I'm Looking For An Algorithm To Optimize Keno Selections
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I used to be pretty good at this stuff, but I just can't get a handle on how to express this properly. Thirty years takes a toll on remembered math, so I hope someone more recently experienced with probability can steer me back on course. In the game of Keno there is a field of 80 numbers, from which 20 are selected for each round. Players pick from n = 3 to 20 numbers and hope that n of them will come up in the next round. In reality, far fewer than n usually pop up, and I'd like to know the probability of each combination occurring. For instance, suppose I pick 6 numbers. How can I calculate the probability that 3 of my numbers will be among the 20 drawn? 5? The simple approach of calculating draws doesn't work, since the classic method I remember determines the probability of all 3 numbers coming up in 3 draws. But given that 20 draws are made regardless of how many I pick changes things considerably. I just can't figure out how to express it. Anyone care to take a crack at it? :-D
Will Rogers never met me.
To get the probability of getting 3 numbers correct just take the total number of selected values divided by the total number of balls. This gives 3 divided by 80 , not 20. The probability of getting 6 numbers correct is therefore 6/80. To determine the probability of your lucky numbers appearing in any number of consecutive draws take this following example. Whats the probability of 3 of your numbers appearing in four draws? Its simply got by multiplying individual probabilities by each other. So the answer should be got by getting (3/80)cubed or raised to power three, ie, (3/80)^3 if you have the zeal and will, i can make you a more elaborate c# software showing the workings of probability in action free of charge but it will take a while.
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To get the probability of getting 3 numbers correct just take the total number of selected values divided by the total number of balls. This gives 3 divided by 80 , not 20. The probability of getting 6 numbers correct is therefore 6/80. To determine the probability of your lucky numbers appearing in any number of consecutive draws take this following example. Whats the probability of 3 of your numbers appearing in four draws? Its simply got by multiplying individual probabilities by each other. So the answer should be got by getting (3/80)cubed or raised to power three, ie, (3/80)^3 if you have the zeal and will, i can make you a more elaborate c# software showing the workings of probability in action free of charge but it will take a while.
Thanks for the kind offer, but no thank you. Your algorithm is seriously flawed, yielding a higher probability of getting 6 right than three. But I appreciate the offer! :-D
Will Rogers never met me.