9 coin flips, 9 tails, chances of 10th flip being heads is still 50/50?

You are correct on both counts. The resolution is in considering all the flips.

How probable / lucky was it to get the previous 9 tails in a row? Very Lucky!

It's so lucky that the difference between 9 and 10 in a row is only a single flip.
As long as it is a fair coin, the probability of 10th trial will be still be 1/2 H, 1/2T. The key here is that the coin is assumed to be fair.

Objective assessment of probability when only two outcomes are possible and both of them have an equal chance is 50/50 as you point out. However the real frequency might not be 50/50 especially  with fewer tries. However if the experiment is repeated multiple times say 100 or even 1000 or even 10,000 then the average will be very close to 500 for H and 500  for T . This is called the Law of Large Numbers and these trials are called Bernoulli trials. Remember that these trials are independent. That is the outcome of the 9th trial doesn’t influence the result of the 10th trial.

Standard deviation (the amount by which the frequency deviates from the expected value of 1/2 and 1/2 ) in these types of Bernoulli trials is inversely proportional to the square root of the sample size. Small sample sizes will have more deviation and large samples won’t have much deviation. 9 trials is a very small sample size . Abraham De Moivre actually threw dice 10,000 times to figure this out and gave to the world the concept of standard deviation which is a big deal in statistics.

That’s why the deviation from the expected result in your small sample trial. Make it 100,1000 and if possible 10,0000 - another law Grand/Strong  Law of Large Numbers kicks in where the average fluctuates only within the standard deviation and doesn’t cross the boundaries at all.

Hope this helps!
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Here's something that helped to convince me when I was younger.

I have a fair coin in my pocket right now, and I'm about to flip it. I'm not a sleight-of-hand artist or anything, so the chances of getting heads or tails are equal, right?

Now, suppose it just so happened that the last few times I flipped this coin, it came up tails. Would that matter? Does this coin really remember or "store" the fact that it came up tails recently?
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>making the chances of getting Heads on the 10th flip higher than 50/50

No, it's still 50% chance to get heads on the 10th flip, because you're flipping 1 coin.

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>Does this have something to do with looking at one flip vs. all the flips?

Yes exactly that, and your instinct that one has a higher chance than the other is correct.  Say you want 10 heads in 10 flips, and can choose whether to start with 9 heads already flipped and flip 1 coin or 0 heads already flipped and flip 10 coins.

1. In one case, you *already* *have* 9 heads and only need **1** more.
2. In the other you have **0** heads, and *need* ***10*** more.

1 unknown at 50% chance vs. 10 unknowns at 50% chance *each*.  Your instinct is correct that *getting 10 heads* in 10 Flips is easier when you already have 9 heads vs having 0 heads. But that's a different question from what's the chance of a single flipped fair coin coming up heads.

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Another way to look at it with the probability formula:

For the probability of a bunch of independent events all occurring, you multiply the probabilities:

= 0.5 \* 0.5 \* 0.5 \* 0.5 \* 0.5 \* 0.5 \* 0.5 \* 0.5 \* 0.5 \* 0.5

or 0.5 \^ 10 = or 0.0976 % chance of flipping 10 coins and all of them being heads.

Once you flip a coin, it's result is *certain*. Certainty, as a percentage is 100% or as a decimal just 1.

So for the coins you've flipped already, you are certain of their result and can replace that with 1 instead of the 50% chance of 0.5:

1 \* 0.5 \* 0.5 \* 0.5 \* 0.5 \* 0.5 \* 0.5 \* 0.5 \* 0.5 \* 0.5  or 1 \* 0.5 \^ 9

which is as we know is the same as the probability of 9 heads.

This is a probability problem : you have all the information about the coin, that is, you know it’s a fair coin. Hence the chance of the 10th flip being heads given that the previous 9 were heads is 0.5

However, if you had no prior knowledge about the heads probability of the coin, then that would be a statistics problem. One approach would be to estimate a range of values for the true heads probability of the coin using the result of 9 tails in a row, this would be called a confidence interval.
Everybody here is right. (This community is great; the level of actual bad advice is astonishingly low. I'm impressed!)

The central question is, are separate coin-tosses really *independent events*?

The various fallacies that are collectively called "gambler's ruin" boil down to not believing in independent events. You've just flipped 9 tails. Your intuition is screaming that 2^(-10) is a really tiny number, and you are forgetting that you are *already* in a really unlikely corner of event space. Intuition is a terrible guide in places like that.

What's funny is that "gambler's ruin" comes in two flavors -- both of them boiling down to superstitious beliefs in different kinds of things. In your case, you are fighting an intuitive belief in a magical *force* that pushes the universe toward average behavior. It takes a long time to grasp that "regression to the mean" happens without any such force!

On the other hand, many gamblers believe in "streaks" and "slumps" -- that is, they feel like they enter periods of exceptionally bad or good luck. A gambler with this kind of belief might think the chances of a 10th tails are *enhanced*, because they are in a "streak". In fact, *this* kind of belief might be more rational than the "magic restoring force" belief, because after tossing nine or ten tails, it might be time to examine the coin more carefully. Some apparently-random events turn out to be not so random! If you have ever been suckered into a game of "three-card Monte", you know what I mean. But in your case, you know the coin is fair, and you should continue to bet even on the next flip.
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I'm going to chime in here -- mostly cuz this is a question I actually have the expertise (such as it is) to answer, which is really cool -- and offer a slightly different perspective from the other responses, one that I've found very useful in the past (especially when faced with complex probabilities)

I'm going to involve multiple universes.  (Stephen Strange may or may not be responsible.)

Consider this:

### You draw a single card from a freshly shuffled deck. What is the probability that you drew the Ace of Spades?

Most people can see, pretty clearly, that there are 52 cards, only 1 is the Ace of Spades, your chance is 1 in 52.

However, another way to look at is like this: There are 52 universes, and in each one, you drew a different card.  Since only 1 of those universes has you draw the Ace of Spades, there is a 1/52 chance that you are in that universe.

Now consider this:

### You draw a single card from a freshly shuffled deck. You show it to someone else and they say it's an Ace. What is the probability that you drew the Ace of Spades?

There are still 52 universes, where 52 different yous all drew 52 different cards.

However, as soon as someone tells you that you drew an Ace, _you can rule out 48 of those universes_, the ones where where you drew 2, or 3, or King, or whatever.  Sure, they may have happened, but _you know you're not in any of those_.  There are now only **four** possible universes you could be in, one for each of the suits.

And indeed, if you do the calculations, you will find that in this situation, you have a 1/4 chance of having drawn the Ace of Spades.

We can now apply this technique to your coin flips: You flipped a fair coin nine times and got T all nine times.  What is the chance that your next coin flip is T?

The chance is still 1/2 (aka 50-50).  Sure, it was really unlikely that you got 9 in a row, but look at the big picture.  For 9 flips:

- 510 other yous got a much less interesting sequence of flips and are now posting on r/learnmath about the Monty Hall problem
- 1 other you got heads 9 times in a row and posting to r/learnmath asking if the chance getting heads again is really 50%

Does that help?

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