Walsingham Posted July 19, 2010 Posted July 19, 2010 New and I hope final question for Dagon: What is the probability of an unlikely event, say one in 10^1005 NEVER happening? "It wasn't lies. It was just... bull****"." -Elwood Blues tarna's dead; processing... complete. Disappointed by Universe. RIP Hades/Sand/etc. Here's hoping your next alt has a harp.
Wrath of Dagon Posted July 19, 2010 Author Posted July 19, 2010 Never as in before the universe ends? I'd say very close to 1. Never as in infinity? Then you've multiplied it by infinity, i.e. you get probability 1 (or actually something infinitely close to 1) that it will happen, but multiplying by infinity is not a good idea. That's why I keep saying "for all practical purposes". "Moral indignation is a standard strategy for endowing the idiot with dignity." Marshall McLuhan
Gorgon Posted July 19, 2010 Posted July 19, 2010 Nice try at evasion Krezack, you still haven't answered my very simple and direct question. Oh, sorry, which one is that? If it's something silly where you misunderstand probability for the 1000th time in this thread, don't even bother waiting for a reply, though. So you don't even read what I post, you're just here to insult me, is that it? My simple question (posted right after your post) is whether you know that the probabiltiy of an event which has already happended is 1? I'm not math whiz, but that question seems completely illogical. What does probability after the fact matter. It happened, and if it was the only thing we were measuring it doesn't have any probability any more. I think what you are thinking is 'what is the probability of the same event happening agian', But that wasn't the question. Na na na na na na ... greg358 from Darksouls 3 PVP is a CHEATER. That is all.
Wrath of Dagon Posted July 19, 2010 Author Posted July 19, 2010 It does matter, as I demonstrated using alanschu's example. You need to use that if you calculate probabilities in which some events have already happened and some haven't. Granted it's a fairly trivial calculation, but it can lead to a lot of misunderstanding if not done correctly. Another example would be trying to calculate the odds of a hand of blackjack winning if you remember which cards have already been discarded. "Moral indignation is a standard strategy for endowing the idiot with dignity." Marshall McLuhan
Gorgon Posted July 19, 2010 Posted July 19, 2010 Well sure, but in a lottery there are no other dependents to hitting the winning number. Na na na na na na ... greg358 from Darksouls 3 PVP is a CHEATER. That is all.
Wrath of Dagon Posted July 19, 2010 Author Posted July 19, 2010 (edited) As I keep trying to explain, it depends on the pool of how many people can win. For someone to win once, you got the pool of all the people playing worldwide. For someone to win twice, only the people who won once already are eligible, in other words a much smaller pool. For someone to win three times, only people who won twice already are eligible, a mere handful. To win the fourth time, it's that exact woman who would have to win again, since to my knowledge she's the only one in the world who had won major lotteries 3 times. Btw, I found out some more about the scratch off lotteries. Turns out the winning number is already on the ticket, and the number you scratch off has to match, depending on how many digits match is how much money you win. The tickets have a bar code scanned at the time of sale, so they can verify the ticket you present isn't fake but matches the ticket that was sold. There seems to be ample opportunity for fraud if you have an insider or can hack into the lottery computers. Edited July 19, 2010 by Wrath of Dagon "Moral indignation is a standard strategy for endowing the idiot with dignity." Marshall McLuhan
Calax Posted July 19, 2010 Posted July 19, 2010 As I keep trying to explain, it depends on the pool of how many people can win. For someone to win once, you got the pool of all the people playing worldwide. For someone to win twice, only the people who won once already are eligible, in other words a much smaller pool. For someone to win three times, only people who won twice already are eligible, a mere handful. To win the fourth time, it's that exact woman who would have to win again, since to my knowledge she's the only one in the world who had won major lotteries 3 times. Wait, WHAT? You're going about that all wrong. The chances of somebody winning again are based on their individual chances compared to every other participant in the second contest. You don't keep "slimming down" the pool/odds so that it's harder and harder to win like that. The chances overall of that sequence of events happening are lower than them winning a contest, but they still have the same chances to win the contest as any other participant. Victor of the 5 year fan fic competition! Kevin Butler will awesome your face off.
Pidesco Posted July 19, 2010 Posted July 19, 2010 Rofl "My hovercraft is full of eels!" - Hungarian tourist I am Dan Quayle of the Romans. I want to tattoo a map of the Netherlands on my nether lands. Heja Sverige!! Everyone should cuffawkle more. The wrench is your friend.
Wrath of Dagon Posted July 19, 2010 Author Posted July 19, 2010 As I keep trying to explain, it depends on the pool of how many people can win. For someone to win once, you got the pool of all the people playing worldwide. For someone to win twice, only the people who won once already are eligible, in other words a much smaller pool. For someone to win three times, only people who won twice already are eligible, a mere handful. To win the fourth time, it's that exact woman who would have to win again, since to my knowledge she's the only one in the world who had won major lotteries 3 times. Wait, WHAT? You're going about that all wrong. The chances of somebody winning again are based on their individual chances compared to every other participant in the second contest. You don't keep "slimming down" the pool/odds so that it's harder and harder to win like that. The chances overall of that sequence of events happening are lower than them winning a contest, but they still have the same chances to win the contest as any other participant. Again, I'm talking about the chances of someone in the world winning, not a specific person winning. A specific (pre-determined) person winning the lottery 4 times has even much lower odds. It would really be septillions, like the article stated. "Moral indignation is a standard strategy for endowing the idiot with dignity." Marshall McLuhan
Amentep Posted July 19, 2010 Posted July 19, 2010 As I keep trying to explain, it depends on the pool of how many people can win. For someone to win once, you got the pool of all the people playing worldwide. For someone to win twice, only the people who won once already are eligible, in other words a much smaller pool. For someone to win three times, only people who won twice already are eligible, a mere handful. To win the fourth time, it's that exact woman who would have to win again, since to my knowledge she's the only one in the world who had won major lotteries 3 times. The only way this works is if you're trying to calculate the odds of someone winning a game of chance in which winning the first time is a pre-requisite for eligibility to play the second time. The lottery doesn't work that way, though, since anyone can play the second time regardless of the result the first time around; in other words the probability of winning doesn't actually change in terms of the 4th one vs the 1st one. I cannot - yet I must. How do you calculate that? At what point on the graph do "must" and "cannot" meet? Yet I must - but I cannot! ~ Ro-Man
Wrath of Dagon Posted July 19, 2010 Author Posted July 19, 2010 As I keep trying to explain, it depends on the pool of how many people can win. For someone to win once, you got the pool of all the people playing worldwide. For someone to win twice, only the people who won once already are eligible, in other words a much smaller pool. For someone to win three times, only people who won twice already are eligible, a mere handful. To win the fourth time, it's that exact woman who would have to win again, since to my knowledge she's the only one in the world who had won major lotteries 3 times. The only way this works is if you're trying to calculate the odds of someone winning a game of chance in which winning the first time is a pre-requisite for eligibility to play the second time. The lottery doesn't work that way, though, since anyone can play the second time regardless of the result the first time around; in other words the probability of winning doesn't actually change in terms of the 4th one vs the 1st one. No, but you can't win the second time if you haven't won the first time already. And we're trying to calculate probabilties of someone winning the second, third, and fourth time, by definition the previous win is a pre-requisite for the next one. That's why you have to take the chances of winning once to the fourth power if we want the probabilities of a specific person winning 4 times, or do you disagree with that? "Moral indignation is a standard strategy for endowing the idiot with dignity." Marshall McLuhan
Calax Posted July 19, 2010 Posted July 19, 2010 (edited) So what you're doing is taking four contests, and calculating the odds of one person winning all four against 16 million people each. Thing is that that requires that people be winning four consecutive times. However, for the most part these contests are independent of one another. Each of the contests has the same chance of winning that the first had for each person. And obviously if they won the first one it's entirely possible that they can win the second, third, and fourth. Basically if all four contests were with the same people, the chances of one person winning all of them would be 1/<however many> Edited July 19, 2010 by Calax Victor of the 5 year fan fic competition! Kevin Butler will awesome your face off.
Gorgon Posted July 19, 2010 Posted July 19, 2010 (edited) As I keep trying to explain, it depends on the pool of how many people can win. For someone to win once, you got the pool of all the people playing worldwide. For someone to win twice, only the people who won once already are eligible, in other words a much smaller pool. For someone to win three times, only people who won twice already are eligible, a mere handful. To win the fourth time, it's that exact woman who would have to win again, since to my knowledge she's the only one in the world who had won major lotteries 3 times. Wait, WHAT? You're going about that all wrong. The chances of somebody winning again are based on their individual chances compared to every other participant in the second contest. You don't keep "slimming down" the pool/odds so that it's harder and harder to win like that. The chances overall of that sequence of events happening are lower than them winning a contest, but they still have the same chances to win the contest as any other participant. We are calculating their chances of winning four lotteries in a lifetime. That's the job, not how hard it was to win the first, second, third and fourth times. It's the same as if all the lotteries they entered in were held at the same time because there is no result without four wins. Edited July 19, 2010 by Gorgon Na na na na na na ... greg358 from Darksouls 3 PVP is a CHEATER. That is all.
Wrath of Dagon Posted July 19, 2010 Author Posted July 19, 2010 Actually I have to admit I just did find a mistake in my calculation, which I think is what Balthamael was trying to get at earlier. I think when you're calculating her specific odds each time you have to divide the chance of the lottery by the amount of money she spent over the period in question, which is something I was having doubts about and convinced myself of the opposite. So if the lottery odds are 1 in 10 million (I think that's about what they are in the Texas scratch off), and she spent 10000 over the period she's been playing, her chances of winning each lottery would be 1 in 1000. So conservatively assuming there are 100 people in the world who won the lottery twice, and that they keep playing each having invested on average 10000, the odds of someone out of them winning the lottery again (that is for the third time) are 1000 / 100, i.e. 1/10. Then that one person winning again would be 1 / (10 * 1000) i.e. 1 / 10000, still really bad odds but I suppose within the realm of possibilty. "Moral indignation is a standard strategy for endowing the idiot with dignity." Marshall McLuhan
Amentep Posted July 19, 2010 Posted July 19, 2010 (edited) No, but you can't win the second time if you haven't won the first time already. And we're trying to calculate probabilties of someone winning the second, third, and fourth time, by definition the previous win is a pre-requisite for the next one. That's why you have to take the chances of winning once to the fourth power if we want the probabilities of a specific person winning 4 times, or do you disagree with that? The problem is that you're approaching it as dependent; ie only someone who has won once can participate (order matters) in the next lottery and there are only four tries within which to complete the objective. In this case, if you actually want to calculate the odds of anyone winning 4 lotteries you have to look at the total number of lotteries played in conjunction with the odds for the lottery (or each lottery if they have different odds), you can't just look at the odds of winning and multiplying them together. The odds of getting two heads Heads in two Tries (HH) is 25% HH - win HT - lose TH - lose TT - lose The odds of getting two Heads in four tries without specifying an order is 11/16 HHHH - Win HHHT - Win HHTH - Win HHTT - Win HTHH - Win HTHT - Win HTTH - Win HTTT - lose THHH - Win THHT - win THTH - Win THTT - Lose TTHH - Win TTHT - Lose TTTH - Lose TTTT - Lose The odds of getting two heads in a row in four tries dips down to 8/16 HHHH - Win HHHT - Win HHTH - Win HHTT - Win HTHH - Win HTHT - lose HTTH - lose HTTT - lose THHH - Win THHT - Win THTH - lose THTT - Lose TTHH - Win TTHT - Lose TTTH - Lose TTTT - Lose Edited July 19, 2010 by Amentep I cannot - yet I must. How do you calculate that? At what point on the graph do "must" and "cannot" meet? Yet I must - but I cannot! ~ Ro-Man
Wrath of Dagon Posted July 19, 2010 Author Posted July 19, 2010 In this case, if you actually want to calculate the odds of anyone winning 4 lotteries you have to look at the total number of lotteries played in conjunction with the odds for the lottery (or each lottery if they have different odds), you can't just look at the odds of winning and multiplying them together. Already did that but through the assumption that there are only a handful of people who have won 2 lotteries. May be it's not a handful, may be it's 1000 or so, hard to tell without knowing the total number of lottery winners in the world and how much they have spent playing lotteries on average. "Moral indignation is a standard strategy for endowing the idiot with dignity." Marshall McLuhan
Reinoc Posted July 19, 2010 Posted July 19, 2010 (edited) The number of people that have won multiple times changes nothing. In a previous statement you say the odds are 1 in 10 millions, and that she bought lets say 10000 tickets in her life. The odds of her winning four times is 10millions^4/10000=10^24. That other people won multiple times doesn't change these odds. Now I guess you'll say 10^24= impossible, but you would be wrong. Edited July 19, 2010 by Reinoc
Tale Posted July 19, 2010 Posted July 19, 2010 I come back and this is what I find? :'( Isn't it awesome? Please, stay. I'm not a one forum man, baby. You can't hold me down. "Show me a man who "plays fair" and I'll show you a very talented cheater."
Gorgon Posted July 19, 2010 Posted July 19, 2010 The number of people that have won multiple times changes nothing. In a previous statement you say the odds are 1 in 10 millions, and that she bought lets say 10000 tickets in her life. The odds of her winning four times is 10millions^4/10000=10^24. That other people won multiple times doesn't change these odds. Now I guess you'll say 10^24= impossible, but you would be wrong. It's like hitting 3 bars on a slot machine except there are millions of symbols and hundreds of slots. Na na na na na na ... greg358 from Darksouls 3 PVP is a CHEATER. That is all.
Oblarg Posted July 20, 2010 Posted July 20, 2010 The number of people that have won multiple times changes nothing. In a previous statement you say the odds are 1 in 10 millions, and that she bought lets say 10000 tickets in her life. The odds of her winning four times is 10millions^4/10000=10^24. That other people won multiple times doesn't change these odds. Now I guess you'll say 10^24= impossible, but you would be wrong. Wrong, this is a binomial distribution. The actual probability if she buys 10,000 lottery tickets (which is an awful lot, it's probably more around 500) would be: (10,000 choose 4)(1/10million)^4(1-(1/10million))^(10,000-4) = 416416712497500 * (1/10000000)^4 * (9999999/10000000)^9996 = 4.16001x10^-14. This is completely ****ing irrelevant, though. If you actually understand the math, you'll see that this is simply the probability of four lotteries being won from a pool of 10,000 tickets, each ticket having a chance of winning of one in ten million. It does not care if the lottery tickets were all bought by the same person. This pool of 10,000 tickets is being arbitrarily decided because we like to think that there is something special about the same person winning each lottery. In actuality, you could construct an identical arbitrary pool of 10,000 tickets for any unique group of four lottery winners and claim that an equally unlikely event has transpired - I'll demonstrate this with a simpler situation below. Perhaps an easier way to demonstrate why this argument is pointless would be to assume that each player buys exactly four lottery tickets. In this case, the probability of one player winning four lotteries is simply (10^-7)^4 = 10^-28 (which, if you notice, is lower than the actual probability we calculated above). Surely this occurrence is impossible, if the chance is so low? Well, no, not really - by symmetry, this is the exact same chance as any outcome in which there are exactly four unique winners, regardless of whether or not the winner is the same person, because each unique combination of four winners results in four successes out of a pool of four tickets - the exact same situation as one person winning four times. Math does not care who holds the tickets - this is simply the human tendency to think that there is something significant about one person winning four times. It's an emotional response, not a logical one. The argument that "that event already transpired, thus the probability is one" is also irrelevant, as one can simply define probability as "the fraction of successes if the experiment were repeated an infinite number of times" instead of "the fraction of the time that the experiment succeeds," as obviously with the latter definition asking the probability of something which has already succeeded or failed is meaningless. This is a pointless debate of definitions which has absolutely no bearing on the issue at hand. Can this discussion please die now? "The universe is a yawning chasm, filled with emptiness and the puerile meanderings of sentience..." - Ulyaoth "It is all that is left unsaid upon which tragedies are built." - Kreia "I thought this forum was for Speculation & Discussion, not Speculation & Calling People Trolls." - lord of flies
Wrath of Dagon Posted July 20, 2010 Author Posted July 20, 2010 (edited) Wrong, this is a binomial distribution. The actual probability if she buys 10,000 lottery tickets (which is an awful lot, it's probably more around 500) would be: It's not a lot, that's only 20 tickets a week for 10 years. Some people spend a lot more than that, especially after she won the first lottery.(10,000 choose 4)(1/10million)^4(1-(1/10million))^(10,000-4) = 416416712497500 * (1/10000000)^4 * (9999999/10000000)^9996 = 4.16001x10^-14. That's the probabilty of a specific pot of $10000 winning the lottery, not the probability of any 10,000 pot in the world winning the lottery.blah blah blah Wrong, all based on your previous incorrect assumptionCan this discussion please die now? I won't post if no one else does, but Krezack still owes me an answer. Edited July 20, 2010 by Wrath of Dagon "Moral indignation is a standard strategy for endowing the idiot with dignity." Marshall McLuhan
Gorth Posted July 20, 2010 Posted July 20, 2010 You guys just ruined my world view. The last 40 years I've been carelessly crossing streets on foot, secure in the knowledge that I have statistics on my side. I mean, nobody gets run over and almost killed by a car twice, right? Well, wrong it seems. Thanks guys “He who joyfully marches to music in rank and file has already earned my contempt. He has been given a large brain by mistake, since for him the spinal cord would surely suffice.” - Albert Einstein
Humodour Posted July 20, 2010 Posted July 20, 2010 So you don't even read what I post, you're just here to insult me, is that it? My simple question (posted right after your post) is whether you know that the probabiltiy of an event which has already happended is 1? *sigh* Oh! It seems I completely missed you ****ting all over probability theory on page 7 when you said: "The chance of someone winning once we know is 1." which is just wrooooooooooong. The probability of the outcome of an event is unchanged by whether or not the event actually occurred. All you do here is state the FACT that someone one. Not the PROBABILITY of that person winning. I won't post if no one else does, but Krezack still owes me an answer. *sigh* If it's something silly where you misunderstand probability for the 1000th time in this thread, don't even bother waiting for a reply, though.
Wrath of Dagon Posted July 20, 2010 Author Posted July 20, 2010 So you agree with alanschu that probability of an event that already happened is not 1? Also in his example the reason I said we know the probability of someone winning is 1 is not because someone already won, but that someone would have to win a lottery eventually, and I didn't mean exactly 1 but close enough to use in the calculation. "Moral indignation is a standard strategy for endowing the idiot with dignity." Marshall McLuhan
Oblarg Posted July 20, 2010 Posted July 20, 2010 (edited) Why are you guys still arguing that definition? It's irrelevant. The fact is that you can point to any four lottery winners and claim "the probability that these four people would win is so small that it could only have possibly happened due to supernatural forces," which is exactly equivalent to saying "the probability that this person would win four lotteries is so small that it could only have possibly happened due to supernatural forces." The math is identical, as I have shown above - all the math cares about is that it is a unique group of four successes. As no one would claim that the fact that four specific individuals won four lotteries is proof of god, this argument is pointless. Edited July 20, 2010 by Oblarg "The universe is a yawning chasm, filled with emptiness and the puerile meanderings of sentience..." - Ulyaoth "It is all that is left unsaid upon which tragedies are built." - Kreia "I thought this forum was for Speculation & Discussion, not Speculation & Calling People Trolls." - lord of flies
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