I Keep Getting Different Answers

I've edited this and added details to the question below to make it clearer.

I've tried this several different ways with different results.

In Blackjack, Assuming a 1% House Advantage....
--What is the probability of a 10 hand losing streak in 30 hands
--Followed immediately by a 10 hand losing streak in the next 120 hands
--Followed immediately by 9 hand losing streak in the next 240 hands.

Appreciate any help..............I have been using the Streak Calculator on Beating Bonuses but when I calculate it myself I get a different answer.

Using the Streak Calculator and multiplying chances I get 1 in 8192. Building my own probablility list, multiplying times the number of events and then multiplying the results of those 3, I get 1 in 299.

The following is REQUIRED for your calculations; irrespective of voodoo nonsense like “streaks” is an exercise in self-destructive futility.

• On average your chances of winning a hand in BJ is 43%
  
• On average your chances of losing a hand in BJ is 48%
  
• On average your chances of pushing a hand in BJ is 9%

Thank you.

But, I am interested in how to get the math correct multiple streaks expected in multiple hands.

Personally, I use Win 42.4%., Lose 49.1% and push 8.5% in my calculations which converts to a 53.7% win and 46.3% lose if you ignore pushes.

Below is my Excel chart for this. I am interested in how to calculate the probability of losing 10 in a row (Green) within a 30 hand cluster followed immediately by losing 10 in a row (Green) within the 120 Hand cluster immediately following the 120 hand cluster, followed by losing 9 hands in a row (Blue) in the 240 hands immediately following the 120 hand cluster.

Does this question now make better sense?



Win Lose
53.7% 46.3%



Streak / Win % / Probability / 1 in X
1 /53.70% /0.2884 /3.5
2 /53.70% /0.1549 /6.5
3 /53.70% /0.0832 /12.0
4 /53.70% /0.0447 /22.4
5 /53.70% /0.0240 /41.7
6 /53.70% /0.0129 /77.7
7 /53.70% /0.0069 /144.6
8 /53.70% /0.0037 /269.3
9 /53.70% /0.0020 /501.5
10 /53.70% /0.0011 /933.9
11 /53.70% /0.0006 /1739.0
















t

-- 10 hand losing streak within 30 hands:
there's 21 ways this can occur:

lose hands 1 through 10,
lose hands 2 through 11,
lose hands 3 through 12,
...
lose hands 20 through 29,
lose hands 21 through 30.

lose rate = 49.1%
let x = the chance of losing 10 hands in 10 hands: .491^10
so what we want is 21x. which is .0171

-- lose 10 consecutive hands in 120 hands:
ok same deal but we want 111x. which is 0.0904

-- lose 9 consecutive hands in 240 hands:
let y = the chance of losing 9 hands in 9 hands: .491^9
so, 232y. which is 0.385


for the combined chance of these 3 events occuring we multiply them and get .000595 (or .0595%). convert that to a fraction and we get 1 in 1681.

please correct me if i am wrong - i am not very good with statistics

i don't see why you would use 46.3%...a "10 hand losing streak" seems to indicate just that - losing 10 consecutive hands

muppet,

Thanks. Your number is close to what I thought the probability should be but I can't seem to get there with any confidence.

Your concept is pretty good, but I think more has to be done to convert from 1 in 933 to x in 30 by dividing 30/933 which then becomes 3.2%.

I think you are correct that then multiply each of the three results by each other to get the probability or it all occurring together.

As to the 46.3%l although that is the chance of me winning a hand. My chance of losing the hand is, correspondingly 53.7% So I have a chance of 53.7% of losing the first hand and to get the chance of losing two hands in a row multiply 53.7% by 53.7% etc. etc. etc.

I found an error in my math and I think when I redo it tomorrow, when my brain is fresh, I will actually come fairly close to your result. I will let you know.

But, I can't figure out how beating bonuses.com sites calculator is coming up with their number. I only wish that were true.

Thanks again for the response. Got my brain running again.

Thanks

If you are looking at for example a streak of 20 hands in 1000 (the example used on beatingbonuses.com), you have 981 possible strings of 20 hands. However, keep in mind that these are not independent. I don't know what math beatingbonuses is using, but that might be part of the discrepancy. If you have a 20 hand streak beginning on hand 100, another beginning on hand 101, and another beginning on hand 102, would you count that as 3 separate 20 hand streaks? Or just one 22 hand streak? By treating each of the 981 strings of 20 hands as independent, you would have to consider that 3 streaks, when really is is one that is slightly longer. Just a thought.

The question you asked was concerning the probability of it happening withing a given number of hands, and not hands that result in a decision.
That said; why on earth would you ignore pushes?

This is more complicated than I thought. Using the above chart I can see I have an error right off the bat.

The first row is incorrect. I will need to start out with the odds of losing once, the first string, as 53.7 and change everything. When I do that the chance of streak of 10 is .2% or 1 in 502.

However, when I go to the calculator at http://www.beatingbonuses.com/calc_streak.htm
and enter 501 in the Hands square, 10 in the streak square and .537 in the probability square I get an answer of 36.9% probability.

So, I did a little more homework and went to (Dead link: http://www.allbusiness.com/health-care-social-assistance/social-assistance-individual/236667-1.html)
and they say, about a third down, I guess I am going to have to table this unless somebody else out there has a clear way to do this
Thanks.

The paragraph that you mention is an incorrect statement. I can barely even begin to count Peter Goodrichs' errors in this article. It just goes to show how much you can trust things you read on the internet.

This is not necessary. Muppets' answer is 100% correct, and it cannot be made any more clearly.

muppet & Sucker

I looked at muppets work with a clear brain in the morning and a cup of coffee. Works for me.

I was using this to compare against my actual results on series of Online games which I thought was giving me way to many losing streaks. Turns out I am only about 20% off expected after about 5000 hands and some sloppy recordkeeping on my part. Acceptable.

No idea, though, how Beating Bonuses Calculator comes up with such different numbers.

Thanks

eh. ok so the events are not independent but is my math correct?

3:45-8:15

Agree so far.

Close. What are the chances of at least one 10 hand losing streak within a sample of 2000 hands? It isn't 1991x=1.62=162%. The calculation you did gives us the expected number of 10 hand losing streaks in 30 hands. The chance of having a 10 hand losing streak in 30 hands would be
1-(1-x)^21 = 0.0170

This is very close to your answer because for this particular example, your chances of one of those streaks within 30 hands is low, so your chances of multiple of those streaks is lower still.

For the same reason, this becomes 1-(1-x)^111=0.0865

And this becomes 1-(1-y)^232=.320 (as you can see, over larger samples, the numbers are moving somewhat lower than what you calculated)


With the corrected numbers above, this becomes 0.000471, or 1 in 2125.



Now...back to that independence thing...

If we take a huge sample (or even a small sample, actually), your calculation will correctly tell us the expected number of 10 hand losing streaks. For example, if we look at a 1,000,000 hand sample, we expect to see 999,991x=814 streaks. And so long as you evaluate each possible string of 10 hands as independent strings, that number is correct. What I mean by that is you must treat a 12 hand streak as a 10 hand streak, followed by a 10 hand streak (starting on the next card), followed by a third 10 hand streak. Following this procedure, you should see that your estimation of the expected number of streaks is good.

The independence thing rears its ugly head when we try to calculate the odds of seeing such a streak (not the expected number) within any given sample size. This is because your streaks will be "clumped" to some degree. If we know that a particular hand is the beginning of a 10 hand losing streak, then the very next hand has a 49.1% chance of being the beginning of another 10 hand losing streak (we know the first nine hands are losses, so we only care about the last hand). That's a lot more likely than 0.491^10.

On the other hand, if a particular hand is a win or a push, you have just eliminated 10 possible hands from contention for being be beginning of a 10 hand losing streak. This causes longer times between streaks than you would ordinarily expect. This longer time between streaks, followed by streaks that are clumped together, results in the correct number of streaks overall but they are not evenly distributed throughout our 1,000,000 hand sample.

I believe that is why the referenced website shows a lower probability of a particular losing streak within a given hand sample size. I don't know the math to calculate the correct odds given the dependence problem, so I would approach it with a simple simulation.

For another look at this topic, please watch this video. The particular part related to this discussion is found from about 3:45 to 8:15, but if you've got 21 minutes, it's worth watching the entire video.

Ok guys, part of the problem is that I did not adequately define string so we get the independence problem

What I meant was a string starting anywhere in the defined fields of 30, 120 & 240 respectively.

Assuming the probablility of losing is 49.1% (.0491) for all 3 problems

#1
Occurrence of a string of 10 losses in within any 30 hands is 49^10 which is .0008 or 1 in 1228. Since we do not have to worry about where the strings start and stop because they can start anywhere in the 30 hands there are 30 possibilities. So, this is 1228/30 or .0244 (1 in 40.9).

#2
Occurrence of a string of 10 losses in any 120 hands is 1228/120 or .0972 (1 in 10.2)

#3
Occurance of a string of 9 losses in any 240 hands is 49^9 or .0017 or 1 in 602.9. 602/240 is .399 or 1 in 2.5. Again, we worry only that the string started in the 240 hands so our possibilities are 240.

Combining the 3 consecutive events
Now since I said the above 3 events occurred consecutively in 1,2,3 order then the possibility of these events occurring consecutively is .0244 x .0972 x .399 = .000952 which is equivalent to a 1 in 1051 that the above 3 events would occur in consecutive order.

The difference between this and other results is due to the fact that there or more string start-points because it is only necessary for the string to start inside the pocket of hands. More strings increases the probability.

Interested in what y'all have to say about the above. Process looks correct to me but result was a lower probability than I guesstimated. Plus, sample data I am using, with some recording errors, is about 5,500 hands and the farthest I have gotten into the scenario is 7 or 8 losses in #2. If the chances are 1 in 1050, we are starting to get at least a couple toes into the Twilight Zone.

Incidentally, I ran the above events on the Beating Bonuses calculator and came up with probabilities of .09, .046, .092 which multiplied together equals 1 in 26,254. I checked their calculator for winning two hands in a row at 50% win/lose and came up with a 1 in 4 possibility.

Cheers!

The independence problem remains so long as we are talking about the frequency of a given streak length and the strings overlap. We can't change the fact that the strings will overlap, so we still have to deal with the independence problem.

This does not change much, really.

This is not correct, as I stated in my previous post. You are calculating the expected number of streaks that will begin somewhere within 30 hands, not the odds of having such a streak.

Ditto for these two calculations.

You are correct to multiply here, but again the numbers you are multiplying are incorrect.

Please reread my above post and change the numbers for the new number of starting points. You'll get a different number, but honestly it still isn't right because of the dependence problem, as I stated before.

As I stated before, the probability is even lower than you would calculate because the "trials" are not independent. The number of streaks will be as expected, but the probability of a streak will be lower within any given sample size. This is true because the streaks are "clumped" together (I hate using that word, but I don't know one better). That means if you do get one, you are more likely to get more than one together. I don't know if the Beating Bonuses calculator takes into account this effect, but the best way to test this yourself is with a simple simulation, whether coded in a simple C++ program or even just done in Excel (since we are talking about a small number of hands). I don't know how big the dependence effect is, but it is definitely there.

Guess I'm having some trouble getting my mind around the independence concept. Let me re-read your postings and chew on it.

As I alluded to in my post there must be more to it than I calculated, if only because my actual results differ so radically from the calculated.

Thanks for your patience.

thanks for the correction nynefingers. i'll need to look at your posts more closely when i have more time

Ditto. I stand corrected also. I have only one question. How did you lose your finger?

LOL i was seriously thinking that same thought when i was writing my last post

What makes you think he doesn't have nine on one hand?