Probability of mystery coin being fair

Let's say I have a bag with 3 coins. 1 coin is a fair coin and 2 coins have a bias (70% chance of heads). I choose a coin at random and flip it 6 times. I observe exactly 3 heads. What is the probability that the coin I chose is the fair coin?

My initial intuition is that my observation doesn't mean anything, and there is a 1/3 chance that it is the fair coin because I chose it at random from the bag with 3 coins.

• No, that doesn't work. Given that the biased coins are a good bit more likely to come up with heads, and given that you had 3 heads out of 6 (which is exactly in line with a fair coin) would suggest that the chances that you are dealing with a biased coin are smaller than that you are dealing with the fair coin. Anyway, use Bayes' Theorem to figure out the exact probability. Feb 22, 2018 at 23:42
• Well, I'd say that the experiment constituted weak evidence that the coin was fair. After all, for the biased coin we'd expect $4.2$ Heads, so $3$ is low (though not astonishingly low). So I'd think the revised probability should be somewhat greater than $\frac 13$. Use Bayes' Theorem to get the exact value.
– lulu
Feb 22, 2018 at 23:43
• @Bram28 That depends on whether you are in the world of Bayesian statistics or not, which is axiomatic. If yes, you are correct. If no, OP is right Feb 22, 2018 at 23:43
• To stress: suppose you tossed your coin $100$ times and saw exactly $50$ heads. I'd say that was very strong evidence that your coin was the fair one. For the biased one, the expected value would be $70$ and the standard deviation would be about $4.58$ so seeing $50$ would be a $4.36 \,\sigma$ event, virtually unheard of.
– lulu
Feb 22, 2018 at 23:47
• Thank you for the comments. I think based on the context that this question was asked, I need to use Bayes' Theorem as suggested.
– ejtt
Feb 22, 2018 at 23:50

$$P(A\,|\,\text{HHH}) = \frac{P(A \cap \text{HHH})}{P(\text{HHH})} = \frac{\frac 1 3(0.5)^3}{\frac 1 3(0.5)^3 + \frac 2 3(0.7)^3} \approx .12 \ne \frac 1 3$$
In Bayesian language, your prior probability of getting Coin A is $1/3.$ Your data is 'three H's in three tosses'. Your posterior probability for Coin A is around 0.12. The prior probability and the data are combined to give the posterior probability.