Combining coin toss probabilities

Question

I came across this question:

From a pool of 84 coins, one of which is 100% biased towards heads, a coin is selected at random. It is tossed 8 times, coming up heads every time. What is the probability that it will come up heads again?

Thinking about it, intuitively the probability seems to be $\frac{1}{84} +\frac{83}{84} *\frac{1}{2}$. This seems to be correct, but I'm still not entirely sure as to why. Can anyone demonstrate what ideas would be used to show that the probability is $\frac{1}{84} +\frac{83}{84} *\frac{1}{2}$, or if that probability is incorrect, then what the correct probability would be?

Answer 1

This answer is not correct. To answer this question let's first think about the first flip, and the probability that we have selected a fair coin.

Now, there are a total of $2*84=168$ equally likely outcomes on the first flip (with the biased coin accounting for two of those outcomes since there are two sides of the biased coin showing $H$)... but we are given evidence that the coin flip is heads. Thus, we can eliminate some of the outcomes on our first flip. For all of the fair coins, of which there are $83$, we can eliminate all of the outcomes in which one of the coins was tails, thus there are $83$ ways we could've gotten to a heads depending on the fair coin we chose. With the biased coin, there are $2$ possible outcomes that yield heads. Then the total number of outcomes that yield heads is $83+2=85$. And the ratio of fair heads to overall heads is $83/85$ so there is a $98$ percent (approximated) chance at this point that we selected a fair coin.

As we flip the coin possible outcomes for the biased coin grow exponentially (as we have yet to flip a tails). Thus, for the $n$th outcome that has not yet yielded a tails we have $2^n$ ways we could've gotten to that outcome given we selected the biased coin. If we were suddenly to flip a tails, we know that we would no longer need to consider the biased coin.

Now, on the $8th$ flip there are $83$ outcomes from the fair coins and $2^8=256$ outcomes from the biased coin. Now, the chance that we have selected a fair coin is $\frac{83}{256+83}=\frac{83}{339}$. We want to calculate the weighted probability of flipping a heads given that we are more likely at this point to have chosen the biased coin.

We calculate this using the rule of sum since our events are mutually exclusive, but use the rule of product to calculate the probability of each separate outcome. Since the fair coin gives us a $\frac{1}{2}$ chance of flipping a heads and there is a $\frac{83}{339}$ chance that a fair coin was chosen... we can say that the probability of selecting a fair coin and flipping a heads is $\frac{83}{339}\cdot \frac{1}{2}$. Likewise, for the biased coin the probability that we flip a heads is $1$ and at this point we have a probability of $\frac{256}{339}$ that the coin is biased, so then the probability of selecting the biased coin and flipping heads is $\frac{256}{339}\cdot 1$. Thus, the answer is

$$P(H\vert HHHHHHHH)=\frac{83}{339}\cdot \frac{1}{2}+\frac{256}{339}\cdot 1.$$

Where this is worded as "the probability of flipping a heads given we have flipped $8$ heads in succession".

Good video to help explain the intuition: https://www.youtube.com/watch?v=Zxm4Xxvzohk&t=132s

Answer 2

Let $8H$ and $C_b$ mean that heads came up 8 times and the biased coin was picked, respectively. We use Bayes' Law.

$$ \mathbb{P}(C_b|8H) = \frac{\mathbb{P}(8H|C_b)\cdot\mathbb{P}(C_b)}{\mathbb{P}(8H)} $$

Total probability on the denominator gives:

$$ \frac{\mathbb{P}(8H|C_b)\cdot\mathbb{P}(C_b)}{\mathbb{P}(8H|C_b)\cdot \mathbb{P}(C_b) + \mathbb{P}(8H|C_n) \cdot \mathbb{P}(C_n)} $$

Finally computation after another Law of Total Probability is…

$$ \mathbb{P}(C_b|8H) =\frac{\frac{1}{84}}{1 \cdot \frac{1}{84} + 0.00390625\cdot \frac{83}{84}} \approx\ 0. 75\\ $$

For the next throw:

$$ \mathbb{P}(H) = \mathbb{P}(H|C_b)\cdot\mathbb{P}(C_b) + \mathbb{P}(H|C_n)\cdot\mathbb{P}(C_n) = 1\cdot0.75+0.5\cdot\left(1- 0. 75\right)=0.875 $$

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If you want the fractional result, $\mathbb{P}(\text{all}\ mH|C_n)\sim \text{Bi}(k,0.5)$, where $k$ is the total number of coin tosses.

In a more general case, the problem could include $k$ throws and a coin with a $\text{bias} = p$, for $p\in[0,1]$.

The process is similar to the one above:

$$ \mathbb{P}(C_b|kH) = \frac{\mathbb{P}(kH|C_b)\cdot\mathbb{P}(C_b)}{\mathbb{P}(kH)} $$

Total probability on the denominator gives:

$$ \frac{\mathbb{P}(kH|C_b)\cdot\mathbb{P}(C_b)}{\mathbb{P}(kH|C_b)\cdot \mathbb{P}(C_b) + \mathbb{P}(kH|C_n) \cdot \mathbb{P}(C_n)} $$

We know that $\mathbb{P}(kH|C_b)=\text{Bi}(k,p)=k$ Finally computation after another Law of Total Probability is…

$$ \mathbb{P}(C_b|kH) =\frac{\frac{1}{C_{\text{tot}}}}{\frac{1}{C_{\text{tot}}} + \frac{\text{Bi}(k,0.5)}{\text{Bi}(k,p)}\cdot\left(\frac{C_{\text{tot}}-1}{C_{\text{tot}}}\right)}\\ $$

From which we can observe multiple effects depending of which quantity is altered.

Answer 3

The probability of tossing eight heads in a row is $$\left(\frac {83}{84}\right)\left(\frac 12\right)^{8}+\frac{1}{84}\times1^8=\frac{83}{21504}+\frac{1}{84}=\frac{113}{7168}$$

The conditional probability that the coin is fair, given that eight heads have been tossed is $$\frac{83}{21504}/\frac{113}{7168}=\frac{83}{339}$$

The conditional probability that the coin is biased, given that eight heads have been tossed is $$\frac{1}{84}/\frac{113}{7168}=\frac{256}{339}$$

Therefore the probability that the next toss results in a head is $$\frac{83}{339}\times\frac 12+\frac{256}{339}\times 1=\frac{595}{678}$$