Why Do The Chances Change?

This is a direct application of Bayes' theorem .

$\mathsf {Pr} (\text{3 heads given that there is at least 2 heads}) \; = \; \dfrac{\mathsf{Pr}(\text{at least 2 heads } \cap \text{ 3 heads} )}{\mathsf{Pr}({\text{at least 2 heads}})} \; = \; \dfrac{\mathsf{Pr}( \text{3 heads} )}{\mathsf{Pr}({\text{2 heads}}) + \mathsf{Pr}({\text{3 heads}})}$

The value of the probability that $\mathsf{Pr} (n\text{ heads})$ can be computed using binomial distribution ,

$\mathsf{Pr} (n\text{ heads}) = \dbinom 3n p^n \cdot q^{3-n} \; ,$

where $p$ and $q$ denotes the probability of obtaining a heads and a coin, respectively, and $p+q=1$ . Since we are given that it is a fair coin, then $p=q= \dfrac12$ .

Hence, $\mathsf{Pr} (n\text{ heads}) =\dbinom 3n \left( \dfrac 12\right)^3$ . And so, our answer is

$\dfrac{\binom 33 \left( \frac 12\right)^3}{\binom 32 \left( \frac 12\right)^3 + \binom 33 \left( \frac 12\right)^3} \; = \; \boxed{\dfrac14} \; .$

• With no info given the ratios of the outcomes are 1 hhh: 3 hht: 3 htt: 1 ttt
• Your friend looked and eliminated the possibility of the 3 htt and 1 ttt
• What's left is 3 chances at hht and 1 chance at hhh.
• Probability of hhh is 1 out of 4

There is a 3/8 chance of getting 2 heads when flipping 3 coins. However, there is only a 1/8 chance of getting 1 head. Therefore the chance of having 3 heads if there are at least 2 heads is 1/4.