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[Buffster]

I'd like some opinions on this matter.


Let's say I went to the casino and played for 250 spins.


First off are 250 spins considered a Long Run seeing it's the maximum spins I saw that night.


Secondly would the results of the last 3 spins be considered a Short Run.


Thirdly would the Short Run results (3 spins) be useful in comparing to the Long Run results (250 spins).


I don't want any answers about the wheel not having any memory and that short or long don't matter.




Thanks in advance


B

[ezmark]

In general, the larger your sample size, the more correct your conclusion of the whole group.  The sample result will be a similar result of the group in total.  What is a proper sample size, I don't know you may want to google that.  A sample size to small may result in a false conclusion. 
I feel a sample size of 3 is to small for a group of 10.  Although, a sample size of 30,000 may be ok for a group of 100,000.
     

[Albalaha]

Buffster,
            Short run and long run are hypothesis and hard to describe in terms of roulette. If you are talking of outside bets, I can take it as "middle run" and not "long run" while for any inside bets it is pretty "short run".

[Number Six]

Here's what I think.

The long run can easily be defined as a point where, let's say in a simulation, any manifested edge of a bet selection stabilises. Whether the edge is negative or positive, doesn't really matter. Once you're into a large sample the edge deviates by smaller and smaller amounts, but the effect of that deviation on any supposed bankroll could get worse and worse up to the point of ruin.

Similarly the short run is any undefined set of outcomes (actually bets) in which the odds are not entirely representative of the long term expectation. You can define and prove its length yourself, and maybe even introduce a degree of predictability using maths or observation alone.

[HarryJ]

  Hi Bufster,
         LONG RUN and SHORT RUN are not mathematical terms and connot be accurately defined.

     I doubt any mathematician would consider 250 spins as sufficient to qualify for the LAW OF LARGE NUMBERS. 3 spins is too small a number to be considered at all.

       However your question does raise a point that is often overlooked. If those members who run tests of hundreds of thousands, even millions, of spins, were to break their results into a large number of very small segments; they would find that the average variations in the segments, whether selected from the beginning or the end, or in fact any where within the test, would remain roughly the same.( Sorry about that sentence. Like Topsy it just grew and grew!!)

       That tends to prove that random remains the same regardless of the number of spins or events. How then does the LAW OF LARGE NUMBERS work??? Or to put it another way, what causes the difference between the long and the short run??

      Regards          Harry

[Dragoner]

what causes the difference between the long and the short run??

"As the number of rounds increases, eventually, the expected loss will exceed the standard deviation, many times over. From the formula, we can see the standard deviation is proportional to the square root of the number of rounds played, while the expected loss is proportional to the number of rounds played. As the number of rounds increases, the expected loss increases at a much faster rate. This is why it is impossible for a gambler to win in the long term. It is the high ratio of short-term standard deviation to expected loss that fools gamblers into thinking that they can win."
Source

[Bayes]

We can quantify the law of large numbers and assert within some probability that the difference between the proportion of successes and the expectation is less than some given error. To clarify and put this on a more intuitive basis, suppose you bet on red for 100 spins, the result could be 40 reds giving a proportion of 40/100 = 0.4, or it could be 60 reds giving a proportion of 60/100 = 0.6. The law of large numbers says as you do more trials, the proportion of successes will converge towards the TRUE probability (or expectation) which in this case (assuming a single zero wheel), is 18/37 or 0.48649 to 5 significant figures. So when you get to 200 trials, the proportion is likely to be closer to 0.48649 than 0.4 or 0.6, and after 300 spins it's likely to be closer still.

But how likely? and how much closer?

It's possible to derive a formula which will give you the answer for any desired probability, error, and expectation. I won't go through the gory details, but just present the formula:

n = x(1 − x) / c²(1 − p)

Where n = the number of trials.
x = the expectation (or probability of a win).
c = the error.
p = the probability that the error will be within the required value.

Example using an EC bet: You want to know, with a probability of 99% (this is p), how many trials are necessary for the proportion of successes to be within 5% of the expectation.

Let's take the expectation to be as above - 0.48649. 5% of this is 0.05 ✕ 0.48649  = 0.024325. This is c in the formula - the error. It means that we want the proportion of successes to be not less than 0.48649 − 0.024325 = 0.46217, and not more than 0.48649 + 0.024325 = 0.51082.

Now plug into the formula to get the number of trials needed:

n = 0.48649 ✕ (1 − 0.48649) / 0.024325² ✕ (1 − 0.99) = 42,220 spins.

The smaller you want the error (c) to be, or the higher the probability required that the error be within this value (p), the more spins you will need.
It's interesting to compare this with the number of trials needed when betting a single number. Again, for c take plus or minus 5% of the "true" expectation which we'll assume to be 0.027027. Take the probability to be 99% as before. Then

c = 0.0013514
x = 0.027027
p = 0.99

n = 0.027027 ✕ (1 − 0.027027) / 0.0013514² ✕ (1 − 0.99) = 1,439,890 spins!!

That's more than 34 times as many spins as were needed for the ECs.

[HarryJ]

    Hi Dragoner,
           You have clearly misunderstod my question. I was referring to Bernoulli's theorem not De Moivre's.

     Let me put it another way.

    Within a Bernoulli system, each event is entirely independent. There can be no connection with what has gone befe or comes after. That being the case, what mechanism allows the 'Law of Large Numbers to accurately predict that the percentage of variance will decrease??? Surely it is more likely that the prcentage of variance would increase the larger the numbe of events??

     What forces random chance to conform to Bernoulli's law??

     Regards           Harry

[Bayes]

     What forces random chance to conform to Bernoulli's law??

   

Harry, it's just what happens if you run a large number of trials. It doesn't seem to make much sense to ask what "forces" random chance to conform to it. It's just a mathematical model. You might as well ask what "forces" a right-angled triangle to conform to Pythagoras' theorem.

Surely it is more likely that the prcentage of variance would increase the larger the numbe of events??

No, because as it says in the Wiki article, the RATE at which the expected loss occurs is higher than that of the SD as you increase the bets. The % of variance does increase, but not as fast as the losses.

[HarryJ]

    Hi Bayes,
                 Pytharoras' Theorem can be proved using other mathematical tools. Not bernoulli's.

      Your neat formulae give an approximate result, but they offer no guarantee.

     Random Chance is under no mathematical control yet it generally conforms within a calculatable margin of error. The answer 'It jut does' is hardly satisfactory.

    Regards      Harry

[Dragoner]

A single coin toss can be either heads(H) or tails(T) with equal probability. 50%-50%


2 coins tosses have 4 outcomes: HH,HT,TH,TT. Notice, that there are two "1T-1H" scenarios. So you have 25% for 2H another 25% for 2T and 50% for 1T-1H. Already seeing the higher probability there at the true probability.


3 tosses: HHH,HHT,HTH,HTT,THH,THT,TTH,TTT. The 8 scenarios have the same probability, but notice, there is only one with 3H and one with 3T, yet there are 3 with 1T-2H and another 3 with 2T-1H. This makes these results 3 times more probable than the others.


10 coin tosses...
There are 2^10 = 1024 scenarios. 252 of those give 5 heads and 5 tails. Yet only one of them will give you all heads and another one will give you all tails. So the 5H5T has 252 times the probability the 10T has.


All tosses are independent, yet the more we play we have a lot more scenarios with results closer to the true probability. It is due to the fact, that out of all the different scenarios a lot more gets you to more equal distributions.


This is why the binomial distribution looks the way it does. And this also explains why the longer we play the more likely we get closer to the true probability.

[Dragoner]

What I quoted from wikipedia earlier:
[attach=1]
I went with 18/37 win probability and 1 unit flat bet.
You can see that at only 10-20 spins EV is very close to 0 and the SD gives the illusion that you can win. Even at 600 spins the EV+SD is above 0, but you can already see, that it will go below 0.


And the reason why the SD isn't increasing fast enough is what I explained with coin tosses.

[Albalaha]

Very Good explanation, Dragoner.

[HarryJ]

   Gentlemen,
           I am well aware of the math and that it generally works. That doesn't answer my question.

    Once the variance has drifted off track, what prompts it to return?? Bear in mind that each event is ompletely independent and has no connection with what has gone before.

      In your 10 coin toss example, if the first 4 tosses all produced heads, the next 6 would have to produce 5 tails to restore equilibrium. An unlikely scenario yet we ve all seen it happen.

      In any long term test the variance can swing from one extreme to thother many times. What causes this search for equilibrium??  Maths can describe it, but can it explain it??

     Regards        Harry

[Dragoner]

In your 10 coin toss example, if the first 4 tosses all produced heads, the next 6 would have to produce 5 tails to restore equilibrium. An unlikely scenario yet we ve all seen it happen.

That's the thing, it doesn't restore equilibrium. That is what most people misunderstand about probability. The most likely scenario after 4 heads is for the remaining 6 to produce 3 heads and 3 tails. (Having a total of 7 heads and 3 tails)
You may think this renders what I wrote earlier wrong, but it doesn't.
If you think about the ratio of heads and tails after the 4 heads, it is 100% heads. Think about what another heads does to that ratio and what a tails does to it.