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# Calculate odds for random picking

Topic closed. 2 replies. Last post 4 years ago by nasp.

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Spain
Member #114615
August 4, 2011
27 Posts
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 Posted: May 2, 2012, 1:05 pm - IP Logged

Hello,

I realize that the following scenario might sound strange but if someone has the answer it would be great. The scenario is as follows:

1. I choose the 100 latest  draws from my lottery which already won the jackpot.

2. I pick 6 random draws from those 100

3. Each of these 6 draws contains one or more numbers, of the winning numbers of the next draw.

4. All  6 draws, have all the numbers of the next winning draw. What I mean is that if I choose 6 draws and all 6 draws have only number 2 included which will be one of the numbers in the following jackpot, it would be pointless.

What are the odds of  trying once to pick out  6 such  draws?

And what are the odds if I try 10 or 20 times to pick out these 6 draws?

If anyone could help it would be great. I tried doing the above in practice but it seems that I never seem to be able to pick 6 draws succesfully.

Thank you,

nasp

cleveland ohio
United States
Member #65897
October 9, 2008
275 Posts
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 Posted: May 2, 2012, 7:07 pm - IP Logged

your odds of doing this would probably be much worse then the actual odds of picking the correct number.

Lets examine your premis.

1st you will pick 6 draws from the past 100 simple enough. 100*99*98*97*96*95/6*5*4*3*2*1 =7152314400. Then you have the problem of having 6 sets of 5 or 6 numbers to choice from discounting doubles.

Its not really confusing on why you cant find the correct path. Your odds are against picking the correct answer and larger then just playing every combo.

Lets take this a step further and say all your numbers are differnt numbers (hence you wont be a 2 in all my 6 picks) in your 6 lines on a pick 5 game. You have 30 numbers. Your combinations would be 30*29*28*27*26/5*4*3*2*1 or 142506 much better but again assuming you have all different numbers.

You can do the math from there if you have doubles or want to add in picks in your question of 10 or 20 picks.

Spain
Member #114615
August 4, 2011
27 Posts
Offline
 Posted: May 5, 2012, 10:25 am - IP Logged

Thank you very much for the reply.  I was thiking that maybe this would have been a good way of i.e. picking 30 to 36 numbers out of the 49.

But as you explained the odds are much worse. Thank you for the explanation! :)

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