is the gamblers fallacy really wrong? (first post)

[coppersurffer]

sorry if this isn't the right area to post this I wasn't sure where it should go. (I didn't want to post in math or something which might have content which the community might think of as more respectable so I put it here so it can be moved somewhere nicer if the admins think it is worthy.)


so recently I've been thinking about the gamblers fallacy.  the gamblers fallacy is the idea that things become more or less likely based on previous results which to most people seems obvious. you can't flip a coin and keep coming up tails forever.
at one point I thought I discovered why the gamblers fallacy is a fallacy that is that any particular combination of 100 coin flips have the same chance of happening. so 100 heads is just as likely as alternating heads tails until the end. or 99 heads and one tails at the end.  but today a new line of reasoning occurred to me.

let us assume that their is some unknown number of times we have to flip a coin before we get a heads. lets call that number x
every time we flip the coin the number of times we have flipped it gets closer to x meaning that everytime we get a tails the odds of the next flip being a tails gets lower. (remember we don't know what x is we just know that the number of flips is approaching x and will reach it eventually) I'm not sure if this qualifies as the gamblers fallacy but it fits the intuitive notion that either good or bad luck can only hold out for so long. 

ps. this line of reasoning occurred when thinking about yellowstone and weather it was a gamblers fallacy to say that the volcano is "due" for an eruption or not

[swansont]

22 minutes ago, coppersurffer said:

 
ps. this line of reasoning occurred when thinking about yellowstone and weather it was a gamblers fallacy to say that the volcano is "due" for an eruption or not

Yellowstone would only be applicable if eruptions were random events, and I don't think that qualifies.

[studiot]

Let us say you want to toss a head- call it H

To analyse the possibility of there being some x number of throws by which an H must have been thrown proceed as follows.

 

If you haven't tossed an H in any throw before x then all the tosses are Ts.

The probability of tossing H is 0.5 and the probability of tossing T is also 0.5. in one throw.

The probability of tossing T in both of  two throws is (0.5) x (0.5) = 0.25

The probabiltiy of tossing T in all of three throws is (0.5) x (0.5) x (0.5) = 0.125

 

can you see the trend?  (we need Two things)

 

1) The probability of tossing T in all of n throws is (0.5) x(0.5) x (0.5) x (0.5)........... n times =  (0.5)ⁿ

2) The probability tossing at least one H is (the probability of tossing all Ts minus 1) = (0.5)ⁿ - 1

To be certain of an H we need this to equal one.  (0.5)ⁿ - 1 = 1

So (0.5)ⁿ must equal zero

So we are looking for an n for which (0.5)ⁿ = 0

To try to find this we need to investigate what mathematicians call a limit, which is a number (zero in this case) which the result of the expression gets closer and closer to the large n gets.

[math]\mathop {\lim }\limits_{n \to x} {\left( {0.5} \right)^n}[/math]

 

Now I hope you can see that this can never be zero since we have

[math]\mathop {\lim }\limits_{n \to \infty } {\left( {0.5} \right)^n} = 0[/math]

That is no matter how large an n we take the result is always greater than zero all the way to infinity.

So there is no x for which the original proposition is true.

 

[Silvestru]

20 minutes ago, swansont said:

Yellowstone would only be applicable if eruptions were random events, and I don't think that qualifies.

I guess coin flips aren't random either. This subject was discussed on a different thread but can't find it .

[Strange]

1 hour ago, coppersurffer said:

let us assume that their is some unknown number of times we have to flip a coin before we get a heads. lets call that number x
every time we flip the coin the number of times we have flipped it gets closer to x meaning that everytime we get a tails the odds of the next flip being a tails gets lower. (remember we don't know what x is we just know that the number of flips is approaching x and will reach it eventually) I'm not sure if this qualifies as the gamblers fallacy but it fits the intuitive notion that either good or bad luck can only hold out for so long. 

It is a form of the gambler's fallacy because it starts with a false assumption.

[studiot]

To analyse Yellowstone or San Andreas or Mt St Helens you need a Bayesian statistical analysis, not the type I posted above.

[Prometheus]

5 hours ago, coppersurffer said:

let us assume that their is some unknown number of times we have to flip a coin before we get a heads. lets call that number x

You can do this but then x is not a number but is itself a random variable, drawn from the geometric distribution. You can work out its expected value (mean), but as it isn't a number the rest of your approach won't work.

I think you were closer with your first line of thought. As you said 100 heads is just as likely as alternating heads and tails or any other particular sequence of 100 flips. Once we reach the end of any particular sequence of 100 the chance of either a head or tail on flip 101 is still 0.5.  I think you may be getting caught out by the fact that in all these possible sequences only 1 leads to all heads, and only one leads to all tails while lots of possible sequences have numbers of heads and tails that cluster around 50 each - there are just more of these possible sequences. But regardless of whichever sequence was realised, the 101 flip is still 50/50.

 

5 hours ago, studiot said:

2) The probability tossing at least one H is (the probability of tossing all Ts minus 1) = (0.5)ⁿ - 1

Oh no, negative probabilities. Isn't that one of the signs of the apocalypse?

 

4 hours ago, studiot said:

To analyse Yellowstone or San Andreas or Mt St Helens you need a Bayesian statistical analysis, not the type I posted above.

I assume were talking eruptions if we're talking Yellowstone? Can eruptions be assumed to be independent? My naive thoughts are that after an eruption the chances of another eruption are diminished as there is a release of pressure.

Apparently there are plenty of non-Bayesian approaches in statistical seismology. Not sure which is considered better in what circumstances though. 

[coppersurffer]

to be clear when i say

15 hours ago, coppersurffer said:

let us assume that their is some unknown number of times we have to flip a coin before we get a heads. lets call that number x

I'm referring to the fact that if we flip until we get a heads that number of flips will be some arbitrary integer(for this particular series of flips, another series of flips will have some other arbitrary integer.).  we can not be sure what that integer is(statistics can help us make some reasonable guess though). we just know that we're perpetually getting closer to it until it happens and we know that the chances of that arbitrary number  being 1000 is far less likely than it being 7 as for it to be 1000 we had to get 999 tails in a row which is highly improbable

 

 

14 hours ago, studiot said:

Let us say you want to toss a head- call it H

To analyse the possibility of there being some x number of throws by which an H must have been thrown proceed as follows.

 

If you haven't tossed an H in any throw before x then all the tosses are Ts.

The probability of tossing H is 0.5 and the probability of tossing T is also 0.5. in one throw.

The probability of tossing T in both of  two throws is (0.5) x (0.5) = 0.25

The probabiltiy of tossing T in all of three throws is (0.5) x (0.5) x (0.5) = 0.125

 

can you see the trend?  (we need Two things)

 

1) The probability of tossing T in all of n throws is (0.5) x(0.5) x (0.5) x (0.5)........... n times =  (0.5)ⁿ

2) The probability tossing at least one H is (the probability of tossing all Ts minus 1) = (0.5)ⁿ - 1

To be certain of an H we need this to equal one.  (0.5)ⁿ - 1 = 1

So (0.5)ⁿ must equal zero

So we are looking for an n for which (0.5)ⁿ = 0

To try to find this we need to investigate what mathematicians call a limit, which is a number (zero in this case) which the result of the expression gets closer and closer to the large n gets.

limn→x(0.5)n

 

Now I hope you can see that this can never be zero since we have

limn→∞(0.5)n=0

That is no matter how large an n we take the result is always greater than zero all the way to infinity.

So there is no x for which the original proposition is true.

 

if I understand this. this means their is no constant value whereupon their has to have been a heads. I think we can still assert that at some arbitrary point(for the paticular sequence) h will happen (assuming the flipper dosn't give up) right?

Edited November 1, 2017 by coppersurffer
Clairfication

[Strange]

7 hours ago, coppersurffer said:

if I understand this. this means their is no constant value whereupon their has to have been a heads. I think we can still assert that at some arbitrary point(for the paticular sequence) h will happen (assuming the flipper dosn't give up) right?

What Studiot is showing (mathematically) is that however long the sequence of flips is, it is possible (50:50) that the next one will be a tail.Therefore there is no upper limit to the number of headless flips. You could flip from now until the end of the universe, and beyond, and get tails every time. And the one after that could be tails as well.

[studiot]

18 hours ago, Prometheus said:

You can do this but then x is not a number but is itself a random variable, drawn from the geometric distribution. You can work out its expected value (mean), but as it isn't a number the rest of your approach won't work.

I think you were closer with your first line of thought. As you said 100 heads is just as likely as alternating heads and tails or any other particular sequence of 100 flips. Once we reach the end of any particular sequence of 100 the chance of either a head or tail on flip 101 is still 0.5.  I think you may be getting caught out by the fact that in all these possible sequences only 1 leads to all heads, and only one leads to all tails while lots of possible sequences have numbers of heads and tails that cluster around 50 each - there are just more of these possible sequences. But regardless of whichever sequence was realised, the 101 flip is still 50/50.

 

Oh no, negative probabilities. Isn't that one of the signs of the apocalypse?

 

I assume were talking eruptions if we're talking Yellowstone? Can eruptions be assumed to be independent? My naive thoughts are that after an eruption the chances of another eruption are diminished as there is a release of pressure.

Apparently there are plenty of non-Bayesian approaches in statistical seismology. Not sure which is considered better in what circumstances though. 

Oh dear thank you Prometheus +1 for spotting that.

My excuse is that this was earlier than I usually post and to adapt the words of Bob Hope

I don't think anything in the morning

I don't think anyhthing until noon

and then it's time for my nap.

 

My apologies to all concerned obviously my equation should have the terms on its left hand side reversed.

1 - (0.5)ⁿ

9 hours ago, coppersurffer said:

if I understand this. this means their is no constant value whereupon their has to have been a heads. I think we can still assert that at some arbitrary point(for the paticular sequence) h will happen (assuming the flipper dosn't give up) right?

NO, it means there is no value, constant or otherwise. Zilch. So no we cannot correctly assert that h will happen, only that it becomes of increasingly greater probability.

 

 

One thing that should be noted is that the axiom that all the probabilities add to 1 means that the possibilities (event) are disjoint.

In other words each throw is entirely independent of the results of any other throw.

This is not the case with geological phenomena, in particular the second comment by prometheus states otherwise.