Oh,all right, cavediver, here goes.
Let's make up an example. Suppose that a lottery is run by choosing three one digit numbers with replacement. That is, there are 10 marbles with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 written on them; on marble is randomly drawn from the cup; it is replaced, the cup is shaken, and the second is drawn; it is replaced, and a third is drawn. How many possible outcomes are there? Well, the first digit can be any of 10, the second can be any of 10, and the third can be any of 10. So the total number of outcomes is 10
3=1000. The possible outcomes are the set U={000, 001, 002, ..., 999}.
Now suppose that the lottery game that Alice chooses is to choose three digits, and she wins $100 if those three digits are drawn in any order. Suppose that Alice chooses the digits 123. Then she will win if the marbles that are drawn are A={123, 132, 213, 231, 312, 321}. That is, there are 6 possible ways she can win; the probability that the ticket is a winning ticket is 6/1000=0.006. Mathematically we express this as P(A)=0.006.
If Alice tries to sell you her ticket for $1, should you buy it? Well, 6 times out of 1000 you will win $100, and 994 times out of a thousand you will win $0, so on average you should win $0.60 (and so far my class has forgotten that I haven't really explained what this means, heh). So you will lose $0.40, on average, if you buy this $1 ticket.
Now suppose that, in order to ratchet up the excitement, the first marble is drawn at 10 a.m., the second at 12 noon, and the third at 2 pm. At 11 a.m., the first marble has been drawn, and it is 2. So we now know that the only going to be 100 possible outcomes: the first digit has been determined, there are still 10 possibilities for the second digit at noon, and 10 possible digits for the third in the afternoon. Furthermore, possible outcomes are B={200, 201, 202, ..., 299}.
Now,
knowing that the first digit is 2, what is the probability that 123 will be the winning number? Well, the only possibilities of winning are (A and B) = {213, 231}; two possibilities, but there are now only 100 possible outcomes. Thus,
knowing now that B has occurred, that is, that the first digit is 2, the probability of winning is P(A|B)=2/100 = 0.02.
This is not a nonsense situation, because now if Alice tries to sell you her ticket for $1, since your winnings now have an expected value of $2 this is a good buy; you will, "on average" earn a $1 profit.
This is how good poker players make money; they apply conditional probabilities, determining the likelihood of completing a good hand based on the information (the cards that are showing) that they have.
Now it is 3 p.m., and we are now told that Alice had the winning ticket all along. Thus, we know that the drawing was either 213 or 231: C={213, 231}. Now we know for sure that Alice's ticket won: P(A|C)=1. It is left as an exercise for the reader to determine whether one should now buy the ticket if Alice still offers it for $1.
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So that is what is going on in the discussion. This is exactly the sorts of things that Pascal and Fermat discussed in their correspondance that invented probability theory; from the very beginning they were discussing various probabilities
given various information that would already be known. They were discussing gambling problems, and, as Modulous the poker player knows, when trying to determine whether one stays in the hand or drops out one has to make decisions based on information that one has at that moment. In fact, even the original probability of 6/1000 can be written as a conditional probability: P(A|U)=0.006, the probability that 123 is a winning ticket number
given that any outcome is possible.
So the question is, what are the odds that my ticket will win Powerball? Well, without any other information, it is 1 in 146 million (assuming that is the correct figure). But the question, what is the probability that my ticket is the winning ticket
given that I already know that it is the winning ticket? Well, that is a completely trivial question, silly really, but nontheless one that is
mathematically reasonable and, in fact, we can even denote it mathematically: P(A|A)=1.
Of course, I think crashfrog is right: if someone says, "Alice won the Powerball yesterday! What are the odds of that happening?" I would interpret that to mean what were the odds before any information was known, or, in more precise mathematical terms, what are the odds of winning
given that all outcomes are possible? I would
not interpret that sentence as asking, "What is the probability that Alice wins
given that we already know that Alice wins?" Mathematically reasonable (as Modulous
et al. have been pointing out), but, as a real-life question that someone might ask, pretty silly.
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From schraf's OP:
The odds of winning the lotto can be a million to one, but if I win it on my first try, then those weren't my odds were they?
Well, the only way those aren't that person's odds are if the assumption
all outcomes are equally likely is incorrect, that is, if the game is actually rigged in some way.
Actually, if their god makes better pancakes, I'm totally switching sides. --
Charley the Australopithecine
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