So - I am no math genius but I do have shower thoughts. And there is one thought about normal distribution that I just couldn't let go. I converted it into a little story to visualize it a little better. Let's see if it makes sense and if it really is a paradox I came up with. Here is the story:

A man is in court. He is said to have murdered someone. There is evidence that stats that he did it - but chances are it is all a coincidence. The judge comes up with a simple solution: "Tomorrow at 8am on market square - you are to toss a coin. Head and your head comes off - tails and you go home a free man. Let the gods decide whether you are to die or not."

The man gladly accepts this offer. You must know - even though it is the middle ages, he is a mathematican - not one to believe in gods. And he also knows probabilites and thinks he has a way of how to manipulate those.

The man takes his fate deciding coin home with him and begins tossing it all night.

The morning comes and everyone is waiting on market square. It is 8 am sharp and the man, as promised shows up with his coin in his hand. He is very confident, because he knows - his chances of dying are at about 0.1 %. In front of everybody, he tosses the coin and: tails. Then man is free to go. Not even the tiniest bit nervous about his fate.

How was that possible? He must have known that his chances where 50 - 50 (assuming the coin cannot land on its edge and will always be tossed and flipped randomly).

Well, here is the thing that I cannot explain:

Last night, the man was home - as I said - flipping his coin over and over again. Since this is a normal distribution, in within the first 10 tosses, the coin showed head 5 times, and tails 5 times. But, after many, many tosses - the coin finally showed head 9 times in a row. This happening comes with a likelihood of 0.2% (according to one of those tree-diagrams). Now - for the 10th time, the chances of head again would be only 0.1% percent if I am not mistaken. Now - in my eyes: All the man had to do was to NOT throw that coin again until his fate was about to be decided - because heads again? That would be insanly unlikely - wouldn't it be?

So, that is my paradox. A random coin toss cannot be manipulated only by waiting for it to be unlikely to show a certain outcome over and over again - or can it?

Thanks for reading my little story :) I hope you guys understand what I am trying to convey here :)

• Sorry...the coin has no memory. Each toss is independent of those that came before. – lulu Apr 10 '18 at 9:48
• Try this: what are the chances of 9 times the same outcome in a row, then the tenth outcome is different? – Joel Reyes Noche Apr 10 '18 at 9:56
• Note that this is an example of the famous gambler's fallacy. en.wikipedia.org/wiki/Gambler%27s_fallacy – gj255 Apr 10 '18 at 13:16
• Before I realized the asker had succumb to the gambler's fallacy, I had assumed the man in the story had just spent all night practicing how to consistently flip a coin to land tails. – Shufflepants Apr 10 '18 at 14:47
• My friend Bob flies a lot, and he is worried about international terrorists blowing up his flights. That's why he always carries a bomb in his suitcase. Bombs are super rare on airplanes, and so the odds that there would be two bombs on a flight are so small as to be infinitesimal. Does Bob have a good strategy or not? – Eric Lippert Apr 10 '18 at 17:00

Since the coin flips are independent, and assuming the coin is fair, the probability that ten coin flips land heads is:

$$P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H) \cdot P(H)\cdot P(H)=(0.5)^{10}$$

The probability that nine coin flips land heads and the tenth lands tails is:

$$P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H)\cdot P(H) \cdot P(H)\cdot P(T)=(0.5)^{10}$$

The probabilities are the same! So he had equal chances of dying or not dying.

All the answers provided explain why there is no "paradox." I would like to provide you an answer that is rather intuitive (hopefully) than formal.

Your question is a good example of what is known as "gamblers' fallacy" (see, Croson and Sundali (2005)). The fallacy occurs when one wrongly assumes that a bin from which a draw is made is finite. In your example, think of having a head as drawing a blue ball from a bin and having a tail as drawing a red ball from the same bin, and the bin contains countably infinite balls half of which are blue and the other half red. Notice that if you draw $9$ blue balls from the bin in a row, the probability of drawing another ball is still $\frac{1}{2}$ since there are still infinitely many blue balls left. this is true even if you draw $1000$ blue balls (in fact, any finite number of balls) from the bin in a row -- meaning that the probability that $1001$st ball is blue is still $\frac{1}{2}$. This is the case with the coin toss: even if heads come up $1000$ times in a row, the probability that a head comes up in $1001$st flip is $\frac{1}{2}$. The reason why you think you have a paradox is that you are mistakenly assuming that you are drawing balls from some bin that contains finite number of balls.

• This is an interesting take I've never seen on the independence of successive tosses - sampling from the future "bin" containing infinitely many balls of each color. I wonder if it will help the OP. I wonder whether it's a useful view in more formal contexts. – Ethan Bolker Apr 10 '18 at 11:50
• The OP claims that the probability of the tenth toss yielding heads, or the tenth ball being blue, is $0.1$% ($2^{-10}$). In your model of the OP's thinking, the probability of the tenth ball being blue is $9.1$% ($1/11$). – John Bentin Apr 10 '18 at 16:35
• @JohnBentin As I indicated, my answer attempts to provide a possible explanation on why the author of OP got into the confusion; and the answer is not a formal one. It seems that the author of the OP thinks: "now that I have 1000 heads in row, the next one must be a tail," a wrong belief that may be triggered by an assumption that the bin contains a finite number of ball. Although his confusion results from comparing a likelihood of having 10 heads in a row to the probability of having a head after the 9th toss, my answer seems to be still valid to explain why he does this comparison. – Green.H Apr 11 '18 at 9:23
• In your model, the probability of drawing ten blue balls in a row is$10!^2/20!\approx 3.65\times10^{-6}$. This does not match either the OP's (approximately correct) estimate of $10^{-3}$ or the sum you gave (which exceeds $1$). – John Bentin Apr 11 '18 at 10:57
• @JohnBentin (the last comment edited, there was a typo-- i missed 0 after 1 and used sum instead of multiplication.) This is why the probabilities that in the OP and in my answer do not match -- thanks though for pointing that out. Probably, I should have added "for example" to my last sentence to show that this is just a possible example. The final point: the likelihood of having 10 heads in a row is $2^{10},$ while the likelihood of drawing 10 blue balls in a row is $\Pi_{k=1}^{10}\frac{k}{k+10}$ – Green.H Apr 11 '18 at 12:47

The answers given so far are all theoretical. Let's be practical.

Exercise #1:

Go get a coin right now, and a piece of paper.

Flip the coin until it comes up heads three times in a row. Now, if the man's theory is correct, it should be more likely that the next toss is tails. Toss the coin one more time and record the result.

Repeat this exercise as many times as it takes to convince you that after three heads in a row, the chances of getting heads are still $50-50$. It should only take you a few minutes.

Exercise #2:

Go to the bank and get a ten dollars in pennies. That's $1000$ pennies. (Or, whatever the cheapest coin is in the country where you live.) Again, get a piece of paper, and two big jars.

Put the pennies in the first big jar and shake them and dump them out. Take every penny that landed "heads" and put it back in the first jar; put the tails in the second jar. There should be about $500$ in each.

Now do it again, but only flip the ones that came up heads. Again, separate out the heads from the tails. Now there should be about $250$ coins that have come up heads twice, and $750$ that came up tails at least once.

Do it again. Keep on doing it until you either have no coins in the heads jar, or you have a group of coins that were flipped and came up heads nine times in a row. If you have none left, start over.

OK, you now have at least one coin that just came up heads nine times in a row. Flip it, and record the results. According to the man's theory, that coin should almost never be heads. What is it in reality?

Again, repeat the experiment until you have convincing evidence that coins do not remember what happened to them in the past.

• The problem with 'intuitive' arguments is that without rigorous proof they could be wrong! :) I always think of Littlewood's proof regarding the difference Ï€(x) âˆ’ li(x) e.g. maths.manchester.ac.uk/~roger/math31022/MSCLee.pdf – Mitch Wheat May 25 '18 at 23:14
• @MitchWheat: I'm not suggesting an intuitive argument, I'm suggesting a practical experiment. Probabilities are not useful in the abstract; they make predictions of observed behaviours. We should always strive to check our probabilistic models against ground truth. The problem with rigorous proofs is that they might be proofs about abstract models whose characteristics do not match the world we wish to make predictions in! – Eric Lippert May 25 '18 at 23:19
• I'm not suggesting you answer is incorrect; merely pointing out the danger of non-rigorous proofs. if you were to observe the difference between Ï€(x) âˆ’ li(x) even for very large numbers you could make the (incorrect) hypothesis (and people did so) that li(x) is always greater than Ï€(x) .....(which it is not) – Mitch Wheat May 25 '18 at 23:25

The probability that a fair coin will come up heads, given that it already has come up heads nine times consecutively, is one half. It would still be one half whatever the history of the way it came up before. This fact is completely consistent with the probability that it will come up heads on the next ten tosses being $2^{-10}$.

Let $E$ be the event that the first nine flips are all heads and $F$ be the event that the first ten flips are all heads. The probability of getting ten conservative heads in ten flips is small. $P(F) =(0.5)^{10}$. But now it is given that the first nine flips are all heads. So the probability that the tenth flip is also a head should be

$$P(F|E) =\frac{P(F\cap E)} {P(E)} =\frac{0.5^{10}}{0.5^9}=0.5$$

If the man had announced at the trial that he was confident in an 0.1% chance of death because he would flip 9 consecutive heads that night, then he would be correct: there is an 0.2% chance of that string of heads occurring. However, he cheated and waited until this monstrously unlikely event happened by exhaustion. Now he's living in the 0.2% universe, so while it's unlikely for him to have gotten there...now that he is there, the past flips are no longer relevant to his future success.

The probability of getting $10$ heads in a row is exactly the same (given a fair coin) as the chance of getting $9$ heads, then a tail. Using your logic, the probability of getting $9$ heads and one tail (in that order) is $\frac{1}{1024}$, so you have only a $\frac{1}{1024}$ chance of surviving.

The probability of an events is the number of ways it can happen divided by the number of results that are currently possible, not the number of ways it can happen divided by the number of results that could have happened.

Suppose you had $512$ hats, and you randomly choose one. Suppose you happen to choose the hat numbered $1$. You could say "The probability of choosing hat $1$ and getting heads is $\frac{1}{1024}$, so I only have a $\frac{1}{1024}$ chance of dying", but that would be fallacious reasoning. The probability of getting the exact combination "hat $1$ and heads" is $\frac{1}{1024}$, but there are $512$ other "hat x and heads" combination. So there's $512$ ways to die out of $1024$ possibilities, giving you a $\frac{1}{2}$ chance of dying.

Similarly, the probability of the exact combination "$9$ heads, then anther head" is $\frac{1}{1024}$, but there are $512$ other combinations that also end with heads as the last flip, so your total probability of dying is $\frac{1}{2}$. You have to be consistent: either divide the number of ways to die once you've gotten nine heads by the number of possibilities that remain once you've flipped the coin nine times, or divide the total ways to die before you start flipping by the number of ways to flip a coin $10$ times. Comparing the number of ways of dying that remain after you've gotten $9$ heads to the number of possibilities before you've flipped the coin is fallacious.

Since this is a normal distribution, in within the first $10$ tosses

Are you using "normal" to mean "usual", or Gaussian? A coin flip is not Gaussian, it's Bernoulli.