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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.

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    $\begingroup$ 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. $\endgroup$
    – Bram28
    Feb 22, 2018 at 23:42
  • $\begingroup$ 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. $\endgroup$
    – lulu
    Feb 22, 2018 at 23:43
  • $\begingroup$ @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 $\endgroup$
    – gt6989b
    Feb 22, 2018 at 23:43
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    $\begingroup$ 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. $\endgroup$
    – lulu
    Feb 22, 2018 at 23:47
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    $\begingroup$ Thank you for the comments. I think based on the context that this question was asked, I need to use Bayes' Theorem as suggested. $\endgroup$
    – ejtt
    Feb 22, 2018 at 23:50

1 Answer 1

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Here's a similar problem you might use as a guide. Suppose the coin you chose got HHH in three tosses. Coin A is fair; Coins B and C both have P(H) = 0.7.

$$P(A\,|\,\text{HHH}) = \frac{P(A \cap \text{HHH})}{P(\text{HHH})} = \frac{\frac 1 3(0.5)^3}{\frac 1 3(0.5)^3 + \frac 2 3(0.7)^3} \approx .12 \ne \frac 1 3 $$

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.

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