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Find the probability that a person tossing 3 coins will get either all heads or all tails for the 2nd time on the 5th toss.

Typically people solve it like:

\begin{align}P &= (\textrm{Probability of getting HHH or TTT in first four trials}) \times (\textrm{probability of getting HHH or Success in the fifth trial})\\ &= { \binom{4}{1} (0.25)^1 (0.75)^3 } \times {0.25}\\ &= 0.1054\end{align}

But if I do Negative binomial with $r=2, ~x=5, ~p=0.25 ~= 0.08899.$

Why this is not treated as negative binomial since we are looking 2nd success on 5 trials? why its different? what understanding am i missing?

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It is negative binomial, but the negative binomial variate $x$ is the number of failures until the second success rather than the number of tosses:

> dnbinom(x=3, size=2, prob=0.25)
[1] 0.1054687

The NB random variable is defined as the number of failures before $r$ successes are obtained.

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  • $\begingroup$ Isn't calculating for 3 failures same as that of 2 successes? I don't understand the params? Thanks $\endgroup$
    – reindeer
    Commented Dec 11, 2022 at 5:32
  • $\begingroup$ You set $x$ to be the number of successes plus the number of failures, but it should be just the number of failures. The definition of NB is the number of failures before $r$ successes are obtained. $\endgroup$
    – Gordon Smyth
    Commented Dec 11, 2022 at 6:07
  • $\begingroup$ OH!! thank you, i got it $\endgroup$
    – reindeer
    Commented Dec 11, 2022 at 10:08

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