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If the probability to win a contest is $0.09$ (and to lose $ 0.91$), and if somebody already lost once, what is the probability to win in the second time? I calculate

$0.91\times0.09= 0.0819 < 0.09$

It follows that each time somebody participates in the contest, his/her chances to win decrease comparing to chance in the previous round.

Is it really so? If not, where am I wrong? I consider each two consequent contests independent.

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    $\begingroup$ Probability does not work that way, at least not typically. If, say, you imagine that are are tossing a (possibly biased) coin each time then the trials are independent. I.e. the probability of a win/loss does not change from trial to trial, regardless of wins and losses that came before. The coin doesn't "remember" past outcomes. $\endgroup$
    – lulu
    Commented Jul 31, 2021 at 10:30
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    $\begingroup$ What you computed was the probability of losing in the first game and then winning in the second game, that is why the probability is smaller. $\endgroup$
    – acat3
    Commented Jul 31, 2021 at 10:34

1 Answer 1

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The general formula for conditional probability is $$ P(A|B)=P(AB)/P(B). $$

Which reads as "Probability of an event A with the condition that an event B happened is the probability that the event A and the event B happened divided by probability that the event B happened". Using this formula, you can get $$ P(\text{Win 2}\ |\text{ Lost 1})=P(\text{Win 2} \ \text{and} \text{ Lost 1})/P(\text{Lost 1}). $$ Events are independent so $P(\text{Win 2 and Lost 1})=P(\text{Win 2})P(\text{Lost 1})$. Thus you have $P(\text{Win 2} |\text{Lost 1})=P(\text{Win 2})$.

To understand the difference between what you did and the general formula, you can draw a table of possible events. $$ \begin{matrix} W & W\\ W & L\\ L & W\\ L & L \end{matrix} $$ You computed a probability of the third line to happen as one event among four. However, what you need is the probability of the third line as an event among the two last lines. This general formula (Bayes probability) represents this second option, and factor $1/P(B)$ is a kind of renormalization of probability, so $P(L|L)+P(W|L)=1$.

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