# Calculate probability of pattern of indipendent events

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?

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

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

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