I have a weighted coin with an unknown $p$ value and some success/failure criteria.

An experiment consists of:

$\text{Pr}($heads$)$ = $p$, $\text{Pr}($tails $)$= $(1 - p)$

The success event is having $3$ heads with less than $4$ tails

The failure event is having $4$ tails with less than $3$ heads.

I would like to maintain an $80\%$ success rate regardless of the number of experiments I carry out.

How do I calculate $p$?

I have tried using a cumulative binomial probability and come up with a $p$ value of $0.41$ but want to confirm that this is the correct method and the figure holds true for any number of experiments.

My assumptions are these:

  1. There is a maximum number of coin tosses, $6$, before an automatic failure. As soon as $3$ heads have been observed before $4$ tails, the experiment is a success.

$n = 6$ and $k < 4$

$P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) = 0.8$

Using the binomial probability model $p \approx 0.41$


1 Answer 1


I’m not entirely sure I understand the question correctly. I don’t know what you mean by “regardless of the number of experiments I carry out”, since that’s part of the meaning of “rate” and it’s not clear what it would mean for the success rate to depend on the number of experiments carried out. I’ll assume that you’re using “rate” in the standard sense of the term and thus the second part of the sentence is redundant. Also I don’t understand why you’re adding the probabilities for low rather than high numbers of heads; I’ll assume that that’s just a mistake.

Your basic approach of looking at the results of $6$ tosses regardless of whether the experiment actually takes $6$ tosses or ends after fewer tosses is a good one. If you add up the right probabilities, you get

$$ \sum_{k=3}^6P(X=k)=0.8\;, $$


$$ \sum_{k=0}^2P(X=k)=0.2\;, $$

which is

$$ \binom60p^0(1-p)^6+\binom61p^1(1-p)^5+\binom62p^2(1-p)^4=0.2\;, $$


$$ (1-p)^4\left(1+4p+10p^2\right)=0.2\;. $$

According to Wolfram|Alpha, the only real root of this equation in $[0,1]$ is $p\approx0.585394$.

If you think about it, this makes sense – it was implausible that with your value of $p\approx 0.41$, which leads to expected numbers of $2.46$ heads and $3.54$ tails in $6$ tosses, you’d have an $80\%$ chance of getting $3$ heads before $4$ tails.

Note that your value is the complement of the correct value, so it may be that the mistake isn’t in the calculation but in the definition of $p$ as the probability of heads (rather than tails).

  • $\begingroup$ Thanks joriki. Makes sense that I was coming at it from the wrong end. My 'number of experiments' comments were about each experiment being discrete and that there can only be one failure in five as an example. I see your point, and this was probably me overthinking the problem. Thanks again! $\endgroup$
    – BuzzBryce
    Feb 26, 2020 at 16:49

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