Woman wins millions in lottery 4 times!

[Oblarg]

Unfortunately, Dagon, that's not how statistics work. It is still a unique sequence of cards, whether or not he predicted beforehand what sequence he would draw. Thus, something with an "impossibly low" probability did occur. He did not only draw "a sequence of cards," he drew "that sequence of cards." The probability of him drawing that sequence of cards might be somewhat meaningless as you can't use it to make any meaningful predictions of conclusions, but it is still the probability of that event occurring, and that event did occur.

That's just wrong, sorry, the event that occurred had a probability of 1 of occurring, as I said.

 

You're going back to the irrelevant debate of definitions. I'm going to define probability in this way: "if I were to repeat the experiment an infinite number of times, the probability of this result is the fraction of the time that this result would occur." This is a perfectly valid frequentist definition of probability, and as you can clearly see, when probability is thus defined it certainly does not "become 1" after an event has occurred. The experiment in this case is drawing several cards from a deck. The result is the exact sequence that I first got when I drew cards from the deck. It should be obvious that if I were to repeat that experiment an infinite number of times, that specific sequence would not come up 100% of the time, thus saying that the probability of that event is 1 because it already happened is complete nonsense with that definition of probability.

 

I've gone over this several times in this thread, and this is the last time.

Edited July 23, 2010 by Oblarg

[Walsingham]

Apologies for the delay.

 

Yes, that is exactly what happened. With multiple witnesses including the professional croupiers, and several members of my old unit. No funny business. No method on my part, besides the feeling I was being mad to bet singles. But of course, knowing probability I knew that it didn't matter how often I won on singles, the odds at each bet stayed the same.

 

Of course I had to joke that I used up a lot of my food luck reserve for no reason. Then that year I

busted my ribs and my family went to hell. But that's just superstitious nonsense.

[Wrath of Dagon]

So you're saying you guessed 1 number out of 37 correctly 9 times out of 10? That is indeed incredible and I have to admit I was wrong about low chance events not happening. Too bad there was no money involved, but of course it never happens then, heh.

 

 

 

Unfortunately, Dagon, that's not how statistics work. It is still a unique sequence of cards, whether or not he predicted beforehand what sequence he would draw. Thus, something with an "impossibly low" probability did occur. He did not only draw "a sequence of cards," he drew "that sequence of cards." The probability of him drawing that sequence of cards might be somewhat meaningless as you can't use it to make any meaningful predictions of conclusions, but it is still the probability of that event occurring, and that event did occur.

That's just wrong, sorry, the event that occurred had a probability of 1 of occurring, as I said.

 

You're going back to the irrelevant debate of definitions. I'm going to define probability in this way: "if I were to repeat the experiment an infinite number of times, the probability of this result is the fraction of the time that this result would occur." This is a perfectly valid frequentist definition of probability, and as you can clearly see, when probability is thus defined it certainly does not "become 1" after an event has occurred. The experiment in this case is drawing several cards from a deck. The result is the exact sequence that I first got when I drew cards from the deck. It should be obvious that if I were to repeat that experiment an infinite number of times, that specific sequence would not come up 100% of the time, thus saying that the probability of that event is 1 because it already happened is complete nonsense with that definition of probability.

 

I've gone over this several times in this thread, and this is the last time.

No, I meant the probability of the event was 1 because the probability of drawing some sequence is 1, not because it already happened, although these things are related. As far as definition of the probability of a past event being 1, that's required by the probability theory I'm familiar with, don't know which one you're talking about, if that's not the case, how do you compute the probability of two independent events, one of which has already happened? Edited July 24, 2010 by Wrath of Dagon

[Oblarg]

So you're saying you guessed 1 number out of 37 correctly 9 times out of 10? That is indeed incredible and I have to admit I was wrong about low chance events not happening. Too bad there was no money involved, but of course it never happens then, heh.

 

 

 

Unfortunately, Dagon, that's not how statistics work. It is still a unique sequence of cards, whether or not he predicted beforehand what sequence he would draw. Thus, something with an "impossibly low" probability did occur. He did not only draw "a sequence of cards," he drew "that sequence of cards." The probability of him drawing that sequence of cards might be somewhat meaningless as you can't use it to make any meaningful predictions of conclusions, but it is still the probability of that event occurring, and that event did occur.

That's just wrong, sorry, the event that occurred had a probability of 1 of occurring, as I said.

 

You're going back to the irrelevant debate of definitions. I'm going to define probability in this way: "if I were to repeat the experiment an infinite number of times, the probability of this result is the fraction of the time that this result would occur." This is a perfectly valid frequentist definition of probability, and as you can clearly see, when probability is thus defined it certainly does not "become 1" after an event has occurred. The experiment in this case is drawing several cards from a deck. The result is the exact sequence that I first got when I drew cards from the deck. It should be obvious that if I were to repeat that experiment an infinite number of times, that specific sequence would not come up 100% of the time, thus saying that the probability of that event is 1 because it already happened is complete nonsense with that definition of probability.

 

I've gone over this several times in this thread, and this is the last time.

No, I meant the probability of the event was 1 because the probability of drawing some sequence is 1, not because it already happened, although these things are related. As far as definition of the probability of a past event being 1, that's required by the probability theory I'm familiar with, don't know which one you're talking about, if that's not the case, how do you compute the probability of two independent events, one of which has already happened?

 

Except you did not draw "some sequence" you drew a specific sequence. Whether or not you predicted that sequence before is irrelevant.

 

As for the probability of a past event being one, that's not required at all - that's only required for a very narrow and rather bad definition of probability.

 

It doesn't matter if one event has already happened, the probability of any given outcome is unchanged. Unless, of course, you're asking the probability of getting a specific result given that one occurs in a specific way, in which case you are defining the probability of the first event as 1. This is a completely different situation.

 

Assuming A and B are independent events:

 

In the first case, you're asking P(A and B)

In the second case, you're asking P(A|B).

 

The two are not the same thing.

 

Now, going back to the deck of cards:

 

Let's call our sequence of cards A.

 

The probability of drawing A, regardless of whether or not A has already been drawn, is P(A). The sequence A is a unique sequence that existed before we drew any cards from the deck - whether or not we specifically predicted A is irrelevant.

 

You, however, are making the mistake of treating the probability of drawing A as P(A|A), because we already know that A was drawn. This is faulty logic . When the cards were drawn, it was not a given that A would be drawn. You cannot revise the probability of A based on your observation of the result of A after you conducted the experiment. That's not how probability works. Once again, read my definition of probability and you'll see that the probability of an event in the past is certainly not 1.

[Wrath of Dagon]

No, I'm saying for two independent events A and B, if you want to calculate P(A and B) before the trial, it's P(A) * P(B). If you know A already happened, then it's 1 * P(B) = P(B). If you have a formula for calculating the chance of a hand in Blackjack, and you need to calculate the odds knowing that certain cards have already been discarded, you would just substitute 0's for the cards already discarded, not derive a brand new formula just for that situation.

 

As far as an event being a particular sequence, that is only true if the sequence can be defined before the trial, because if you can define it, the odds can be calculated. Otherwise it's just "some sequence" and it only becomes defined after the trial, and the chance of getting that sequence during that particular trial is 1, because it already happened. (and the chances of "some" sequence are also 1, which is how they're related)

Edited July 24, 2010 by Wrath of Dagon

[Oblarg]

No, I'm saying for two independent events A and B, if you want to calculate P(A and B) before the trial, it's P(A) * P(B). If you know A already happened, then it's 1 * P(B) = P(B). If you have a formula for calculating the chance of a hand in Blackjack, and you need to calculate the odds knowing that certain cards have already been discarded, you would just substitute 0's for the cards already discarded, not derive a brand new formula just for that situation.

 

As far as an event being a particular sequence, that is only true if the sequence can be defined before the trial, because if you can define it, the odds can be calculated. Otherwise it's just "some sequence" and it only becomes defined after the trial, and the chance of getting that sequence during that particular trial is 1, because it already happened.

 

Nope. The probability of P(A and B) is always P(A) * P(B). If A has already occurred then the question is now P(B|A). P(A) remains unchanged.

 

And the sequence existed before the trial. It makes no difference if I had it written down on a piece of paper or not. It is still a unique sequence.

 

You don't seem to quite grasp the idea that the probability of "some sequence being drawn" is 1 because the probabilities of all unique sequences add up to that, and they are not independent. If you draw a sequence, the probability that you would draw that particular sequence is tiny, albeit somewhat meaningless as, as we have established, it's not a particularly significant sequence as it was not defined beforehand. The fact remains that a very unlikely event did occur.

[Wrath of Dagon]

So every time you draw five cards a very unlikely event occurs. You have a very strange definition of "unlikely".

[Oblarg]

So every time you draw five cards a very unlikely event occurs. You have a very strange definition of "unlikely".

 

Yes.

 

It's a concept you have to be able to grasp if you're going to understand statistics.

 

You have many mutually exclusive outcomes with tiny probabilities that add up to 1. Every time you get a specific outcome, it was a very unlikely outcome, yet you will still *always* have an outcome.

 

Now, admittedly, in a situation like this the probability of each specific outcome isn't particularly useful - most of the useful information you can gain from this system revolves around operating with a specific class of outcomes.

Edited July 24, 2010 by Oblarg

[Wrath of Dagon]

OK, so you say probability of P(A and B) is always P(A) * P(B). So given that A has already happened, how do you compute P(A and B)?

[Oblarg]

OK, so you say probability of P(A and B) is always P(A) * P(B). So given that A has already happened, how do you compute P(A and B)?

 

The same way you always do. P(A and B) means: "If I were to repeat A and B an infinite number of times, what fraction of the time would both A and B occur?"

 

You are confusing P(A and B) with P(A|B). The two are not the same thing.

[Whipporwill]

Christ, what an imagination I've got.

[Wrath of Dagon]

OK, so you say probability of P(A and B) is always P(A) * P(B). So given that A has already happened, how do you compute P(A and B)?

 

The same way you always do. P(A and B) means: "If I were to repeat A and B an infinite number of times, what fraction of the time would both A and B occur?"

 

You are confusing P(A and B) with P(A|B). The two are not the same thing.

OK, so it is a matter of definitions as you say. I can have just as consistent a framework if I define any past event as having probability 1. And you can say that unlikely events are occuring every time you draw cards by your definition. None of it matters in practice though, because the only events we should consider as likely or unlikely are the ones it's possible to bet on, for the lack of a better definition. For example, saying some unlikely sequence of cards will happen does nothing to enable you to bet on that sequence. Saying a particular sequence (or a set of sequences) will happen enables you to compute odds and bet on that sequence and see if you've won or lost once the trial happens.

[Humodour]

OK, but you're assuming that it's I who's misunderstanding and not Calax

 

That's because it is you. The whole thread has been the Obsidian Forums vs Wrath of Dagon on matters of probability from the start, and Hurlshot doesn't have to be a maths guru to be able to observe that.

[Wrath of Dagon]

I'd bet you never took a probability class either.

[Humodour]

I'd bet you never took a probability class either.

 

Either? What, now you're admitting you haven't taken one?

 

Besides covering lots of combinatorics in maths extension 2 for my year 12 HSC, I took the following stats courses at ANU:

 

http://studyat.anu.edu.au/courses/STAT2001;details.html

http://studyat.anu.edu.au/courses/STAT2008;details.html

 

Why does this always come back to credentials for you Wrath? This is an Internet forum. You have no way to verify what I just posted, and we have no way to verify your claims you've gone to university. What we do have to go on, however, is a thread chock a block full of your posts containing elementary mistakes about probability.

[Wrath of Dagon]

I normally assume regular posters on here are truthful, unless there's a reason to believe otherwise. And what elementary mistakes have I made?

[Whipporwill]

If the odds of Divine intervention are low enough, that too must be impossible.

[Walsingham]

I normally assume regular posters on here are truthful, unless there's a reason to believe otherwise. And what elementary mistakes have I made?

 

I think the latter part of this statement physically hurt my face as I ran into it.

 

I liked the first part tho.

[alanschu]

So every time you draw five cards a very unlikely event occurs. You have a very strange definition of "unlikely".

 

I think this is where you might be confusing yourself a little bit.

 

 

Yes, an unlikely event occurs every time you draw 5 cards.

 

However, it is typically a different unlikely event that occurs every time you draw 5 cards. It is unlikely that I would draw the cards that I listed a page ago. You can agree with this because it's just as unlikely that I would draw those 5 cards again.

 

Two events happen when I draw 5 cards from a deck of cards, and one (as Oblarg points out) is part of the probability mass function of the other. One event is the probability that I draw any 5 cards (the probability of this is 1). The other event is the probability that I draw 5 specific cards in a specific order. This event is very small, and a part of the probability mass function of a discrete random variable. The sum of all points on a probability mass function is 1, since it's a sum of all the probabilities of all outcomes.

 

 

 

OK, so you say probability of P(A and B) is always P(A) * P(B). So given that A has already happened, how do you compute P(A and B)?

 

Oblarg, I'll need you to verify my math here, as it's been a looong time, but I hope this might clear things up:

 

So Dagon asks "Given that A has already happened, how do we compute P(A and B)." Which I read and interpret as, effectively, what is the probability of B happening, given A has happened. In probability notation, this would be P(B | A).

 

To clarify, given that A has already happen, what is actually being asked is what is the P(B | A), since we know A has happened.

 

If I am not mistaken, this would be:

 

P(B | A) = P(A and B) / P (A) [A and B are indepedent events, so P(A and B) = P(A) * P(B) which is, for easy math sakes, lets say 0.01. P(A) = P(B) so 0.01 * 0.01 = 0.0001]

 

= 0.0001 / 0.01

= 0.01

= P(B)

 

Given that A has already happened, we're calculating P(B|A) in order to find out a solution to P(A and B) where A has already occurred.

 

 

It's late, so I think this may be a bit off.

 

Let me know.

[Wrath of Dagon]

Here's from a guy who claims to be a statistician: http://wmbriggs.com/blog/?p=2597 What I don't get is why the odds of the lotteries are so low, I thought the odds are usually twice the jackpots. Here's a guy who claims he got hit by meteorites 6 times: http://www.metro.co.uk/weird/835482-man-hi...geted-by-aliens (I saw all this on Wargamer forums).

Edited July 30, 2010 by Wrath of Dagon

[Calax]

I've had family members struck by lightning a couple times.

[Wrath of Dagon]

Lightning strikes aren't that unusual though, also depends on the person's behavior.

[Calax]

http://www.lightningsafety.com/nlsi_pls/probability.html

 

1/280,000

 

Family member was struck twice, old teacher was struck three times.

Edited July 31, 2010 by Calax

[Wrath of Dagon]

That's just based on assumptions though. People who are outside in thunderstorms for some reason will be struck a lot more often than the average person, who will almost never be struck.

[Calax]

That's just based on assumptions though. People who are outside in thunderstorms for some reason will be struck a lot more often than the average person, who will almost never be struck.

And somebody who buys a number of tickets at a Lotto will have a higher chance of winning than the next guy.