO EFFECT the probabilty for the next card. Suppose this hand is heads up with no burn cards. Prior to seeing the AsKs in your hand and the QsJs2h on the board you would have said that the probabilty that the 8th card in the deck is a spade is 13/52. The other difference with the Twins is that in this case you know how the probabilty for the 8th card has been affected by what's been revealed. Seeing the AsKsQsJs2h has changed the probabilty for the 8th card to a known 9/47. That's the conditional probabilty for the 8th card Given your hand and the flop cards. The twin with the tony is the same idea only the original 50% probabilty is changed to an Unknown conditional probabilty when the Tony twin is revealed. In fact the Tony twin completely Determines which twin is behind door 2, we just don't know what the determination is.

Also, as far as the past future thing. What exactly do you mean when you say the probabilty is 9/47 that the 8th card will be a spade? I believe you mean that if you arrived in this same situation - a randomly shuffled deck and cards dealt as they are - repeatedly over numerous trials you would then see a spade on average 9/47 ths of the time. Whether or not this is actually done doesn't really matter. But that's what you Mean when you state the probabilty.

However, what if I told you this. We are going to keep the cards just as they are and shuffle in numerous players to sit in your seat to have the fixed 8th card revealed to them. Let the random Suit S denote the outcome of the revealing of the fixed but unknown 8th card to a random player put in your seat. Would you still insist the probabilty that (S is a spade) is 9/47?

PairTheBoard

punter11235

08-30-2005, 10:37 AM

PairTheBoard, thank you for your cruciate to educate people here. These are interesting problems and I needed some time to understand them at one point in my life.
I think there is another way to look at these problems. Here it is :
a)2 envelope problems : strategy "switching is best" assumes that number N was chosen equiprobably at random from all integers which is of course nonsense. If we assume that number wasnt chosen at random, we cant assume that there is 50-50 probability that another envelope contains higher amount cause we dont have information about chosing amounts in envelopes. This is why we cant determine proper strategy here.
b)Twins with tony haircut problem : again, to give proper strategy we need to know probability of older twin having tony haircut.
c)Another problem for all nonbelievers: Someone gave us rigged coin, say with probability of 70% to flip head but we dont know this. We take this coin and flip. If we agree with all this people :

[ QUOTE ]

After reading this thread, it is quite clear that your understanding of the term probability is far worse than the people you describe, especially DS.

I'm going to be really disappointed if the answer is shown to be pointless and a huge waste of my time

To PairTheBoard: I read the whole thread, and I've come to the conclusion that you are a moron. Please never again use the word probability, you have no idea what it means.

[/ QUOTE ]

We can tell that probability that our flip will be head is 50% which is of course nonsense.

Basic intuition is this : if in any given situation there are outcomes A and B and we say that A have probability of say 40% we should expect that if this situation reoccures many times our distribution will be in fact 40% for A and 60% for B. This is not the case in all abovementioned problems.

By the way I am amazed that so many posters here dont have respect for people who try to educate them and so many of them while lacking basic understading of the subject are sure of being correct.

Best wishes