Request for you

[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.