Request for you

[Frank]

(09-20-2009 05:22 PM)mario Wrote:  

Quote: If you are able to have a pairs for strictly 4 drawselections it's not "wonderful pairs with one good number in 4 sessions!"- it's a miracle-but I have my experience and know that probably it das not happen too often or maybe not any (100%) of the pairs is sucsesfull in 4 session.

Why this is a miracle, a pair with a good one number in 4 session?
I calculated in this way with BB:

-use =HYPGEOMDIST(0,2,7,49) in Excel, the probability to hit 0 numbers from 2; 0.732142857, so 73% for a bad results!
- use =BINOMDIST(4,4,0.73,FALSE) for pobablility to repeat for 4 consecutive sessions this bad results; 0.287331552, so 28% to wait 4 sessions and don't obtain any good number for that pair!
- so 100-28=71% to have a GOOD RESULT!

It's not a miracle I think. Any pair has a 71% chance to give us a good number in 4 sessions!

And the chances increase: for 8 sessions 91%, for 16 sessions 99%; so any pair in 16 sessions can give us a good numbers; even so exist pair that grow up 20 draws, but its fall in 1%, very, very bad luck!

What do you think?

Then exist some strategy, we don't need to play any pair! So the chances increase!

Perhaps Ido must begin another tread called 49s? I want to post some pairs with good results for that game and to explain how I obtained its.

Im familiar with using Excel for functions like this, especially the HYPGEOMDIST function, but the way you use it to decide that there is only a 73% chance of a bad result of a named pair doesn't add up.

Had you said this:- =HYPGEOMDIST(2,2,7,49) in Excel, for the probability to hit 2 numbers from 2 you would have got 0.017857143 which is your 1 in 56 odds. You then infer from this that the probability of not matching 2 balls is 1- 0.017857143 equals 0.982142857 or 98% ! You can then correctly use the 0.982142857 in the BINOMDIST function to get the probability of a non matching pair in 4 consecutive draws = 0.93046 = 93%. By using different values of the coefficients in the BINOMDIST function one can plot multiple curves for the different possibilities of both NOT matching 2 balls in N draws, and the probabilty of getting one, two or three such matches of pairs over time.



In the above chart, the black curve shows the decreasing chances of NOT matching a pair as the number of tries increases. The 4 tries example is at the left hand side up at 93%. Obviously this curve falls the more tries you have , you are more likely to succeed sometime.

Notice that the green curve which is the probability of having just one match in the given time -rises initially to a maximum value after 56 draws (familiar figure ?) and here the curve crosses the black curve where the probability of matching a pair ( once) equals the probability of NOT matching a pair at all in the given number of draws. After this point the prob of matching the pair just once over the time falls because the chance of matching it twice or 3 times increases to replace it. Ominously, the blue curve rising as we increase the number of tries is the probability of ANY success at all - matching the pair once , twice or more only gets to certaintly levels after about 300 draws ! So yes it would be a miracle to do it in just 4.

Out of interest, the same curves for 6/49 are shown (ignore the bonus or booster ball) below.



Its essentially the same, but the relevant figure is 78 , I'm sure you know why.


In the real world in the UK lotto, this chart below (for consecutive pairs only) illustrates how long you have to wait for pairs in a 6/49 lotto. 30% of pairs (287 of 833) turn up before youve waited 30 draws, 55% (287 +188 of 833) turn up before you've waited 60 draws etc (exactly as per the blue curve above).




It resembles the black curve from above, which is no coincidence as it follows the same law. OK thats enough, I think you get the picture