An example: coin-tossing
Simulation of coin tosses: Each frame, a coin is flipped which is red on one side and blue on the other. The result of each flip is added as a colored dot in the corresponding column. As the pie chart shows, the proportion of red versus blue approaches 50-50 (the Law of Large Numbers). But the difference between red and blue does not systematically decrease to zero.
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). In general, if we let Ai be the event that toss i of a fair coin comes up heads, then we have,
\Pr\left(\bigcap_{i=1}^n A_i\right)=\prod_{i=1}^n \Pr(A_i)={1\over2^n}.
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. Given that the first four rolls turn up heads, the probability that the next toss is a head is in fact,
\Pr\left(A_5|A_1 \cap A_2 \cap A_3 \cap A_4 \right)=\Pr\left(A_5\right)=\frac{1}{2}.
While a run of five heads is only 1⁄32 = 0.03125, it is only that before the coin is first tossed. After the first four tosses the results are no longer unknown, so their probabilities are 1. Reasoning that it is more likely that the next toss will be a tail than a head due to the past tosses, that a run of luck in the past somehow influences the odds in the future, is the fallacy.
так там о чем? о том что после 4 орлов вероятность следующего орла все равно 50%, так как броски действительно независимы. Что вероятность 5 орлов маленькая - пофиг, если уже выпали 4 подряд то вероятность следующего орла все равно 50%.
Это совсем-совсем не то же самое что говорить "будут 2 броска Маши и Даши, результатов пока нет, какая вероятность что будет хотя бы одна решка"
Еще раз - откройте учебники, ваша фэлласи никак не противоречит терверу просто разговор о разных ситуациях которые вы сводите в одну
