A box contains two coins: a regular coin and one fake two-headed coin (P(H)=1). One coin is choose at random and tossed $n$ times.

If the first n coin tosses result in heads, What is the probability that the $(n+1)^{th}$ coin toss will also result in heads?

My solution:

$$\text{Required Probability} = \frac{1}{2}\times \left(\frac{1}{2}\right)^n \times \frac{1}{2}+\frac{1}{2} \times 1^n \times 1$$

  • $\begingroup$ Your solution is correct. $\endgroup$ – Mark Fischler Aug 15 '16 at 18:08
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    $\begingroup$ This does not look correct to me. If $n$ is very large then you are sure that you must have the weighted coin, in which the desired probability is $1$. Your formula, however, tends to $\frac 12$. Indeed, your value decreases as $n$ increases which makes no sense (to me). $\endgroup$ – lulu Aug 15 '16 at 18:11
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    $\begingroup$ @MarkFischler I expect you misread the question. The OP correctly calculates the probability of getting $n+1$ consecutive heads, but that is not at all what the question was asking. $\endgroup$ – lulu Aug 15 '16 at 18:16

The probability of seeing $n$ heads in $n$ tosses of a fair coin is $2^{-n}$. Thus the total probability of seeing $n$ consecutive heads is $$\frac 12\times 2^{-n}+\frac 12 \times 1$$ Therefore, given that you have in fact observed $n$ consecutive heads the new estimate for the probability that you have the fair coin is $$\frac {\frac 12 \times 2^{-n}}{\frac 12\times 2^{-n}+\frac 12 \times 1}=\frac {1}{1+2^n}$$ and the probability that you have the fake coin is therefore $$\frac {2^n}{1+2^n}$$ It follows that the probability that the next toss is also heads is $$\boxed {\frac 12\times \frac {1}{1+2^n}+\frac {2^n}{1+2^n}=\frac {1+2^{n+1}}{2+2^{n+1}}}$$

Note: as $n$ goes to $\infty$ this tends to $1$, as it clearly should.

  • $\begingroup$ why is that the probability of a fair coin? $\endgroup$ – Jorge Fernández Hidalgo Aug 15 '16 at 18:21
  • $\begingroup$ @CarryonSmiling That's the portion of the winning results which is explained by the fair coin. Worth remarking that my final answer coincides with yours, so I expect we are saying the same thing in different words. $\endgroup$ – lulu Aug 15 '16 at 18:22
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    $\begingroup$ @CarryonSmiling On inspection, we are saying the same thing in the same words. $\endgroup$ – lulu Aug 15 '16 at 18:26

First we find the probability that the coin is fair.

To do this we use Bayes theorem, we want $P(A|B)$, where $A$ is the probability the coin is fair and $B$ is the probability the first $n$ flips are heads.

By Bayes theorem:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$, the numerator is $\frac{1}{2^{n}}\times \frac{1}{2}$ and the denominator is $\frac{1}{2}+\frac{1}{2^n}\times \frac{1}{2}=\frac{2^n+1}{2^{n+1}}$. hence $P(A|B)=\frac{1}{2^{n}+1}$.

Therefore the final answer is:

$\color {green}{\frac{1}{2^n+1}\times \frac{1}{2}}+\color{\purple}{(1-\frac{1}{2^n+1})}=1-\frac{1}{2(2^n+1)}$

The green part is the probability that it is heads and that the coin is fair, and the purple part is the probability that it is heads and the coin is fake.


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