Gambler's Fallacy vs Probability Distribution

[Andrew2]

Hi!

My question has to do with my misunderstanding with Gambler's Fallacy and probability distribution. My university knowledge on probability and statistical analysis has faded quite a bit, mostly due to the difficulty I have with understanding some of the concepts I faced there, so please be gentle with me!
Within video games and in gambling, I've always had a preconceived idea that I later found out was called Gambler's Fallacy. For example, in a coin toss scenario, if I toss 3 heads in a row, I would *feel* as though there is a higher chance that I will then toss a tail. Another example would be in a video game like League of Legends, where a character with 50% critical strike chance can attack something without crit-ting (by chance) and so one may believe that there is a higher chance to crit on their next attack. Another real-life example would be in slot machines, where a player may bet on Red because he has seen Blacks go past 4 times in a row in previous turns. And finally, a funny joke that I read included a man who was afraid of bombs in his airplane flight, and so he brought his own bomb to his flight because surely the probability of having two bombs on the plane is way less than having one!

So my initial question is, why isn't the history of these events affecting the chance of what's coming next?

Now I understand that in the coin toss example, although I *feel* as though there is a greater chance to toss a tail, I know that the chance is 50%. No matter how many heads in a row I get, on my next turn, the chance to get a heads or tails will still be 50%. But, I also know that the probability distribution of this scenario looks like this (taken from: https://ned.ipac.caltech.edu/level5/Berg/Berg2.html) :


What I get from this graph is that there is a higher chance to get 2 heads out of 4 tosses compared to every other situation (like 1 heads in 4 tosses, or 0 heads). So, if I was to get 2 tails in a row, surely I should be able to expect a higher chance to get at least 1 head in the last 2 tosses because the number of events getting at least 1 heads is greater than the number of events getting 0 heads.

After rolling 2 tails, P(2 heads) + P(1 heads) = 2/16 + 1/16 = 3/16
Compared with no heads: P(0 heads) = 1/16

As I'm writing this and recollecting my thoughts, I'm catching a glimpse of what I may be misunderstanding. Perhaps the Gambler's Fallacy has to do with things inside an event (on a microscale), whilst the probability distribution is more about the placement of the events (on a macroscale)? For example, there's a low chance for us to be born in 1st world countries (probability distribution), but the chance of a new quiet baby is not affected by how many crying babies in a row are introduced (fallacy)?

Sorry if I may not have expressed myself well in that last paragraph. Anyhow, I want to know your explanations and I want to deepen my knowledge on this matter.

So, my final question is: what am I misunderstanding?

 

[Dr.Peterson]

Perhaps the Gambler's Fallacy has to do with things inside an event (on a microscale), whilst the probability distribution is more about the placement of the events (on a macroscale)?

That's a big part of it.

A probability distribution is about long-term behavior; it says nothing about what will happen next.

So it is correct to think that, if you toss heads several times in a row, eventually you will get back to having equal numbers of heads and tails; but that will most likely happen because, over the next thousand tosses, you'll get approximately equal numbers, which will simply swamp the temporary imbalance.

It is wrong to think that something is forcing the next toss to be more likely tails, in order to bring the balance back right now.

 

[Steven G]

After rolling 2 tails, P(2 heads) + P(1 heads) = 2/16 + 1/16 = 3/16
What you said above is not correct at all. The coin has no memory, so it does not matter that you first tossed to tails.
P(2 heads) + P(1 heads) = 1/4 + 1/2 = 3/4

The flaw with a man who was afraid of bombs in his airplane flight brought his own bomb to his flight because surely the probability of having two bombs on the plane is way less than having one!

Yes, the odds of two random passenger bringing a bomb onto a plane is much less that just one random passenger bringing a bomb. The flaw is that this passenger who is afraid on bombs being on his plane is NOT randomly bringing a bomb.

Think of this problem which is similar to the plane problem. Suppose you bring your car, which is not a Jaguar XKE, to an intersection and watch the traffic for an hour everyday for a year. You notice that on average you see one Jaguar XKE per hour. Now suppose that the car you are in while watching the traffic daily for an hour is a Jaguar XKE (you bought one just to sit in while you watch cars). Do you now expect to see an average of 0 Jaguar XKE per hour (not including the one you're in)? Do you see why the average should still be one?

For the record, you roll dice and toss coins.

 

[Andrew2]

Thank you Dr. Peterson and Steven for your posts and the time you've taken to answer here! I think I understood the concepts you both have illustrated.


For Steven, let me just digest this Jaguar XKE example you gave.
For example, let's say I saw the same Jaguar XKE per hour (what a madman!). If I bought a different Jaguar XKE, I would still see the same madman Jaguar XKE per hour. If I bought the madman's Jaguar XKE, I would see 0 Jaguar XKE per hour. Ah! I guess in the plane problem, it would be similar to me buying the other guy's bomb off him haha.

Otherwise, let's say I saw a Jaguar XKE per hour that were all different compared to the previous cars. If I bought my own Jaguar XKE to watch cars...YES. I would still expect to see an average of 1 Jaguar XKE per hour. Though if there was another person watching cars just like me (but without having their own Jaguar car), on that particular day, maybe they would also see my Jaguar car going through which ends up with a rate of: 25Jaguars/24hours ≈ 1.04 Jaguars per hour. But that was not the point of your example. Understood!

Dr. Peterson and Steven, this is what my viewpoint was, and now this is how it changed from interpreting your messages:
My old viewpoint - In a game of 4 coin tosses, if I toss 2 tails in a row, I am more likely to toss 2 heads in the next two turns. Why? Because it is more likely that I have the highest probable result than having the lowest probable result of 4 tails in a row.

My new viewpoint - First of all in the above case, the chance of rolling TT and rolling HH in the next two turns is the same; 1/4.

Secondly, if I change the statement to "I am more likely to toss at least 1 head in the next two turns", then the probability of that is 3/4 as Steven has demonstrated. No wonder! It's very possible that I conditioned myself into thinking wrongly because I was rewarded for this wrong viewpoint. E.g. in slot machines, yes, 4 Reds may have passed in succession and I may think that at least 1 Black will come very very soon (let's say...in the next two turns!). But that Black may come through because there is a high 75% chance in guessing correctly for at least once in the next two turns. NOT because I was right in assuming that the balance of 50-50 is occurring now.

Finally, the history of what happened doesn't affect what happens next because we can't be sure where we are on the probability distribution graph. If I toss 2 tails in a row, I can't reside in the fact that I will more likely toss two heads in the next two turns with the faulty logic that having 2 heads and 2 tails is the most common result. The probability distribution graph doesn't show how likely I am to get 2heads and 2tails, but rather it just shows the "long-term behavior". It's possible that the specific case in the coin game example may be the more uncommon end result of 4 tails. In other words, yes it's possible to find the probabilities of what I'm tossing next, but I can't say that the result that I get will be the most likely result shown in long-term behavior.

Dr. Peterson and Steven, is there anything wrong with what I've said in this post?

 

[Dr.Peterson]

My new viewpoint - First of all in the above case, the chance of rolling TT and rolling HH in the next two turns is the same; 1/4.

Secondly, if I change the statement to "I am more likely to toss at least 1 head in the next two turns", then the probability of that is 3/4 as Steven has demonstrated. No wonder! It's very possible that I conditioned myself into thinking wrongly because I was rewarded for this wrong viewpoint. E.g. in slot machines, yes, 4 Reds may have passed in succession and I may think that at least 1 Black will come very very soon (let's say...in the next two turns!). But that Black may come through because there is a high 75% chance in guessing correctly for at least once in the next two turns. NOT because I was right in assuming that the balance of 50-50 is occurring now.

Finally, the history of what happened doesn't affect what happens next because we can't be sure where we are on the probability distribution graph. If I toss 2 tails in a row, I can't reside in the fact that I will more likely toss two heads in the next two turns with the faulty logic that having 2 heads and 2 tails is the most common result. The probability distribution graph doesn't show how likely I am to get 2heads and 2tails, but rather it just shows the "long-term behavior". It's possible that the specific case in the coin game example may be the more uncommon end result of 4 tails. In other words, yes it's possible to find the probabilities of what I'm tossing next, but I can't say that the result that I get will be the most likely result shown in long-term behavior.

The reality is that once you've tossed two coins, you aren't "on the distribution graph" for tossing four coins at all! You're on the distribution for the number of heads in tossing two coins given that the first two are tails. This eliminates all but four outcomes:

​

Of the four remaining cases, 1 has a total of 0 heads, 2 have a total of 1 head, and 1 has a total of 2 heads. The probability of getting at least one head in the following two tosses is 3/4, just as it is for any two tosses. This is what you said second.

And the probability that the number of heads will come to the expected 2 is only 1/4, the same as the probability of getting two heads on any two tosses. This is what you said first.

 

[Andrew2]

Dr. Peterson: "The reality is that once you've tossed two coins, you aren't "on the distribution graph" for tossing four coins at all! You're on the distribution for the number of heads in tossing two coins given that the first two are tails."

I think that was the nail in the coffin for me!
So, I would like to edit something that I've said previously in:

Andrew2: "The probability distribution graph doesn't show how likely I am to get 2heads and 2tails, but rather it just shows the "long-term behavior". It's possible that the specific case in the coin game example may be the more uncommon end result of 4 tails. In other words, yes it's possible to find the probabilities of what I'm tossing next, but I can't say that the result that I get will be the most likely result shown in long-term behavior."

In the red lines from the above, I mean to say that the probability distribution graph DOES show how likely I am to get a net result of 2 heads and 2 tails (whichever order), so that I CAN say that the result that I get will be the most likely individual result shown in long-term behavior - given that I am at the beginning of the coin toss game where I haven't tossed any coins yet. But also, it's more likely that I would not get a net result of 2 heads and 2 tails:

P(2H & 2T; any order) = 3/8
P(not 2H&2T) = 5/8

I think I'm steering away from the point, so I'll stop here. Thank you Dr. Peterson for your insight!

 

[Dr.Peterson]

But also, it's more likely that I would not get a net result of 2 heads and 2 tails:

P(2H & 2T; any order) = 3/8
P(not 2H&2T) = 5/8

That's true; although 2 heads is the most likely result of tossing 4 coins, it is less than 50%, so you are more likely to get more or less than 2 heads than to get exactly two heads. What is true is that if you toss 4 coins at a time, millions of times, the average number of heads per toss will probably be close to 2. (That's the Law of Large Numbers.)

And you're right that this is not directly relevant to your original question.