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I'm a laymen, and the Gambler's fallacy has bothered me for decades. I'm looking for someone to help me look at it from another perspective.

Here's why it confuses me:

  1. Of course, if you were to flip a coin 100 times, none of those results should be affecting the 101th flip. The 101th flip is 50/50
  2. Statistically, if you flipped a coin 100 times, you would expect around 50 heads and 50 tails.

So here's the question... why is it if you flipped 99 coins: 50 Heads and 49 Tails, why are we allowed to say that the 100th coin is still 50/50? Or put in another way, if we flipped 99 coins, and they were all 99 heads, why are we not allowed to say that the 100th coin is going to likely be a tails?

I logically understand that it will always be a 50/50 chance, but where I get confused is when we introduce a set of coin flips (like say we are going to flip 100 coins total), my mind wants to believe that the flips will be distributed 50/50 across the set.

I dug around this stackexchange and did find this answer which sort of helps: https://math.stackexchange.com/a/845405/1276792

His third point i think is the closest I've been to understanding it, that there is no expected difference between the number of heads and tails... but it's still hard for me to reconcile the fact that the 100th toss is 50/50 still. Maybe just a rephrase would help me

Mods feel free to close this question if it is not in the correct spirit of the community

A O
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  • The result of me flipping a fair coin and it being heads vs tails is going to be the same chance whether I ate pancakes for breakfast vs cereal. Similarly, its going to be the same whether I am wearing a red shirt vs a blue shirt. Similarly, its going to be the same whether previously I flipped some large number of heads or not. None of these priors affect anything about what is yet to occur. – JMoravitz Jan 08 '24 at 16:37
  • If anything... having flipped 99 heads in a row should have made us wary about the initial claim that the coin was in fact a fair coin and not rigged. At some point along the way I would have asked to look at the coin and make sure it wasn't in fact a double-faced trick coin. Otherwise, no... the universe doesn't care and doesn't do some cosmic rebalancing... If we were to repeat this repeatedly over and over, of the situations where we happen to have flipped 99 heads in a row, the next coin flipped will again be heads again just as often as the next coin flipped would be tails. – JMoravitz Jan 08 '24 at 16:40
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    It's funny that you say “my mind wants to believe that the flips will be distributed 50/50 across the set” when in fact it's much more likely that they will not be distributed 50/50 across the set. You have only about an 8% chance of seeing an exactly 50/50 distribution of heads and tails in a sequence of $100$ flips. – MJD Jan 08 '24 at 17:19
  • This has been discussed here many, many times already. Here's another: https://math.stackexchange.com/q/1863453/25554 – MJD Jan 08 '24 at 17:21
  • (Observe for example that even in four flips, there is only a $\frac6{16}=37.5%$ chance of flipping two heads and two tails, and $62.5%$ chance of flipping something else. Even three-and-one alone is more likely than two-and-two.) – MJD Jan 08 '24 at 17:26
  • Thanks for the links @MJD, give me a moment to read through them!

    JMoravitz thanks for the comment too, let's say that the coin is definitely fair. I get that every coin flip is independent, and is still 50/50, but i just have a hard time reconciling a 50/50 probability over a set of 100 flips. Actually what kind of helped is MJD's comment that there's only an 8% chance that it's exactly 50 heads 50 tails

     – A O Jan 08 '24 at 17:27

1 Answers1

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If you like the linked post with the conditional probability, maybe it helps to actually lay out all possibilities for two coins:

$$ \begin{array}{c c} (T,T) & (T, H)\\ (H, T) & (H, H) \end{array} $$ It is clear, that if we know the first coin, we know in which row we are, but not in which column. So conditional on the first coin being Tails (T), we know we are in the first row for example. And here we have a 50%/50% chance for either outcome.

Now if you had three coins, then you could turn this two dimensional table into a 3 dimensional array. Here I put both layers next to each other (but imagine them being behind each other

$$ \left[\begin{array}{c c} (T,T,T) & (T,T, H)\\ (T, H, T) & (T, H, H) \end{array}\right] \qquad \left[\begin{array}{c c} (H,T,T) & (H,T, H)\\ (H,H, T) & (H,H, H) \end{array}\right] $$ Knowing the first two coins (e.g. both tails (T,T, ?)) puts you in the first row of the first layer. You still have a 50%/50% distribution.

You can continue doing this with n-dimensional arrays.

Meta discussion

Knowing a lot about your sample so far (e.g. 99 coins are tails), puts you into a very small remaining set, which is unlikely itself. But you already know you are in this small set. So it does not matter how unlikely it is. What matters is the distribuiton of the remaining coin. And that does not change.

Now count the number of cases, where there are 3 tails in our 3 coin toss example. Compare with the cases where there are 2 tails and the case where there are only heads. You will notice that extreme outcomes are less likely.

This is because we essentially stop caring about the order they have appeared in and thus throw different results together into a bin.

So while every particular sequence (H, T, T, H, T,...) is equally likely, there are more of them with an even distribution of heads and tails, so we expect an even distribution. But if you already know the sequence so far, you are simply in this very peculiar subset and conditionally do not know anything about the next element.

Felix B.
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