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11/28/2014 10:25:18 AM
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This is incorrect, whilst the immediate probability of the item appearing again (or number appearing again in roulette) is not affected by the previous outcome, the odds of something happening repeatedly are the odds of it happening multiplied by the power of the number it times it happens. If your math was correct then betting that something happened once would give you the same odds as betting that the same number on the roulette board appeared 4 times in a row. I'm sorry but your logic simply isn't logical.
English
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Edited by Frano89: 11/28/2014 10:35:45 AMYou're now confusing yourself. Probabilities are a totally different thing to odds. Odds are made up of previous outcomes. Probabilities are totally different. Tell me this. If I tossed 10 coins and they all landed on heads. What would you bet came up next? Surely a tails is the more likely to come up? But say for instance I tossed those coins heads came up ten time but a new person came in. That person never seen my first 10 tosses. For him he sees odds of the coin being heads or tails as 50/50.
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Furthermore the wikipedia page of the exact fallacy you mention illustrates this use of maths:"The gambler's fallacy can be illustrated by considering the repeated toss of a fair coin. With a fair coin, the outcomes in different tosses are statistically independent and the probability of getting heads on a single toss is exactly 1⁄2 (one in two). It follows that the probability of getting two heads in two tosses is 1⁄4 (one in four) and the probability of getting three heads in three tosses is 1⁄8 (one in eight)." I rest my case, you are wrong
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Did you read on? Now suppose that we have just tossed four heads in a row, so that if the next coin toss were also to come up heads, it would complete a run of five successive heads. Since the probability of a run of five successive heads is only 1⁄32 (one in thirty-two), a person subject to the gambler's fallacy might believe that this next flip was less likely to be heads than to be tails. However, this is not correct, and is a manifestation of the gambler's fallacy; the event of 5 heads in a row and the event of "first 4 heads, then a tails" are equally likely, each having probability 1⁄32.
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I completely understand that the probability of the next toss is unaffected by the outcome of the previous ten, however, the probability that the same outcome of those tosses will appear 10 times in a row is not 1/2 it is 0.5 to the power of 10 and this is the situation we are discussing. I would refute your suggestion that I am confusing myself good sir, (regardless of whether you believe me or not I am currently in an engineering lecture at the university of Cambridge which is a very maths heavy course) Google how to work out the probability of a coin producing heads 10 times in a row and you will get the same answer I have given you
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That's the probable outcomes of future tosses though. Where as what I'm saying is the probability of the one toss or in this case the probable outcome of xur's inventory on any given Friday. (Not talking his probable inventory weeks ahead)
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That does not change the fact that the probability of the item appearing 4 times in a row (which is what has happened) is 1/6 to the power of 4 as illustrated by the wikipedia quote
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Well this should be more in reference to odds rather then probability then, just because of the nature of it being RNG
