# two coin (normal & both heads) toss probability

I have two coins, one normal and one having both heads. Randomly picked one and tossed it 3 times. All three were HEADs. What is the probability of seeing a tail on next toss?

Going with independent event approach,

normal: 1/2 x 1/2 = 1/4
biased: 1/2 x 0 = 0
total: 1/4 + 0 = 0.25


Is this correct? Or is there a different approach?

• Think for the two cases. For the first toss, you can either get T/H or H/H. So 1/4 is correct. However, not for the following tosses. Consider simply that you get 100000 H's, now clearly having the biased coin is not 50/50 event anymore. May 8 '17 at 20:24

If the coin is fair then the number of heads in a row before you flip has no affect on the probability of another head so the the answer for a fair coin is $\frac{1}{2}$.

If this is a coin with both heads then you clearly has no chances to see tail at all.

Using Bayes' theorem you can say that $$P(\mbox{your coin is double-headed}) = \frac{1\cdot \frac{1}{2}}{1\cdot \frac{1}{2} + \frac{1}{8}\cdot \frac{1}{2}} = \frac{8}{9}$$

Just to add more generality, using the same approach you can show that the probability that the coin is double-headed after $N$ heads in a row and no tails is $$P = \frac{p}{p+(1-p)\cdot2^{-N}},$$ where $p$ is the probability you think the coin is double-headed before you receive any information via flips. After some rather big $N$ it will be very close to $1$ for any reasonable $p$.

\begin{align} P(\text{T| HHH}) &= \dfrac{P(\text{T \cap HHH})}{P(\text{HHH})} \\ &= \dfrac{P(\text{fair coin}) \cdot P_{\text{fair coin}}(\text{T \cap HHH}) + P(\text{trick coin}) \cdot P_{\text{trick coin}}(\text{T \cap HHH})}{P(\text{fair coin}) \cdot P_{\text{fair coin}}(\text{HHH}) + P(\text{trick coin}) \cdot P_{\text{trick coin}}(\text{HHH})} \\ &= \dfrac{\dfrac{1}{2} \cdot \dfrac{1}{2^4} + \dfrac{1}{2} \cdot(0)}{\dfrac{1}{2} \cdot \dfrac{1}{2^3} + \dfrac{1}{2} \cdot(1)} \\ &= \dfrac{1}{18} \end{align}

This is not correct. If you didn't know what had happened on the three previous tosses, you would have equal chances of having either coin, so you would have a probability of $1/4$. However, after seeing three heads, the chance that the coin you are tossing is the double-headed one has increased.

You can work out the chance that you are tossing the normal coin using Bayes' rule: $$P(\text{normal coin}\mid HHH)=\frac{P(\text{normal coin and }HHH)}{P(HHH)}$$ then use this to find the probability of a tail next.

You have a result of tossed three heads in three tries. What is the probability that the coin is fair?

• To obtain the result, you have either picked a fair coin and tossed two head, or else you have picked a double header and done so. \begin{align}\mathsf P(F\cap R)~&=~\tfrac 12\cdot\tfrac 12^3&&\text{Probability for Fair and Result}\\\mathsf P(F^\complement\cap R)~&=~\tfrac 12\cdot 1^3 &&\text{Probability for Unfair and Result}\end{align}

• So the (conditional) probability for having a fair coin given the results is:

\begin{align}\mathsf P(F\mid R) ~&=~ \dfrac{\mathsf P(F\cap R)}{\mathsf P(F\cap R)+\mathsf P(F^\complement\cap R)} \\[1ex] &=~ \dfrac{2^{-4}}{2^{-4}+2^{-1}} \\[1ex] &=~ \dfrac 19 \end{align}

• Then, by the law of total probability, the (conditional) probability for seeing a tail on the next toss given the result is: \begin{align}\mathsf P(T\mid R)~&=~\mathsf P(T\mid F)~\mathsf P(F\mid R)+\mathsf P(T\mid F^\complement)~\mathsf P(F^\complement\mid R)\\[1ex] &=~ \tfrac 12\cdot \tfrac 19+0\cdot \tfrac 89 \\[1ex] &=~ \dfrac 1{18}\end{align}

This is sensible, since seeing three heads in a row should make you more confident that you have selected the double headed coin.