Request for you

[mario]

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(10-14-2009 08:21 PM)Frank Wrote:  Had you said this:- =HYPGEOMDIST(2,2,7,49) in Excel, for the probability to hit 2 numbers from 2


Frank not hit 2 from 2, only 1 for 2! just 1 number from a pair not both numbers!
And I said about the process of obtained probability in 2 steps using the contrare phenomen:
- =HYPGEOMDIST(0,2,7,49) in Excel not =HYPGEOMDIST(2,2,7,49) for hit 0 from 2 numbers if in 49 drawn 7 = 0.7321
- to repeat this phenomen in 4 consecutive draws is
=BINOMDIST(4,4,0.7321,FALSE)= 0.28

So for a favorable process have probability 1-0.28=0.72!! so 72% for a win (at least a number for pair)!

What is it wrong? You illustrated another situation with hit 2 from 2 so whole pair!
I don't need the whole pair, only one and alone number for pair!!

Frank, very interesting but you analyzed a pair or 2 from 2 and I analise 1 from 2 not all pair, only one number for it!! Where in this history of draws you must wait hundred of draws for a one good numbers from a pair???
So I think that probability is good? I'm wrong?

[Frank]

(10-15-2009 03:19 AM)mario Wrote:  :
Frank, very interesting but you analyzed a pair or 2 from 2 and I analise 1 from 2 not all pair, only one number for it!! Where in this history of draws you must wait hundred of draws for a one good numbers from a pair???
So I think that probability is good? I'm wrong?

The maths you used told me that you were looking at the probability of not matching 2 from 7 balls ! hence HYPGEOMDIST(0,2,7,49) where 2 is the number of balls you are intersted in and 0 is your success rate you are looking for. Nothing in that function tells the maths about looking for one of a pair ! Thats why it is wrong.

[mario]

(10-15-2009 03:46 AM)Frank Wrote:  

(10-15-2009 03:19 AM)mario Wrote:  :
Frank, very interesting but you analyzed a pair or 2 from 2 and I analise 1 from 2 not all pair, only one number for it!! Where in this history of draws you must wait hundred of draws for a one good numbers from a pair???
So I think that probability is good? I'm wrong?

The maths you used told me that you were looking at the probability of not matching 2 from 7 balls ! hence HYPGEOMDIST(0,2,7,49) where 2 is the number of balls you are intersted in and 0 is your success rate you are looking for. Nothing in that function tells the maths about looking for one of a pair ! Thats why it is wrong.

Yes, but I calculate for contrarely process, which is the probability that if choose a pair and hit 0 in 4 consecutive draws! This is situation for lost and contrarely for a win that it mean at least one number for pair = 1-0.28=0.72 or 72%!

I think that it may be calculate in this mode starting with a contrarely phenomen. Do you think that?

[Frank]

(10-15-2009 02:58 PM)mario Wrote:  

(10-15-2009 03:46 AM)Frank Wrote:  

(10-15-2009 03:19 AM)mario Wrote:  :
Frank, very interesting but you analyzed a pair or 2 from 2 and I analise 1 from 2 not all pair, only one number for it!! Where in this history of draws you must wait hundred of draws for a one good numbers from a pair???
So I think that probability is good? I'm wrong?

The maths you used told me that you were looking at the probability of not matching 2 from 7 balls ! hence HYPGEOMDIST(0,2,7,49) where 2 is the number of balls you are intersted in and 0 is your success rate you are looking for. Nothing in that function tells the maths about looking for one of a pair ! Thats why it is wrong.

Yes, but I calculate for contrarely process, which is the probability that if choose a pair and hit 0 in 4 consecutive draws! This is situation for lost and contrarely for a win that it mean at least one number for pair = 1-0.28=0.72 or 72%!

I think that it may be calculate in this mode starting with a contrarely phenomen. Do you think that?

Ahh, now I see Mario what you are getting at. Its the probability of EITHER of two numbers not appearing in 4 consecutive draws being 28% hence the probability that either one WILL appear is 72%. I was thinking that you had one ball fixed in your mind and you were waiting for the other, which is the same as matching both. In that case, I apologise - I think you are correct.

Just to be sure I have looked back over some sample 7/49 results to check and got :-



Each row shows how likely to appear after that exact number of draws, so if you add up the percentages of all the draws before that point you get the cumulative chance (e.g cumulative chance within 4 draws is 25+19+14+9=68% . (i would probably be 72% I f I had the time to do all the pairs).

About 68% chance (only from a sample of pairs not all of them) for one of the pair to appear within 4 draws. About 86% chance for one of them to appear within 14 draws.
Its been an interesting excercise for me Mario, allowing me to do some well needed revision !

By the way you don't need to use HYPGEOMDIST to work out that probablility of one of a pair in any draw, its quite straightforward.

We are waiting for 2 numbers so the probability of the first number coming out in any draw is 7/49 = 1/7 = 0.142857143. The probability of the second number coming out in any draw taking into account that the other one failed is 6/48= 0.125. The probability of either turning up is 0.142857143 + 0.125 = 0.267857143. This means the probability of neither turning up in any draw is 1- 0.267857143 = 0.732142857, same answer as HYPGEOMDIST.

[mario]

(10-15-2009 08:34 PM)Frank Wrote:  Just to be sure I have looked back over some sample 7/49 results to check and got :-



Each row shows how likely to appear after that exact number of draws, so if you add up the percentages of all the draws before that point you get the cumulative chance (e.g cumulative chance within 4 draws is 25+19+14+9=68% . (i would probably be 72% I f I had the time to do all the pairs).

68% is not 72% perhaps because in the first step you calculated 25% and I thought this:
in the first step I have the formulas:
- =HYPGEOMDIST(1,2,7,49) for one number and must add
- =HYPGEOMDIST(2,2,7,49) for hit 2 numbers and also is a favorable case

so I obtain in first step more than you 0.25+ 0.017857 = 0.267857

and so on in final for step 4 obtain 0.7126 or almost 72%
similarly for step 14 the probability grow up.

In fact this formulas show us a good probability and a pair can play for 16 steps with 99% probabilities!!!

Incredible and even so in this condition is hard to remain with moneys!
Why?
So I saw can exist series of win with length 8-10 and in that time you can acumulate a good amount but even so appear an exception and lost a big part of amount. It's a hard play. Perhaps exist a strategy and a management of stakes, this is what we must discover!

[Frank]

(10-16-2009 05:34 AM)mario Wrote:  68% is not 72% perhaps because in the first step you calculated 25% and I thought this:
in the first step I have the formulas:
- =HYPGEOMDIST(1,2,7,49) for one number and must add
- =HYPGEOMDIST(2,2,7,49) for hit 2 numbers and also is a favorable case

so I obtain in first step more than you 0.25+ 0.017857 = 0.267857

and so on in final for step 4 obtain 0.7126 or almost 72%
similarly for step 14 the probability grow up.

In fact this formulas show us a good probability and a pair can play for 16 steps with 99% probabilities!!!

Incredible and even so in this condition is hard to remain with moneys!
Why?
So I saw can exist series of win with length 8-10 and in that time you can acumulate a good amount but even so appear an exception and lost a big part of amount. It's a hard play. Perhaps exist a strategy and a management of stakes, this is what we must discover!

Mario the image I posted was not a calculation, it was ACTUAL results from an example pair in UK lotto history. Had I used a different pair that 68% might have been 75% and had I done 1176 pairs they would have averaged out to 72%.

My calculation uses the probability of 'not success' for either of a pair (0.732143 calculated earlier) and then uses BINOMDIST(S,S,0.732143 ,FALSE) in the middle column ( to work out the diminishing chance of a fail)
where S is the step number row by row, so you get:-


so the last row middle formula would be =BINOMDIST(20,20,0.732143 ,FALSE)

The right hand red column is 1-(P) in each middle column and is the increasing probability of success of at least one of a pair, which are the figures we are intersted in draw by draw corresponding to actual results. My calculated results are the same as yours as far as yours go...

Edit

Quote:So I saw can exist series of win with length 8-10 and in that time you can acumulate a good amount but even so appear an exception and lost a big part of amount. It's a hard play. Perhaps exist a strategy and a management of stakes, this is what we must discover!

I can't remember now what strategy you are using (please remind me so I don't have to search for it), but all this analysis only covers the case of betting on 2 numbers separately (for each