Mathematics (Jackpot) - Number drawn X times in last Y draws

[lottoarchitect]

TRANSFERED FROM OLD FORUM

Here is another equation to determine the chance to have a number appear X times in the last Y draws of a lottery game (Pick 5/6/7). This uses the Bernoulli formula which I'll describe here. Again, assume we have a game a of b (eg 5 of 45, a=5, b=45) and we want to find out the chance to have a number to appear x times in the last y draws (obviously 0<=x<=y).

First, we have to determine the chance p to have a number appear in a draw. This is equivalent to find how many combinations contain our number among the whole possible combinations.
The combinations that contain a particular number are C(b-1,a-1) - the combination formula. In our example of a=5, b=45 we have exactly 135751 combinations from a total of C(45,5)=1221759 combinations. So, p=C(44,4)/C(45,5)=135751/1221759=0,111.

Now, we want to find out the chance P to have a particular number drawn x times in the last y draws.
The Bernoulli formula suggests that the result is
P=C(y,x) * (p^x) * [(1-p)^(y-x)] where k^l is the power of k to l.

An example: a=5, b=45 (game 5 of 45). What is the chance of a number to appear x=3 times in the last y=5 draws?
We've determined p=0.111 before.
So, the chance is C(y,x) * p^x * (1-p)^(y-x) =
= C(5,3) * 0.111^3 * 0.888^(5-3) =
= 10 * 0.00136 * 0.78854 = 0.0108 or 1.08% approx.

As you can see, this rate is rather low, so we can "safely" remove any numbers that have appeared 3 or more times in the last 5 draws.
Note that this rate is to have a number appear 3 times in the last 5 draws, which means that if we have a number appeared 2 times in the last 4 draws, then the chance to appear in the next draw is 1.08% (and therefore have the number appear 3 times in the last 5 draws).

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The following tables are the results of the above equation.

Horizontal axis - X parameter
Vertical axis - Y parameter

Game type 5 of 45 (a=5, b=45)
p=0,111111111

Code: Select all

      0       1       2       3       4       5
01  88,89%  11,11%  ------  ------  ------  ------
02  79,01%  19,75%  01,23%  ------  ------  ------
03  70,23%  26,34%  03,29%  00,14%  ------  ------
04  62,43%  31,21%  05,85%  00,49%  00,02%  ------
05  55,49%  34,68%  08,67%  01,08%  00,07%  00,00%
06  49,33%  37,00%  11,56%  01,93%  00,18%  00,01%
07  43,85%  38,37%  14,39%  03,00%  00,37%  00,03%
08  38,97%  38,97%  17,05%  04,26%  00,67%  00,07%
09  34,64%  38,97%  19,49%  05,68%  01,07%  00,13%
10  30,79%  38,49%  21,65%  07,22%  01,58%  00,24%

Game type 6 of 40 (a=6, b=40)
p=0,15

Code: Select all

      0       1       2       3       4       5       6
01  85,00%  15,00%  ------  ------  ------  ------  ------
02  72,25%  25,50%  02,25%  ------  ------  ------  ------
03  61,41%  32,51%  05,74%  00,34%  ------  ------  ------
04  52,20%  36,85%  09,75%  01,15%  00,05%  ------  ------
05  44,37%  39,15%  13,82%  02,44%  00,22%  00,01%  ------
06  37,71%  39,93%  17,62%  04,15%  00,55%  00,04%  00,00%
07  32,06%  39,60%  20,97%  06,17%  01,09%  00,12%  00,01%
08  27,25%  38,47%  23,76%  08,39%  01,85%  00,26%  00,02%
09  23,16%  36,79%  25,97%  10,69%  02,83%  00,50%  00,06%
10  19,69%  34,74%  27,59%  12,98%  04,01%  00,85%  00,12%

Game type 6 of 49 (a=6, b=49)
p=0,12244898

Code: Select all

      0       1       2       3       4       5       6
01  87,76%  12,24%  ------  ------  ------  ------  ------
02  77,01%  21,49%  01,50%  ------  ------  ------  ------
03  67,58%  28,29%  03,95%  00,18%  ------  ------  ------
04  59,30%  33,10%  06,93%  00,64%  00,02%  ------  ------
05  52,04%  36,31%  10,13%  01,41%  00,10%  00,00%  ------
06  45,67%  38,24%  13,34%  02,48%  00,26%  00,01%  00,00%
07  40,08%  39,15%  16,39%  03,81%  00,53%  00,04%  00,00%
08  35,17%  39,26%  19,17%  05,35%  00,93%  00,10%  00,01%
09  30,86%  38,76%  21,63%  07,04%  01,47%  00,21%  00,02%
10  27,08%  37,79%  23,73%  08,83%  02,16%  00,36%  00,04%

Game type 6 of 53 (a=6, b=53)
p=0,113207547

Code: Select all

      0       1       2       3       4       5       6
01  88,68%  11,32%  ------  ------  ------  ------  ------
02  78,64%  20,08%  01,28%  ------  ------  ------  ------
03  69,74%  26,71%  03,41%  00,15%  ------  ------  ------
04  61,84%  31,58%  06,05%  00,51%  00,02%  ------  ------
05  54,84%  35,01%  08,94%  01,14%  00,07%  00,00%  ------
06  48,63%  37,25%  11,89%  02,02%  00,19%  00,01%  00,00%
07  43,13%  38,54%  14,76%  03,14%  00,40%  00,03%  00,00%
08  38,25%  39,06%  17,45%  04,46%  00,71%  00,07%  00,00%
09  33,92%  38,97%  19,90%  05,93%  01,13%  00,14%  00,01%
10  30,08%  38,39%  22,06%  07,51%  01,68%  00,26%  00,03%

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Lotto Architect,

Thanks for these tables they are a great resource and aid in determining numbers to play - or not to play.

I think I prefer this statistical approach rather than relying on some mystical, psychic or divine prediction.

draughtsman

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Mr Lotto Architect,

The Bernoulli tables in this thread are a great resource in assisting one decide on which numbers might be avoided at the next draw where Bernoulli's probabilities are based on occurrences so far and the chance of a nunber occurring at the next draw if it has already had some recent occurrences. However I am wondering if these Bernoulli tables might be able to be prepared in the reverse so that they might express a probability for a number to occurr at the next draw based on that numbers non occurrence over the past few draws. Given that I quite confidently reject a number if its chance of occurrence at the next draw is at 1% or less I presume I could choose to include a number if Bernoulli suggestsed its chance of occurence was up near or in excess of 99%. Would such a table be possible please? Or perhaps I should ask first can such a table be prepared?

regards

relowe

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The tables already describe this. You have simply to observe the 1 column. To identify the chance of a number not to appear in the next draw (given it has not appeared yet), read the value at column 0. To identify the chance the number to appear at the next draw, given it hasn't appeared in the previous draws, read the value in column 1.
What might look stange is that the rates are in decreasing order after a certain amount of draws (lower bound for 0/1 columns) because statistically a number owes to appear a "certain" amount of times in a given range of draws and this is reflected in the Bernoulli tables. Thus, we cannot really expect to have a number not to appear at all in the last eg 30 draws and thus the chance for this not to appear again in the next draw is very low (the number normally should have appeared already sometimes at previous draws). But, as luck controls everything, we cannot exclude that too. This is why we'll never have an 100% or 0% for an event to occur.
In the above tables, the most dominant areas in a 5/45 game for a number to occur (given it hasn't appeared in previous draws) is between 6 & 10 draws later (rates around 37-38% in column 1).
Perhaps you would expect a table that says eg if we haven't a success till draw eg 30, then the draw 31 should have a huge % to occur (as it has to occur sometime of course!). I'll have a look at this property and if I come out with something useful, I'll post it. There is no distribution that describes what you ask for. Even the geometric distribution, shows the chance to have a number occur after n draws of non-hit. This follows the Bernoulli tables above as well.

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Mr LA Man,

Thanks for your time on this - so easy when you know how Those Bernoulli tables now take on even more meaning. And certainly appreciated if you find any other statistical clues on numbers to come at the next draw.

relowe

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Hi Mr LA,

Could I trouble you please for an extra Bernoulli table - for a game type of 6 from 45.

Thanks

relowe

---

relowe, you can use the formula above to find any particular game you want. Even better, you can use Excel as a template to generate quickly Bernoulli tables for any type of game you want. If I have some time I'll provide those tables you ask but unfortunately at the moment I lack free time.

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Mr LA.

Thanks for the challenge. Have included below a Bernoulli table for a 6 from 45 type lottery; I believe it to be correct. Perhaps any other more competent mathematician than I (and Excel exponent) might care to verify this.

Bernoulli Table

Lottery Type 6 of 45 (a=6, b=45) where p = 0.133333

Chance of a ball appearing X times (horizontal) in Y draws (vertical).

Code: Select all

......0......1......2......3......4......5......6..

01 86.67% 13.33% ------ ------ ------ ------ ------
02 75.11% 23.11% 01.78% ------ ------ ------ ------
03 65.10% 30.04% 04.62% 00.24% ------ ------ ------
04 56.42% 34.72% 08.01% 00.82% 00.03% ------ ------
05 48.89% 37.61% 11.57% 01.78% 00.14% 00.00% ------
06 42.38% 39.12% 15.04% 03.09% 00.36% 00.02% 00.00%
07 36.73% 39.55% 18.25% 04.68% 00.72% 00.07% 00.00%
08 31.83% 39.17% 21.09% 06.49% 01.25% 00.15% 00.01%
09 27.58% 38.19% 23.50% 08.44% 01.95% 00.30% 00.03%
10 23.91% 36.78% 25.46% 10.45% 02.81% 00.52% 00.07%

Use this table such that if a ball has been drawn 2 times in the last
3 draws what is the chance of it being drawn 3 times in 4 draws (for example).
Or for any other draw occurrence and draw history ie if 4 times in 8 draws what
is the chance of 5 times in 9 draws? My choice is to avoid numbers where the chance is about 1% or less. Others may consider avoiding numbers where the chance of appearance may be say upto 5%. These tables have been quite accurate for me although I did have a recent occurrence where 1 number appeared 5 times in 7 draws. I think even Mr Bernoulli would describe that as unusual.

(Apologies on the table formatting - the column numbers just won't line up without the dots.)

And note that in Lotto Architect the user can create Draw Filter algorithms to allow study of numbers drawn for any number of requisite past draws - other than the draw filters pre-provided with the program.

regards

relowe

---

Hi Mr LA Man,

Can you confirm my thinking on the use of these Bernoulli tables please.

My usual use of these tables is to observe for balls to avoid ie what is the chance of appearance at the next draw after the ball has already appeared so many times in the last x draws. (And if it is below say 1% I avoid such a ball.) However if I observe that a ball has not been drawn for say 6 draws (in a 6 from 45 type lottery) the odds of it being drawn at the next draw ie the potential 7th draw of delay for the ball, are at some 39.55% (read from the zero times draw column at 6 draws and observe in the 1 times drawn column at draw 7 = 39.55%) Now given that a single ball in such a lottery has a chance of appearance of 13.3% does this mean that if it has not been drawn in 6 draws it has virtually three times the odds of appearing at the 7th draw?

Am I interpreting this correctly?

Many thanks

And incidentally thanks for the run down in another thread on your proposals for version 2.3 of Lotto Architect. It looks substantial, comprehensive and terrific. Can't wait.

relowe

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This is the exact interpretation problem I mentioned in the previous post.
The Bernoulli tables show the following: "What is the chance of a number to appear x times in the last y draws".

Thus, this comment you mention

My usual use of these tables is to observe for balls to avoid ie what is the chance of appearance at the next draw after the ball has already appeared so many times in the last x draws. (And if it is below say 1% I avoid such a ball.)

has an error in logic. What this 1% shows here (we are in decreasing ratio in that column) is a rather good indication that we should expect the number to appear actually. Just observe for example, the column 0. It starts with a high ratio and gradually, reduces to almost 0%, after e.g. 20 draws. This is totally normal, as we obviously expect the number to appear at least sometimes within the last 20 draws, and thus this is why we have such a low rate at 20 draws. Obviously, if we haven't a number in the last 20 draws, then on 21 draw (in column 0) we have an even smaller ratio, closer to 0%. That simply means, the chance for this number NOT to appear again (column 0) is almost 0%!!! In this situation, if you notice the column 1, it still has a low rate (but higher than column 0). This is still normal as statistically the number should have appeared not only once but more times. I would consider this number for inclusion for all the following draws actually. We do not remove this number, we include this number in our selection as it definitely owes to appear! These are extreme cases to observe but if we are in this position, very low %, then we definitely expect the number to appear (thus we move to the next higher columns to keep up with statistics results). The usual transition from one column to the next one is expected to occur at the highest ratios displayed on the next column. This is the comment I made to observe the column x+1, at position y+1 if the number has appeared x times in the last y draws.
On the other hand, if we face 1% or less, whilst in that column we are in increasing order, then we can easily ommit this number from our selection. Thus, take care of interpreting this 1% limit. Interpretation works differently, depending on if we are in increasing order or decreasing order. In decreasing order, we include the number, in increasing, we can ommit it. Probably this is my fault as I din't mention the difference of interpretaion of increasing/decreasing order in the initial posts.

Now, this comment

Now given that a single ball in such a lottery has a chance of appearance of 13.3% ...

refers to percentage p, used by the Bernoulli tables. Thus, comparing these ratios is not valid, as they show different things. They do not conflict with each other of course! Conclusion is that we do not have three times the odds of appearing at the 7th draw in your example.

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Mr LA,

Thanks for your detailed reply. Clearly I have not been using the tables as well as I could have, but I now expect to be able do so. Not too difficult to actually produce the table, but using it properly....

Thanks again - appreciated

relowe

[Crocodile23]

An example: a=5, b=45 (game 5 of 45). What is the chance of a number to appear x=3 times in the last y=5 draws?
We've determined p=0.111 before.
So, the chance is C(y,x) * p^x * (1-p)^(y-x) =
= C(5,3) * 0.111^3 * 0.888^(5-3) =
= 10 * 0.00136 * 0.78854 = 0.0108 or 1.08% approx.

As you can see, this rate is rather low, so we can "safely" remove any numbers that have appeared 3 or more times in the last 5 draws.

"Safely" remove?
What "safely" means?
Does it mean that maybe( ) no matter what we would do our probabilities to achieve any goal we have remains the same?
That is with other words: History don't counts!

Well theoretically of course and history doesn't count at all! But what is happening in practice and also not in the long run but in our short life?
In practice means: Does exist a true randomness in the lotteries of the games? I guess not, so history may count. It may not of course but the fact it may count, gives us hopes that we can increase with all these statistics and systems, our probabilities to win. Fake hopes? Maybe but not for sure.....
In our short life means, although if 25 e.g appears 4 times in a row, then it has the exact same chance to appear in the next lot. But inside a huge number of times(drawings) for example for 8000 years this situation is common that is 25 to appear 5 times in a row, but in our short life it's much more rare so we have good hopes that it will not happen indeed.

[lottoarchitect]

Hi Crocodile23,

"safely" means that we can use this information and we expect this to be 99% correct. If we are so unlucky and fall in the 1% case, we can blame luck or the neighbor. If however you want to bet on the 1% and include this, it is up to you. Would you bet on a number that you know it has a chance of 1% to appear? This is a mathematical equation and has nothing to do with fixed draws or whatever. Maths say that number would have 1% chance to occur. If the lottery is truly random, then it will indeed have 1% chance to occur on the previous example. If the lottery is not random, then we possibly can observe something else to use.
Its up to you to use any approach you believe could be suitable based on your beliefs. However, it is better to have something to use, even this equation instead of nothing and relying on pure luck.

cheers
lottoarchitect

[Crocodile23]

Sorry but i will be harsh. I have to defend Mathematics.

"safely" means that we can use this information and we expect this to be 99% correct. If we are so unlucky and fall in the 1% case, we can blame luck or the neighbor. If however you want to bet on the 1% and include this, it is up to you. Would you bet on a number that you know it has a chance of 1% to appear? This is a mathematical equation and has nothing to do with fixed draws or whatever. Maths say that number would have 1% chance to occur. If the lottery is truly random, then it will indeed have 1% chance to occur on the previous example. If the lottery is not random, then we possibly can observe something else to use.

All the above are a complete load of crap. Period.
Sorry but it's true. You might want to take a look at an elementary probabilities book.

Even if a specific number, let's say 12, appears 8 times in a row it still has the EXACT same probability to appear in the 9th draw with every other number let's say the 23.

It is true that if a specific number, let's say 12, appears 8 times in a row then the probability to appear 9 consecutive times, that is to appear also in the 9th draw(AND to have been appeared in the last 8 draws), is 1/9^9 = 1/387420489 = 0.00000026 %

But what many people make as a mistake, like you did, it to forget about the bold red parenthesis i've mentioned above.

So your:
"What is the chance of a number to appear x=3 times in the last y=5 draws? Answer: 1.08%
So if a number appears 3 times in the last 5 draws, then it has 1.08% to appear again in the next draw."


Is wrong and the reality is that:
•This number has the exact same probability with all other numbers to appear in the next draw. And the probability is 1/9 = 11.1 %.
•The 1.08% says that this is the probability of this number to appear in the next draw AND this number to have been appeared 3 times in the previous 5 draws.
The fact that we know this number appeared 3 times in the last 5 draws, does not change anything.

Its up to you to use any approach you believe could be suitable based on your beliefs. However, it is better to have something to use, even this equation instead of nothing and relying on pure luck.

Sorry but this is wrong again. We always are based on pure luck.
Wheels and better lower limits of wheels does not increase our probability to win, they just decrease the vigorish for the companies that offer lotto, so they decrease our loss, so we can bet more times with the same money. Just that!

[lottoarchitect]

I'm not into opening a war for mathematics. You are free to believe anything you want. As a matter of fact, you don't say something different to what I say. I never wrote anywhere that if you observe individually each test, that a number will have 1.08% to occur. Of course it will have the p probability. However, here we talk about the probability of obtaining exactly r events in n trials, which is what the Bernoulli formula describes and I displayed in the examples. Your red parenthesis is only a subset of what Bernoulli formula can calculate, or in simple words n-1 events in n trials.

Is wrong and the reality is that:
•This number has the exact same probability with all other numbers to appear in the next draw. And the probability is 1/9 = 11.1 %. <- This is the p
•The 1.08% says that this is the probability of this number to appear in the next draw AND this number to have been appeared 3 times in the previous 5 draws. <- This is what I have displayed here

You say the same thing I say. I have nothing more to add in this. I don't see where the 'violation' of mathematics has occurred.

lottoarchitect

[Numbers_Guru]

I know this is old but it is interesting.

Crocodile 23 is correct, no one number has a higher degree of probability for dropping than any other from a statistical perspective of all things being equal.

However I do not believe this to be the case. I recently ran some analysis on what I call number intensity. Essentially I was trying to see if specific numbers had an intensity cycle attached to them which effected how often they appeared / dropped.

What I found was a resounding NO. No intensity or probability build up that is predictable, but on the other hand it also seemed that if I picked the top 10 numbers with the highest intensity then I would often end up with 2 or 3 of them as winning numbers, mixed in with seemingly random results for numbers that where least due to hit or had lower probability. Unusual hey ? Spent the next 6 weeks trying to figure out if there was a way to predict which ones would "relieve intensity" vs. those that would hit at low intensity, all to no avail.

So whilst I could not find a prediction pattern based on intensity or probability increase I did discover a pattern attached to analysis of "past" numbers. Does this mean the same will hold true for future numbers? I am waiting and collecting data to make that determination.

[lottoarchitect]

The topic discusses "number drawn X times in last Y draws" and not "chance to have a specific number drawn". What crocodile indicates is correct of course but is irrelevant to interpreting percentages for the "number drawn X times in last Y draws" case which is a different thing.

[Numbers_Guru]

Hi Any and all,

Out of curiosity I ran the equations against past Lotto win results for a period of 20 years.

Also varied the depth of x and y magnitude to see what effect these varying limits would have on the remaining "unremoved" or included numbers.

Next was to check the number of winning number hits against remaining numbers...

In essence:

> By varying x and y you end up with a change in the number of remaining numbers to pick your "winning" numbers from.
> The increase in the depth of above variables results in the hit rate of actual winning numbers from the draw against the predicted remaining number dropping. So you get less strikes of actual winning numbers.
> The only way to get more winning number strikes to decrease the variables, it seems that each lotto game (Mon, Wed, Thurs, Sat etc.) has a goldilocks zone of x and y which differs for each one. That is a range within the x and y variables that gives the best range of available numbers in which the actual numbers hit.
> Big downer with the above point is that NO combination of x and y variables results in the remaining numbers containing all the actual winning numbers for any individual game.

Essentially you will never win using this technique, you will if you select the goldilocks zone of x and y values for a specific Lotto game increase your chances of minor 2, 3 and 4 number hits but in reducing the overall quantity of numbers being played you are also playing into random exclusion (I just made that up).

Random exclusion is what I would call exclusion of the possibility that every number has an equal chance of falling.

Now having said that I am going to contradict myself....

Even though each and every number appears to have an equal and fair chance of falling, you can still increase your number of hits of minor 2, 3 and 4 numbers using the above technique.

This would be how many lottery programs have testimony pages full of winners, they secure enough users playing enough increased odds and they are bound to have a higher rate of "winners" that they can parade across their web pages.

In summary, use this technique to improve potential to win smaller prize pools.

Do not expect that this technique will help you win the big one.

Last comment..... Most lottery programs can increase your potential of winning minor prizes as per above. However the shear fact of removing numbers or including certain numbers immediately removes them from you chances of winning.

My recommendation ... find the best numbers, the numbers with the highest hit combinations across time and then play these numbers consistently.

All care...

Numbers.

[suzanne]

I want to respond to what you wrote in the context of the method of choosing lottery numbers
Lotto This is not a method is a way of life I follow lottery for several years and purchased tickets online there are many sites that have Calculator's lotteries mean they take over 100 lotteries back and put all the numbers that come up and then you can see statistically whether your number will happen this time ... but I do not believe this method I follow stories that tell about the people and most of the winners when they filled their tickets they chose the quickest method it says that the computer choose the numbers for them no numbers lucky numbers and birthdays so I really believe in luck.
There are some really good web sites, but one which is basically a site that compares all the companies and collect all the data on the lottery and showing all results is
called: <--- MOD EDIT: Link removed. Links generally permitted only to official lottery commission sites, or to other sites if they aid discussion to support some argument. --->

Strongly recommends

[DC_Treybil]

Below are two tables. The first shows the odds of getting "n" or more consecutive hits with odds of 2 : 1. "n" is in the first column. The second table shows the actual values of powers of 20 to 221 next to the corresponding result from the first table. The two results track nicely up to 2 15. After that, they vary from slightly to considerably. This is the result of a sample size of 10,000,000. While odds < sample_size / 300 or so, it's all good. So that is a good rule of thumb for a viable sample size for such simulations.

Upshot: You don't have to dwell on these tables very long before you see that indeed, a "hot" number is just displaying normal random behaviour. Its odds of hitting again are unchanged by the number of time it has occurred consecutively.

Note: Won't you kindly pardon the less-than-perfect column alignment?

0 5,000,099 . . 5,000,099 10,000,000 . . . . . . . . .1.00
1 2,499,916 . . 7,500,015 .4,999,901 . . . . . . . . .2.00
2 1,249,877 . . 8,749,892 .2,499,985 . . . . . . . . 4.00
3 . 625,428 . . 9,375,320 .1,250,108 . . . . . . . . . 8.00
4 . 313,070 . . 9,688,390 . . 624,680 . . . . . . . . .16.01
5 . 155,913 . . 9,844,303 . . 311,610 . . . . . . . . .32.09
6 . . 77,933 . . 9,922,236 . . 155,697 . . . . . . . . .64.23
7 . . 38,986 . . 9,961,222 . . .77,764 . . . . . . . . 128.59
8 . . 19,371 . . 9,980,593 . . .38,778 . . . . . . . . 257.88
9 . . . 9,731 . . 9,990,324 . . .19,407 . . . . . . . . 515.28
10 . . . 4,841 . . 9,995,165 . . . .9,676 . . . . . . .1,033.48
11 . . . 2,413 . . 9,997,578 . . . .4,835 . . . . . . .2,068.25
12 . . . 1,219 . . 9,998,797 . . . .2,422 . . .4,128.82
13 . . . . .605 . . 9,999,402 . . . .1,203 . . .8,312.55
14 . . . . .299 . . 9,999,701 . . . . . 598 . .16,722.41
15 . . . . .164 . . 9,999,865 . . . . . 299 . .33,444.82
16 . . . . . .71 . . 9,999,936 . . . . . 135 . .74,074.07
17 . . . . . .33 . . 9,999,969 . . . . . . 64 . 156,250.00
18 . . . . . .18 . . 9,999,987 . . . . . . 31 . 322,580.65
19 . . . . . . .8 . . 9,999,995 . . . . . . 13 . 769,230.77
20 . . . . . . .3 . . 9,999,998 . . . . . . . 5 2,000,000.00
21 . . . . . . .2 . 10,000,000 . . . . . . . 2 5,000,000.00




0 . . . . . . 1.00 1.00
1 . . . . . . 2.00 2.00
2 . . . . . . 4.00 4.00
3 . . . . . . 8.00 8.00
4 . . . . . 16.00 16.01
5 . . . . . 32.00 32.09
6 . . . . . 64.00 64.23
7 . . . . 128.00 128.59
8 . . . . 256.00 257.88
9 . . . . 512.00 515.28
10 . . . 1,024.00 1,033.48
11 . . . 2,048.00 2,068.25
12 . . . 4,096.00 4,128.82
13 . . . 8,192.00 8,312.55
14 . . 16,384.00 16,722.41
15 . . 32,768.00 33,444.82
16 . . 65,536.00 74,074.07
17 . 131,072.00 156,250.00
18 . 262,144.00 322,580.65
19 . 524,288.00 769,230.77
20 1,048,576.00 2,000,000.00
21 2,097,152.00 5,000,000.00