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I have a long list of independent events. Of these, $71\%$ are WINS and $29\%$ of them are LOSSES. I have calculated the probability of losses with this formula : \begin{align} 0.29^2 &= P(\text{two losses}) \\ 0.29^3 &= P(\text{three losses}) \\ 0.29^4 &= P(\text{four losses}) \end{align} How I can calculate a probability of this specific ordination pattern : loss, loss, win, loss, loss, loss?

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As they are independent events,

The probability of any pattern is simply multiplication of events' individual probabilities.

So, here pattern is loss, loss, win, loss, loss, loss

$P = 0.29 * 0.29 * 0.71 * 0.29 * 0.29 * 0.29 $

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The easiest way to make sense of this question is to assume you are talking about particular patterns of length $k$ on the next $k$ trials.

Then, the probability of $A = \{LL \text{ on next } 2\}$ has $P(A) = (.29)^2 = 0.0841,$ as you say. Similarly, $B = \{LLL \text{ on next } 3\}$ has $P(B) = (.29)^3 = 0.0841 = 0.024389.$ Notice that $B \subset A$ and $P(B) \le P(A),$ as required.

Then letting $M = \{LLWLLL \text{ on next } 6\},$ we have $P(M) = (.29)^5(.71) = 0.0015.$ This is the answer to your question (with the interpretation that we are talking about these specific $k = 6$ outcomes in this specific order during the next $k = 6$ trials. (Without such a precise interpretation, I don't know how to answer your question. The probability is $1$ of getting this sequence of six outcomes sometime in the future if trials stretch to infinity.)

.29^5*.71
[1] 0.001456292

Also, the probabilities of $WLLLLL,\; LWLLLL,\; \dots LLLLLW$ on the next six are all the same (six events in all). Thus, the probability of five $L$'s and one $W$ anywhere among the next six trials is the binomial probability ${6 \choose 5}(.29)^5(.71) = 0.0087,$ which is six times the probability of any one of the constituent events.

b = dbinom(5, 6, .29); b
[1] 0.008737749
b/6
[1] 0.001456292
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