# Unfair Coin Toss probability, Independence

We have two coins. The first one is a fair coin and has $$50\%$$ chance of landing on head and $$50\%$$ of landing on tails. for the other one it is $$60\%$$ for head and $$40\%$$ for tails. We choose one of them randomly (with equal chance) and toss it three times.The result is Tails-Head-Head. If we toss it again for the 4th time, what is the probability of the coin landing on tails.

At first, it seems like a bayes theorem question, but after some thought I think because the result of every toss is independent of others, the answer is just:

$$\frac{1}{2} . \frac{1}{2} + \frac{1}{2} . \frac{4}{10} = \frac{9}{20}$$

The first $$\frac{1}{2}$$ is because we choose the coins randomly.

I want to know whether my argument is correct for this probability question or not.

• I think your answer is correct. – Kavi Rama Murthy Oct 3 '19 at 23:36
• Intuitively, even though initially the probability is 50-50 for each coin type, by observing THH the odds increase that the coin comes from the biased coin with higher probability of heads. As a limit case, if one of the coins has heads on both sides, by observing tail you would be sure you have chosen the other coin. – Momo Oct 3 '19 at 23:42
• No, your thought that every toss is independent leads you to the wrong result. The point is that the a posteriori chance that you have chosen the unfair coin, given you THH experiment thus far, is now a tad more than $\frac12$. So the probability of the next toss being headss is a bit more than $\frac9{20}$. This is indeed a Bayesian question. – Mark Fischler Oct 3 '19 at 23:44

The answer, giving two terms corresponding to the actual coin being fair, then the actual coin being biased, will be $$P(T) = \frac{\left(\frac12\right)^3}{\left(\frac12\right)^3+\frac25\left(\frac35\right)^2} \left(\frac12\right) + \frac{\frac25\left(\frac35\right)^2}{\left(\frac12\right)^3+\frac25\left(\frac35\right)^2}\left(\frac25\right)\\$$

Doing the arithmetic, $$P(T) = \frac{1201}{2690} \approx 44.65\%$$ The probability that your coin is biased is about 53.5%, which is why the tails on the next throw is less likely than you might think.

Call the fair coin $$A$$ and the biased coin $$B$$.

$$P(THT|A) = (.5)^3$$

$$P(THT|B) = (.4)(.6)(.4)$$

$$P(A| THT) = \frac{P(THT|A) P(A)}{P(THT|A) + P(THT|B)}$$

$$P(B| THT) = \frac{P(THT|B) P(B)}{P(THT|A) + P(THT|B)}$$

(Note that $$P(A|THT) + P(B|THT) = 1$$, as it must)

Once you have the probability it is $$A$$, you can easily calculate the probability the 4th coin is $$H$$ ($$0.5$$). And likewise if the coin is $$B$$ ($$0.6$$).

You can plug in the numbers.