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Suppose I toss a coin with $0.25$ chance of $H$. I toss it until I get $k+1$ times heads. What is the probability that I will have to toss the coin $4k$ times?

The last throw must be an $H$ so I have $4k-1$ options to choose $k-1$ throws where I get $H$. So $$\ {\ 4k-1 \choose k-1 }\cdot 0.75^{3k-1} \cdot 0.25 ^{k}$$

But I'm wrong about something here as the answer is $\ {\ 4k-1 \choose k } \cdot 0.75^{3k-1} \cdot 0.25^{k+1} $

I know I can just plug the variables into negative binomial formula but still I don't understand what am I missing here? what I need $\ 0.25^{k+1} $ successes? and why I need to chose $\ k $ places instead of $\ k-1 $ places as I know the last throw must be $H$ ??

Thanks!

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    $\begingroup$ You are almost correct. Note that in total $k+1$ heads (successes) are needed (not $k$): exactly $k$ in the first $4k-1$ throws and $1$ at the $4k$-th throw. $\endgroup$ – drhab Dec 6 '18 at 15:11
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If you need $k+1$ heads overall then you must succeed $k$ times from the first $4k-1$ throws - if you throw $k-1$ heads in the first $4k-1$ and then another head then that is only $k$ heads overall. The probability that you succeed $k$ times in the first $4k-1$ throws is

$$ P(\textrm{there are } k \textrm{ heads out of the first } 4k-1 \textrm{ throws})= p = \binom{4k-1}{k}\cdot 0.75^{3k-1} \cdot 0.25^{k} $$

Now, the probability that it takes you $4k$ throws to succeed $k+1$ times is the probability that you throw $k$ heads in the first $4k-1$ throws times the probability the last throw is a head (which is 0.25). So

$$ P(\textrm{it takes 4k throws to see } k+1 \textrm{ heads}) = 0.25p = \binom{4k-1}{k}\cdot 0.75^{3k-1} \cdot 0.25^{k+1}$$

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Since your last throw results in a head and you need a total of $k+1$ heads, you have $4k-1$ throws to get the rest of $k$ heads.

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