Is the Gambler's Fallacy Really a Fallacy?

[Ignea_unda]

A string of a certain length, of ANY composition will have the same probability, (1/S)^n, correct?

Then, any of those strings can be taken, and the probability of a string of length n+1 with either a black OR red at the end, would be (1/S)^n+1.

 

[uart]

Hi Vokl. Let's define "Volkl's Ratio" as the the (coin tossing) ratio of [marginal probability of head following immediately after another head] divide [overall probability of heads].

I've never seen this ratio published so you could make quite a name for yourself by experimentally measuring and documenting it. So get yourself a coin and get tossing.

 

[paulfr]

you go on to believe that when we are not talking about infinity that the expected outcomes is somehow similar to the outcome as if we were talking about infinity.

I think it is YOU who is doing this, not me. Otherwise where do you get this idea that the number of R's must = the number of B's ?
That is why you expect the next event to be R after a string of B's.
Unless you are postulating a new theorem of Probability, your expectation comes from the Law of Large Numbers. But you violate its foundation by limiting the events to 50K. Thus you can not invoke it and the next roll is determined by Probabilities from Physics, not those from Statistics.

Also, you frequently talk about randomness. True, the discrete variable of the outcomes is random. But this only insures/predicts the P(#B's) has a Gaussian Probability Density. It says nothing about a single, given event which is the focus of your assertions here..

Your idea would work for a HyperGeometric Distribution. That is the case say for drawing Red and Black colored balls of equal number from a bag, but NOT replacing them after each event. Then if you saw 5 B's you could conclude that the P(R) on the next draw is > P(B) because the Sample Space is changing and you know how.

Look at this another way. Your thesis implies that if you see a string of 30 Blacks, that the Probability that next one will be a Red is very high. At least higher than P(B). That is the same as saying that P(31 B's) is much different than P(30). But we know this is not true because we know the PDF is continuous and Gaussian. You can not know that you have just seen the end of the string. It could be a string of 60 that is coming. Or a string of 59, or 58, or 57, etc, etc. You add all those possibilities together and you can see P(B) is higher than it appears.

 

[Well_InTheory]

The logic here does not require a computer. The point is that there is a higher probability that smaller sets of like numbers occur than larger sets of like numbers, so there is a tendency for the next value to oppose the previous string of like values.

Baye's Theorem (conditional probability) solves this one.

The chances of red (R) and black (B) are the same, so

P(R) = P(B) = 0.5

The same goes for any string of reds or blacks, with probability

P(nR) = P(nB) = \frac{1}{2^n}

You want the probability of the (n+1)th result being black given that the previous n results were all black, which is

P((n+1)=B|\sum_{i=1}^{n} (i = B)) = \frac{P(\sum_{i=1}^{n} (i = B)|(n+1)=B)P(B)}{P(\sum_{i=1}^{n} (i = B)|(n+1)=B)P(B)+P(\sum_{i=1}^{n} (i = B)|(n+1)=R)P(R)}

or in words, "the probability that the next one is black, given that the previous n were black, is equal to

the probability that the previous were black given that the next is black,

multiplied by the probability of any being black

divided by

the probability that the previous were all black given the next is black plus the probability that the previous were all red given the next is red."

This simplifies, since as per the first equation, the probability on any given try is the same for red and black. Also, the probability of n reds is the same as the probability for n blacks.

What you end up with is

P((n+1)=B|\sum_{i=1}^{n} (i = B))) = 0.5

as everyone else in this thread has already told you.

Where you're getting confused is in the text I quoted above - it's true that you're less likely to find the string of (n+1) black than you are to find the (n) black within a larger string, but just because you've found the (n) it doesn't mean that *this time* the next one will be the longer string. Equally so, it doesn't mean that this time it will be the shorter one.

Formally, your confusion is ascribing some forcing on the part of http://en.wikipedia.org/wiki/Regression_toward_the_mean#Other_statistical_phenomenon". That is a result of the unlikeliness of having a long string, not the cause.

 

[verty]

Volkl, saying that the next 10 will tend away from being all black is not the same as saying the next 1 will tend away from black. 10 will tend away from black, yes. The odds will be 1 in 1024 in general. But 1 won't tend away, it'll be black in 1/2 of cases generally.

 

[Ryan_m_b]

If I was a serious gambler there is a small chance that I could place 50,000,000 bets physically at a casino. Pretending that the game we are playing offers fair odds, the chances of one particular outcome coming up 1000 times in a row within the set compared to the same particular outcome coming up 100 times must be much less. If this is true the probability of 20 particular outcomes in a row must be less then the probability that 10 of the same particular outcomes can come up. Doesn't this prove that there is a tendency towards randomness meaning that there is a tendency to have less of the same particular value coming up in a row. To me, this logic proves that the gamblers fallacy is in itself a fallacy. Or do you believe that all 50,000,000 could be the same value for anyone living on earth? I had a roulette wheel with no greens in mind.

Volkl it's quite simple in practice. Let's imagine you had a fair coin. The chance of heads or tails is 1/2. If you flip it once more the chance is the same. The difference comes when looking at the combination you are getting. Take a look at this image

Each coin toss has the same probability because there are two outcomes with equal likeliness;

H

or

T

However by the second coin toss there are 4 different possible combinations we could have had

HH

or

HT

or

TH

or

TT

So the chance of getting any specific combination at this point is 1/4. Do you see? And to work out the possibility of all the combinations of the next coin toss all we have to do is times 1/4 by 1/2 which will give us 1/8. If you check that on the image you will see that it's right, by the 3rd coin toss there are 8 possible and equally probably combinations.

 

[HallsofIvy]

If I was a serious gambler there is a small chance that I could place 50,000,000 bets physically at a casino. Pretending that the game we are playing offers fair odds, the chances of one particular outcome coming up 1000 times in a row within the set compared to the same particular outcome coming up 100 times must be much less. If this is true the probability of 20 particular outcomes in a row must be less then the probability that 10 of the same particular outcomes can come up. Doesn't this prove that there is a tendency towards randomness

I wouldn't say a "tendency" to randomness. The whole point of a roulette wheel is to have randomness right from the start.

meaning that there is a tendency to have less of the same particular value coming up in a row.

Again, "tendency" is the wrong word . The probability of, say, "222222" is exactly the same as "128435". Of course, the probability of "222222" is far smaller than the probability of "anything other than "222222" or even "not all 6 the same result", just as the probability of "128435" is far smaller than the probability of "anything other than "128435".

To me, this logic proves that the gamblers fallacy is in itself a fallacy.

Please explain how that proves it. Are you clear on exactly what the "gambler's fallacy" is?

Or do you believe that all 50,000,000 could be the same value for anyone living on earth? I had a roulette wheel with no greens in mind.

Well, I presume you can show that the probability of 50,000,000 runs of the same thing is much less that 1 over the number of people on Earth so it is extremely unlikely- but the whole point of "random" is that it could happen.

(On practical note, if a roulette wheel started giving the same result time after time, I am sure the casino would take it out of action long before it got to 50,000,000!)

 

[Volkl]

Well in theory
Thank you for identifying the heart of the concept I am getting at I.e. Small stings of blacks in a row occur more frequently than large strings of blacks in a row.

Formally, your confusion is ascribing some forcing on the part of http://en.wikipedia.org/wiki/Regression_toward_the_mean#Other_statistical_phenomenon". That is a result of the unlikeliness of having a long string, not the cause.[/QUOTE]

How does this unlikeliness you speak of get implemented in reality? This is the dicotomy and why it is ultimately related to the gamblers fallacy because the way this unlikeliness is implemented is to change to red and ultimately this acts like an added force - to the 50/50 equation that everyone loves prooving but is leaving out - that changes the likelihood of a black coming next to less than 50% I.e. A slightly safer bet on red.

 

[Volkl]

In a set of 50,000,000 what percentage (of the groups of blacks in a row) would be one black in a row? What percentage would be two blacks in a row. What percentage would be three blacks in a row? What percentage would be four blacks in a row? What percentage would be five blacks in a row?

Would red's percentages look the same as blacks?

 

[Fuz]

In a set of 50,000,000 what percentage (of the groups of blacks in a row) would be one black in a row? What percentage would be two blacks in a row. What percentage would be three blacks in a row? What percentage would be four blacks in a row? What percentage would be five blacks in a row?

Would red's percentages look the same as blacks?

We havn't tried this experimentally, so how could we know? Also, why does it matter whether the string of outcomes are of the same color? The roulette table doesn't know or care whether black shows up or red shows up. The permutation of N consecutive outcomes has the exact same theoretical probability of "showing up" as a different permutation of N consecutive outcomes, as stated a few times before.

If I draw a number out of a hat with values ranging from 1 to 100,000,000,000, drawing the number "444,444,444" is just as "random" as drawing "349,912,955".

 

[Hurkyl]

I just generated 100 random numbers from 0 to 1, and I got the following sequence:

1110010110001111001111111010111110000111110001111010011011111111001101010011001110110101100100111101
​

I couldn't believe it! The odds of getting that sequence are simply astronomical -- roughly one chance in 1,000,000,000,000,000,000,000,000,000,000.

After I saw the 99th digit, I was certain the last one had to be a zero -- after all, it was just so incredibly unlikely it would have been a 1 to give me the above sequence. But there you go, it was a 1 anyways!

 

[Volkl]

I'll bet you that there are more cases of single blacks than cases of two blacks in a row. Would you take that bet?

 

[Fuz]

I'll bet you that there are more cases of single blacks than cases of two blacks in a row. Would you take that bet?

Yes, I'll take that bet, because that is equivalent to saying RBR will show up more often than simply BB.

 

[Volkl]

That sequence of a hundred, displays the tendency for the groups of consecutive like numbers to be small. Whatever the reason, it means that a change is not only inevitable, but most likely frequent. Thanks for displaying it!

 

[Volkl]

You took that bet even after the post of the 100 random numbers? -you loose - now pay up ;)

 

[Fuz]

You took that bet even after the post of the 100 random numbers? -you loose - now pay up ;)

Explain please...

 

[micromass]

You took that bet even after the post of the 100 random numbers? -you loose - now pay up ;)

If Black = 1, then I counted 55 non-isolated blacks and 8 isolated blacks. I think you need to pay.

 

[Volkl]

Yes, I'll take that bet, because that is equivalent to saying RBR will show up more often than simply BB.

Imagine a longer set? You would be even more incorrect as the length approaches infinity.

 

[micromass]

Imagine a longer set? You would be even more incorrect as the length approaches infinity.

Quite the contrary.

Please post proof of your statements: theoretical or experimental.

 

[Volkl]

If Black = 1, then I counted 55 non-isolated blacks and 8 isolated blacks. I think you need to pay.

Three blacks in a row does not count as two combinations of two blacks in a row I.e. Only two blacks in a row count as two blacks in a row.

 

[micromass]

Question:

Would you believe that the following sequence of numbers is random:

010101010101010101010101010101010101010101010101010101010101010101010101010101

Why (not)?

 

[Fuz]

Imagine a longer set? You would be even more incorrect as the length approaches infinity.

The probability of drawing a red marble, then a black marble, then a red marble out of a bag is 1/8. The probability of drawing 2 consecutive black marbles out of a bag is 1/4. I was taught this when I took pre-algebra in 6th grade.

 

[Volkl]

Do you think there are just as many sets of 100 blacks in a row as there are individual blacks? Do you really need a simulation for this?

 

[micromass]

Do you think there are just as many sets of 100 blacks in a row as there are individual blacks?

Nobody is claiming this.

 

[Volkl]

The same concept holds true when comparing to 99, or 98, or 97...or 2. Doesn't it?

 

[micromass]

Please answer this:

Question:

Would you believe that the following sequence of numbers is random:

010101010101010101010101010101010101010101010101010101010101010101010101010101

Why (not)?

 

[Volkl]

Sure it is likely to be random because the values change, values tend to like doing that when things are random.

 

[paulfr]

No one has mentioned the Binomial Theorem and the Bernoulli Trial Formula.
For any finite sequence with known probability, these laws govern the outcomes.
I don't have time today to give a detailed example, but maybe someone can pick up the ball here for me.

Probability is fascinating partly because one single word or one misinterpretation can completely change a problem solution from correct to incorrect.

And like all Fallacies, The Gambler's Fallacy "seems to be true" only because the language obscures the whole truth of the situation.

 

[micromass]

So you believe that a sequence of 500000000 alternating occurences of 01010101 could occur in nature?? Wow...

 

[Fuz]

Sure it is likely to be random because the values change, values tend to like doing that when things are random.

Why? That's not at all what true randomness is.

Go here: http://www.random.org/cgi-bin/randbyte?nbytes=20&format=b and hit refresh a bunch of times.