A Tricky Coin

Let $F$ and $H_4$ be the events of having the fair coin and flipping 4 heads, respectively, and denote the event of having the unfair coin as $\overline{D}.$

Via Bayes' Theorem , $P(F|H_4) = \frac{P(F)\cdot P(H_4|F)}{P(H_4)} = \frac{P(F)\cdot P(H_4|F)}{P(F)\cdot P(H_4|F) + P(\overline{F})\cdot P(H_4|\overline{F})} = \frac{\frac{8}{9} \cdot \frac{1}{16}}{\frac{8}{9} \cdot \frac{1}{16}+\frac{1}{9}\cdot 1} = \frac{1}{3}.$

Since there is a $\frac{1}{3}$ probability of having chosen the fair coin given that 4 heads were flipped, the probability that the next flip is heads is $\frac{1}{3}\cdot \underbrace{\frac{1}{2}}_{P(H|F)} + \frac{2}{3} \cdot \underbrace{1}_{P(H|\overline{F})} = \frac{5}{6}.$

(Trying to avoid relying on Bayes Theorem)

For the first four coin tosses there are 9x16 possible outcomes. (9 for the choice of coin and 16 for the outcome of 4 tosses.) But we can ignore most of these. Only 8 of the outcomes will produce four heads from a fair coin. 16 outcomes will produce four heads from the fake coin. Each of these 24 outcomes is equally likely and hence the probability that Zeb has picked a fair coin is only 8/24 = 1/3.

Now it is simply a matter of calculating the probability of a head as 1/3 x 1/2 plus 2/3 x 1 = 5/6.

There's a $\frac {8}{9}$ chance that one of the fair coins is chosen. $\frac {8}{9} \cdot (\frac {1}{2})^4 = \frac {1}{18}$ .

There's a $\frac {1}{9}$ chance that the coin with 2 heads is chosen.

The probability that 4 heads is tossed in a row is $\frac {1}{18} + \frac {1}{9} = \frac {1}{6}$ .

Given that 4 heads was chosen, the probability that one of the fair coins is chosen is $\frac {\frac {1}{18}}{\frac {1}{6}} = \frac {1}{3}$ .

The probability that the unfair coin was chosen is $\frac {\frac {1}{9}}{\frac {1}{6}} = \frac {2}{3}$ .

So the probability of the next toss being heads is $\frac {1}{3} \cdot \frac {1}{2} + \frac {2}{3} = \boxed {\frac {5}{6}}$

$\begin{aligned}\textnormal{P(Picked biased | 4 heads in a row)}&=\frac{\textnormal{P(4 heads in a row | Picked biased)}}{\textnormal{4 heads in a row}}\times\textnormal{P(Picked biased)}\ \left(\textnormal{by }\href{https://brilliant.org/wiki/bayes-theorem/}{\textnormal{Bayes' Theorem}}\right )\\&=\frac{1}{\frac{8}{9}\times\left(\frac{1}{2}\right)^4+\frac{1}{9}}\times\frac{1}{9}\\&=\frac{2}{3}\end{aligned}$

Picking the biased coin and picking an unbiased coin are exhaustive events since we are given that Zeb picks a coin. $\begin{aligned}\therefore \textnormal{P(Picked unbiased | 4 heads in a row)}&=1-\frac{2}{3}\\&=\frac{1}{3}\end{aligned}$

$\begin{aligned}\textnormal{P(Next flip is a head | 4 heads in a row)}&=\textnormal{P(Picked biased | 4 heads in a row)}\times 1+\textnormal{P(Picked unbiased | 4 heads in a row)}\times\frac{1}{2}\\&=\frac{2}{3}\times 1+\frac{1}{3}\times\frac{1}{2}\\&=\frac{5}{6}\\&\approx {\color{#20A900}{\boxed{0.833}}}\textnormal{ (3 s.f.)}\end{aligned}$

P(Real and 4 heads)=8/9 * 1/16 = 1/18 P(Fake and 4 heads) = 1/9 P(Real and 4 heads and then a fifth head) = 8/9 * 1/16 * 1/2 = 1/36 P(Fake and 5 heads) = 1/9

(1/36 + 1/9)/(1/18 + 1/9) = 5/6

I'm not sure if this is a real solution or whether it is just an accident, but that's how I got the answer. I built the entire probability model first and this was what applied to the problem.

Let f be the number of fair coins.

Let u be the number of unfair coins.

Formula for calculating the probability of Head for any given nth flip would be.

$\frac{f + u \times 2^n}{2 \times f + u \times 2^n}$

So, to answer the original question.

f = 8

u = 1

n = 5

Probability for Head in the fifth flip would be: $\frac{8 + 1 \times 2^5}{2 \times 8 + 1 \times 2^5}$

: $\frac{5}{6}$