Calculation Of Odds

[Shanga]

I have a question about how to calculate the odds for a 649 game, of getting 6 correct, in a set greater than 6. For example, if a set of 20 were selected for, how do you calculate the odds of getting 6 in that set of 20?

 

[Rob50]

In your example the probability of having all the six numbers in a set of 20 is C(20,6)/C(49,6), C(20,6) meaning the possible number of groups of 6 numbers over a set of 20 etc. A simple calculations gives a probability 0f ~3/1000 (exactly 0.0027717).

For different sets, say of N numbers, the probability would be:

C(N,6)/C(49/6).

You can calculate this easily using EXCEL worksheet function COMBIN(number, number_chosen).

 

[johnph77]

Concur.

There are 38,760 possible combinations in a 6/20 matrix, 13,983,816 in a 6/49 matrix.

Another way of calculation is by using factorials.

6/49: (49!/43!)/(6!) = (49x48x47x46x45x44)/(1x2x3x4x5x6) = 13,983,816

6/20: (20!/14!)/(6!) = (20x19x18x17x16x15)/(1x2x3x4x5x6) = 38,760.

 

[Shanga]

Tks for your help Rob50. That explains why from a mathematical standpoint, it is still so difficult to capture 6 from a selection of 20.

 

[First Prize]

In your example the probability of having all the six numbers in a set of 20 is C(20,6)/C(49,6), C(20,6) meaning the possible number of groups of 6 numbers over a set of 20 etc. A simple calculations gives a probability 0f ~3/1000 (exactly 0.0027717).

For different sets, say of N numbers, the probability would be:

C(N,6)/C(49/6).

You can calculate this easily using EXCEL worksheet function COMBIN(number, number_chosen).

Do you believe that maths can solve winning numbers.IF that is possible, then probably you could be a winner everydays.I don't think it possible.

 

[Rob50]

Splendid first post First Prize and welcome to this BB

My questions for you are:

1. Why do you believe that I believe that math can solve winning numbers?
(By the way I think that there are no ways, mathematical or not, that one can predict, I prefer this instead of “solve”, the lottery numbers, but I wanted to help answering Shanga’s question therefore I had to use some elementary math)
2. Do you know any way else to answer Shanga’s question totally ignoring math?
3. Do you really understand why the maths, as you call it, can’t solve the winning numbers?

And last, how much math do you understand other than 1+1 = 2?

 

[kosteczki]

heh you want to talk about difficult, from a group of 29 selected numbers we choose 20 random combinations. Not only did none of our combinations not hit anything out of the 29 numbers only 2 hit.

 

[Bertil]

calculation of odds

Concur.

There are 38,760 possible combinations in a 6/20 matrix, 13,983,816 in a 6/49 matrix.

Another way of calculation is by using factorials.

6/49: (49!/43!)/(6!) = (49x48x47x46x45x44)/(1x2x3x4x5x6) = 13,983,816

6/20: (20!/14!)/(6!) = (20x19x18x17x16x15)/(1x2x3x4x5x6) = 38,760.

johnph77 seems to know probability theory. Can you also
give me a formula for calculating the expected variance
for the mean frequency of the numbers drawn in any
lotto in terms of matrix n/N and total draws D? I've been
given a formula that does not fit the actual data for the
17511 total draws world wide of 6/49.

Stig Holmquist

 

[kosteczki]

Hey Stig, I was wondering if you are sure that's only 17511 draws world wide??? I was figuring about 9500+ just for all the Canadian 649's. I am sure the USA have a lot more and other countries do to.

But I may be wrong.

 

[Brad]

You're prob right Kosteczki, the Czechs alone have had a 6/49 running since 1957 with more than 4,500 draws to date.

So adding that to your Canuck estimate makes it about 14,000 draws just for those two countries ... surely there must be more than 17K draws worldwide ...

Cheers

 

[Bertil]

Czeck 6/49

You're prob right Kosteczki, the Czechs alone have had a 6/49 running since 1957 with more than 4,500 draws to date.

So adding that to your Canuck estimate makes it about 14,000 draws just for those two countries ... surely there must be more than 17K draws worldwide ...

Cheers

Brad, Please tell me where I might find a table of total
frequencies for each integer in the Czech 6/49 game.

Stig Holmquist

 

[Brad]

Re: Czech 6/49

Brad, Please tell me where I might find a table of total
frequencies for each integer in the Czech 6/49 game.

Stig Holmquist

Hi Stig, I don't have the table you ask for but here is the thread with a link to the game's history from day one.


Cheers

 

[johnph77]

Re: calculation of odds

johnph77 seems to know probability theory. Can you also
give me a formula for calculating the expected variance
for the mean frequency of the numbers drawn in any
lotto in terms of matrix n/N and total draws D? I've been
given a formula that does not fit the actual data for the
17511 total draws world wide of 6/49.

Stig Holmquist

***************

Redid the probability table for 6/49. Probability of any duplicate occuring before or at 17,511 draws is 0.0012529464, or 1::798.12. Probability for a duplicate occuring on Draw 17,512 is 0.0000000736.

 

[Bertil]

variance for frequencies of 6/49

Thank you john for the odds. But my primary interest is in the
variance for the mean frequency for any lotto n/N after D draws.

After we have the frequency we can convert it to chi-square
and determine if the actual value indicates a bias. But there
is a problem about the proper chi-square calculation. Are
you familiar with it?

Stig Holmquist

 

[johnph77]

stig holmquist -

There's a formula for it but I'm not sure what it is. The frequency tables on my site were originally calculated on a spreadsheet with progressive formulas based on occurances in smaller lotteries and are position-oriented. The number of occurances of each number has to be the same for any lottery except for bonus balls.

I believe the formula for the number of occurances of each number in a lottery is (nP/N), where n is the number of balls drawn, P is the number of total possibilities in the draw and N is the number of balls to be drawn. I haven't tested this formula, but it's logical, at least from where I view it. You would then take that result and divide it by D (draws to be analyzed) to determine how many times any possible number should appear.

Addition - that formula works for a 6/49 lottery.

gl

John

 

[W Kaenzig]

I am a lotto junkie and started to publish a book a few years ago,
but made mistakes and am correcting.
Ea single # repeats the same # of times. Ea pair of #'s repeat the same # of times. Ea triple # repeats the same # of times. Ea quad repeats the same # of times, and ea Quint repeats the same # of times.
The 6th # is occurs in quantity exactly opposite of the first #. The 5th # occurs in quantity exactly opposite the 2nd #. The 4th # occurs exactly opposite the 3rd #.
This is simple math and is not questionable. Odds for any combo can be calculated using simple math without ringing one's hands regarding odds of probability. W

 

[W Kaenzig]

occurances

In ALL games of 6/49 each nr appears 1,712,304 times.
(13,983,816x6) divided by 49. W