Which Twin has the Tony?

[PairTheBoard]

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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.

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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]

"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]

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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.

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owned

[PairTheBoard]

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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[img]/images/graemlins/spade.gif[/img]K[img]/images/graemlins/spade.gif[/img]. The board comes Q[img]/images/graemlins/spade.gif[/img]J[img]/images/graemlins/spade.gif[/img]2[img]/images/graemlins/heart.gif[/img]. 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.

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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]

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 :

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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.


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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

[TomCollins]

So in this example, suppose you can see the back of the turn card that is dealt. The back of the turn card has a giant red dot on it. The other cards in the deck do not have the red dot on it.

Therefore, it is completely determined what the next card will be. The probability is no longer 9/47. It either will be a spade or won't be a spade!