A Mathematical Paradox About Probabilities [duplicate]

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 :)

Solution 1:

Something to think about:

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.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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 $2^{-10}$.

Solution 5:

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

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