Bob and Alice play with a coin. Together they have thrown $30$ heads and $70$ tails.

Bob says the coin is not fair. Alice disagrees and proposes to keep tossing. They toss another $100$ times and end Up with $100$ heads and $100$ tails.

Bob admits he is surprised but argues about the fairness anyway. He says there are $2^{200}$ possible outcomes and $\binom {200}{100} $ of them are $100$ heads and $100$ tails. So the probability that you are correct in claiming it has a probability of heads over tails between $\frac {99}{200}$ and $\frac {101}{200}$ is very close to $A = 1 - \frac{\binom {200}{100}}{2^{200}}$.

Cindy comes along , draws a Pascal triangle and points out that with a fair coin you have a probability of getting 100 heads equal to $A$. Danny joins and claims that - because of what Cindy Said - it works in reverse too , if your probability is $A$ then your coin must be fair.

Emmet joins the discussion and says Danny is wrong , because he does not take into account the unfair coins.

Who is right , who is wrong ? What is the truth ? What are the probabilities ? How certain are we that the coin is fair ?

  • $\begingroup$ What is the meaning of $\binom{100}{200}$? It seems that Bob had a lapsus. By other hand, what does it mean "...the probability that you are correct [...] is very close to A..."? $\endgroup$ – Carlos Mendoza Nov 3 '15 at 22:54
  • $\begingroup$ I think i wrote the binomium upside down. Sorry. $\endgroup$ – mick Nov 4 '15 at 0:06

(Edited. The first answer was wrong.)

The probability $P$ that a fair coin shows $\leq30$ heads or tails in the first $100$ throws is given by $$P={2\over 2^{100}}\sum_{k=0}^{30}{100\choose k}\doteq0.0000785014\ ,$$ and the probability that this occurs two times in a row is $P^2\doteq6.16\cdot10^{-9}$. Rejecting the null-hypothesis ("This coin is fair") after such an outcome means that you err in $0.000000616\%$ of the cases where actually a fair coin was presented to you.

| cite | improve this answer | |
  • $\begingroup$ This is incorrect. By this thinking, you would reject "The coin is fair" even if you threw $50/50$ and $50/50$ (the most likely outcome) , since probability of that is $\approx (\frac 1 {\sqrt{50\pi}})^2\approx 0.006=0.6\%$. $\endgroup$ – A.S. Nov 4 '15 at 3:31
  • $\begingroup$ @A.S.: Thank you for spotting my error. Hope it's acceptable now. $\endgroup$ – Christian Blatter Nov 4 '15 at 9:28
  • $\begingroup$ It's canonical now. It's interesting that the new $P$ is only order of magnitude larger than the previous one - which shows how quickly normal distribution dies off. $\endgroup$ – A.S. Nov 4 '15 at 9:42

I would suspect sleight of hand (or a biased throw mechanism).

Using the normal approximation to the binomial distribution, and assuming we have a fair coin, we'd have $\sigma=5$ (for each block of $100$ throws). Thus (under the assumption the coin is fair) our team has produced two $4\,\sigma$ events in a row.

At a guess, I'd look at the independence hypothesis. From the numbers I'd predict that the distribution of results following a $T$ is very different from the distribution following an $H$. Different enough to permit our team to reject the hypothesis of independence.

Note: as remarked by a commenter, @A.S., failure of independence really doesn't get the job done. At least no straight forward model simulates this plausibly. That leaves us with sleight of hand. The coin switched (from one weighted coin to another) in between blocks of throws...or some other external factor changed.

| cite | improve this answer | |
  • $\begingroup$ I think dropping identical distribution of throws (i.e. sleight of hand - biased one way in the first half and the other way in the second half) is a much better explanation compared to dropping independence. Why do you predict the throws are not independent? I tried a simple model (of constant correlation between successive throws) and a very rough MLE estimate only slightly spreads the distribution. $\endgroup$ – A.S. Nov 3 '15 at 18:29
  • $\begingroup$ Yeah, I tried that also, with similar results. So, I agree. Different coins. Or, some other factor changed. $\endgroup$ – lulu Nov 3 '15 at 19:45
  • $\begingroup$ Since a coin can't be biased beyond a few percent, must be a professional thrower who is careful above initial position of the coin (H or T) and throw technique (see Perci Diaconis's machine). $\endgroup$ – A.S. Nov 4 '15 at 3:25
  • $\begingroup$ I was thinking of that machine earlier! In my recollection he got a very profound bias...what sort of "weighting" could he manage? $\endgroup$ – lulu Nov 4 '15 at 11:15

We cannot know the truth about the coin from these 200 tosses (experiments). However we can evaluate the evidence and calculate probabilities given certain assumptions. That is we could ask and easily answer certain questions.

For example: assuming all of the throws are iid what is the probability of seeing the throws we've seen if the coin had 60% chance of landing tails.

| cite | improve this answer | |
  • $\begingroup$ Except having seen what happened, you may now suspect that Alice is controlling the result of each throw, or swapped a coin biased one way for a coin biased the other way. $\endgroup$ – Henry Dec 3 '15 at 12:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.