A fair coin is flipped $100000$ times and you get $100000$ Heads in a row. What is the probability that you get Heads on $100001$th flip?

Question

My Answer: $1/2$
My Reasoning: Each coin flip is an independent event, outcome of which doesn't depend on preceding flips. So, even though the probability of $100000$ heads in a row is very low, the probability of next coin to be a head is $1/2$ itself.

An answer by a peer: $1$
Their reasoning: Its very naïve to say that the probability will be $1/2$ without considering the Bayesian approach which considers the information of both data as well as the prior distribution of parameter, here one can easily check the posterior probability to be $1$ with simple Bayesian analysis.

So, what is the correct acceptable answer? What is the correct reasoning here? If their answer is correct, please explain me how Bayesian approach is used here.

Answer 1

If it's actually a fair coin, then the flips are independent and you would be correct.

However, there is pretty strong evidence that it isn't actually a fair coin

Answer 2

The assumption of a coin being fair is not a property of the coin but a feature of the whole coin flipping system. It is easy to flip a perfectly symmetrical and balanced coin with unequal chances of heads and tails. Theoretically speaking if you are sure that the whole coin flipping system is fair, it is true that the chances of heads in the next trial is 1/2. But how would you ensure that? I don't know.

Under a Bayesian perspective with, for instance, an uniform prior (Beta(1,1)) the posterior pdf for $p$, the probability of heads, will be a Beta$(100001,1)$, with $E(p)=0.99999$. An answer that appears closer to common sense in a situation like this.

I would be willing to bet a good deal of money on heads in the next flip.

Answer 3

"A fair coin is flipped 100,000 times in a row and you get 100,000 Heads in a row".

Clearly you have a fair coin with "Heads" on both sides, so the probability is 1.

Answer 4

The problem states that a fair coin is flipped a hundred thousand times, and comes up heads each time. By definition, a fair coin is a coin such that every toss is independent from every other toss, and the probability of coming up heads on any particular toss is exactly $\frac{1}{2}$. If the problem states that this coin is fair, then the fact that the first hundred thousand tosses are all heads is completely irrelevant.

Therefore, as the problem is stated, I would argue that the correct answer is "The next toss will be heads with probability $1/2$."

Answer 5

You are correct, the answer is 1/2.

Ask your friend this question: If $2^n$ people were flipping a fair coin $n$ times each, then the probability $p(n)$ of at least one of them getting $n$ heads in a row is

$$p(n) = 1 - \left(1-\frac{1}{2^n}\right)^{2^n}$$

Observe that $$\lim_{n\rightarrow\infty}p(n) = 1-\frac{1}{e}>63\%$$ In fact $p(100,000)>63\%$ thus it is more likely than not that someone, say it was Alice, actually did flip 100,000 heads in a row with a fair coin. Would your friend still conclude that the probability is 1 that Alice's next coin flip is heads?

Answer 6

If you start with the definition of a fair coin as IID yielding heads or tails with 50% probability each time you throw it, then obviously the answer is that the probability is still 50% after 100 000 throws, regardless of what the outcome of those throws was. It's literally just there in the definition.

However, if this actually happened to you then you could at this point be perfectly certain that whoever gave you the coin did not mean the above by “fair coin”. In practice in such a situation you might well assume they just lied to you. One way of modelling this Bayesianically is to assign prior probabilities regarding whether the provider of the coin was an honest person.

But with a result that crass, the concrete explanation to go with is as Suzu Hirose and gnasher729 wrote: the coin appears to have heads on both sides.

How can we pin that to a statistical computation? Well, so we admit that it's not god-given that every coin has heads on one side and tails on the other. We do still have a heavy prior on this being the case. To formulate that precisely: the images on both sides of the coin are random variables, but side A showing heads is highly correlated with side B showing tails, vice versa. Say, 99.99%, i.e. initially we're basically sure that this is a bog-standard fair heads&tails coin.

	Side A is heads	Side A is tails
Side B is heads	$P(HH) = 0.01\%$	$P(TH) = 99.99\%$
Side B is tails	$P(HT) = 99.99\%$	$P(TT) = 0.01\%$

But then we have we have this 100000-heads result. We still assume that the coin lands with equal probability on either side, but we can now compute posterior probabilities for the actual images on the coin: the conditionals are $$\begin{align} P(100000H | HH) =& 1 \\ P(100000H | TH) =& 2^{-100000} \\ P(100000H | HT) =& 2^{-100000} \\ P(100000H | TT) =& 0. \end{align}$$ We have $$\begin{align} P(100000H) =& \sum_{c\in\{HH,TH,HT,TT\}}\!\!\!\!\!P(c)\cdot P(100000H | c) \\=& 0.0001 + 2\cdot 0.9999\cdot 2^{-100000} + 0 \\\approx& 0.0001 + 2^{-99999}. \end{align}$$ Then $$\begin{align} P(HH | 100000H) =& \frac{P(100000H | HH)\cdot P(HH)}{P(100000H)} \\=& \frac{0.0001}{0.0001 + 2^{-99999}} \\\approx& \frac{10^{-4}}{10^{-4} + 10^{-30102}} \\\approx& 1 - \frac{10^{-30102}}{10^{-4}} \\\approx& 1 - 10^{-30998} \end{align}$$ IOW, we can at this point be almost completely sure that the coin has heads on both sides.

(Feel free to repeat the calculation with as many 9-digits appended to the $HT$-prior as you fancy, it won't change the result meaningfully.)