ed84c Posted August 6, 2005 Posted August 6, 2005 Right, Aparantly the chance of England not winning the test today is approx 2-1 Now, in theory that means in say 3 matches England would not win 2 and win 1. But what are the chances of the afformentioned situation occuring. In theory they are certain, but evidently that is not the case.
danny8522003 Posted August 6, 2005 Posted August 6, 2005 Im pretty sure you can't get probabilities of probability because they would be completely usless. The odds of 2-1 describe the likelihood as well as they can, they are obviously not certain. We could play 2.7 trillion games at 2-1 odds and lose every single time. It's highly improbable but not impossible. Lets say the fact that we lose the first one has no effect on the next game. For each individual game the odds will still be 2-1 independant of whether you won or lost the last one. If this isnt the case then if we talk about probabilities of probability then what's stopping us talking about the probabilities of the probabilities of probability and so on until your brain melts into a gooey pink gunge...
matt grime Posted August 7, 2005 Posted August 7, 2005 you certainly can ask what is the probabilty that a situation occurs in a given game such that 1/3 of the time outcome X occurs *from here*. to estimate it you need to decide at what points you're going to check the bookies' odds. Say every interval, or at the start of the day. Then look back over the history of the game and see at how many times, at the start of the interval or day, a team had a 1/3 chance of winning (presumably according to the bookies here). at least that's one interpretation of it.
mezarashi Posted August 7, 2005 Posted August 7, 2005 Let me make this clear by using another example. The chances of you passing your math exam is 2-1 or better stated as 2/3, meaning that if you were to take the math exam 3 times, you would pass 2 and fail 1, "theoretically". This theoretically means that on average. If you took one billion trillion trillion trillion exams, most likely you would have passed two thirds the time. The statistics would normalize to this probability. Now the question is, what is the probability of you passing twice and failing once in 3 tests. Logically, lay out the outcomes in taking three tests. Pass (P) and Fail (F). PPP PPF PFP FPP PFF FPF FFP FFF Out of the 8 possible outcomes, there are 3 in which the 2pass-1fail predicament occurs: (in parenthesis the probability that each would happen) PPF (2/3 x 2/3 x 1/3) PFP (2/3 x 1/3 x 2/3) PPF (1/3 x 2/3 x 2/3) Summing that up, you have a 4/9 chance, which is the answer I believe you were looking for. This (4/9) is the probability of the 2-1 pass-fail happening in three tests, but the probability of you passing on any particular test is still 2/3. Note that the total is lower than your probability on each particular test. Does this result makes sense. What if the probability that you pass a particular test is 50%. Maybe try it with only 2 trial tests. That's something to think about.
matt grime Posted August 7, 2005 Posted August 7, 2005 i was answering a different question, apologies, misread the request.
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