Getting at least $2$ heads when flipping a coin $3$ times but the coin is biased so that heads are $3$ times more likely than tails.

Can anyone explain how uneven probability works?

Thank you


2 Answers 2


Let $P(H) = p$ be the probability of one head. In many scenarios, this probability is assumed to be $p = \frac{1}{2}$ for an unbiased coin. In this instance, $P(H) = 3P(T)$ so that $p = 3(1-p) \implies 4p = 3$ or $p = \frac{3}{4}$.

You are interested in the event that out of three coin tosses, at least 2 of them are Heads, or equivalently, at most one of them is tails. So you are interested in finding the likelihood of zero tails, or one tails.

The probability of zero tails would be the case where you only received heads. Since each coin toss is independent, you can multiply these three tosses together: $P(H)P(H)P(H) = p^3$ or in your case, $(\frac{3}{4})^3 = \frac{27}{64}$.

Now we must consider the case where one of your coin flips is a tails. Since you have three flips, you have three independent opportunities for tails. The likelihood of two heads and one tails is $3(p^2)(1-p)$. The reason for the 3 coefficient is the fact that there are three possible events which include two heads and one tails: $HHT, HTH, THH$. In your case (where the coin is 3 times more likely to have heads): $3(\frac{3}{4})^2(\frac{1}{4}) = \frac{27}{64}$.

Adding those events together you get $p^3 + 3(p^2)(1-p) = \frac{54}{64}$. Note that the 3 coefficient is a result of the binomial expansion as used by @Back2Basic


$a = P(H), b = P(T) \to a+b=1, a=3b \to a = 0.75, b = 0.25 \to P( x \geq 2) = P(2) + P(3) = \binom{3}{2}a^2b + a^3$. That's all I can offer....

  • $\begingroup$ $P( x \geq 2) = P(2) + P(3) = \binom{3}{2}a^2b + a^3$, I don't understand that part. Can you link on something that I can read about this? And it seems you use binomial expansion. Not sure I'm familiar with using it in this context. @Back2Basic $\endgroup$
    – Stupid
    Jan 16, 2015 at 14:28
  • $\begingroup$ I used binomial distribution $P(x=k) = \binom{n}{k} p^kq^{n-k}$ $\endgroup$
    – DeepSea
    Jan 16, 2015 at 14:30

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