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I have two identical looking coins, one is fair and has an equal chance of coming up heads or tails, but the other is weighted and will always land on heads. You pick one of the coins at random, toss it three times and get three heads. Given this, what is the chance that you've picked the weighted coin?

My way of solution is the following one.

With the fair coin, the probability that you get heads triple times is $1/8$.

With the biased coin, it is $1$. What's the total probability that you get three heads on three tosses? $1/8 + 1 = 9/8$

Now, assume you got your three heads. $\dfrac{\dfrac{1}{8}}{\dfrac{9}{8}} = 0.1111$ for fair coin and $0.888$ for biased one.

So the chance that I pick up biased coin is $0.888$.

Is my solution is correct?

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  • $\begingroup$ How could the total probability be $>1$? $\endgroup$ – lulu Jul 14 at 11:05
  • $\begingroup$ Note: your approach is mostly good, but you have to take into account the probabilities of choosing the various coins. Nowhere in your calculation do you use the fact that there were two coins. In this case the factor you require cancels out so, in the end, you do get the correct value. $\endgroup$ – lulu Jul 14 at 11:07
  • $\begingroup$ The total probability is 1. $\endgroup$ – vasiop Jul 14 at 11:07
  • $\begingroup$ I try with Bayes rule and end with the same solution. Should I show you. $\endgroup$ – vasiop Jul 14 at 11:08
  • $\begingroup$ As I say, the answer is correct because the factor you left out cancels out. Still, what you wrote is incorrect. The statement "What's the total probability that you get heads on three tosses? 1/8+1=9/8" is obviously wrong. $\endgroup$ – lulu Jul 14 at 11:09
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Let $W$ denote the event that you pick the weighted coin.

Let $E$ denote the event that you get $3$ heads by $3$ tossings.

To be found is $P(W\mid E)$ and this can be calculated on base of:$$\left[P(W)P(E\mid W)+P(W^{\complement})P(E\mid W^\complement)\right]P(W\mid E)=P(E)P(W\mid E)=P(W\cap E)=P(W)P(E\mid W)$$

Substitution on LHS and RHS gives:$$\left[\frac12\cdot1+\frac12\cdot\frac18\right]P(W\mid E)=\frac12\cdot1$$leading to: $$P(W\mid E)=\frac89$$

(This is application of Bayes rule but I try to avoid fractions with a probability in the denumerator as most as I can)

Here $\frac12$ is the factor that cancels out mentioned by lulu in his comments.

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Using Bayes rule.

$F$: denote the event you picked the fair coin

$B$: denote the event you picked the biased coin

$D$: Data collected, i.e., observed 3 heads in 3 tosses

We want to calculate: $P(F|D)$ and $P(B|D)$

Assume that the trials are independent. This is equivalent to saying: $P(D|F)=1/8$

$P(D|B)=1$

Let us also assume that $P(F)=P(B)=0.5$, implying that "you randomly pick coin" with equal chances of picking either one.

Apply Bayes theorem, which basically starts from:

$P(B|D)P(D)=P(D|B)P(B)$

Bring over the second term on the left hand side to the right hand side and expand by considering total probability as:

$P(D)=P(D|F)P(F)+P(D|B)P(B) = 1/8*1/2+1*1/2$

$P(B|D) = [P(D|B)P(B)]/[(P(D|F)P(F)+P(D|B)P(B))]$

I have got $\dfrac{8}{9}$ for biased coin.

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