Posted by Ion Saliu on December 22, 2000.

In Reply to: * Lottery player thankful for Keno software, systems* posted by Fatdade on December 21, 2000.

• I received a while ago a Keno-related message:

"I've been attempting, with some success, to apply you formulae. Unfortunately, I've come across a situation where I'm not quite certain what to do, and write in hopes you can help. I'm trying to solve for the following situation: What are the odds in Keno, at any given degree of certainty, that a single number WOULD NOT be selected at a given number of consecutive draws. I sat in a Keno parlor in Reno over the weekend, and tracked numbers. To make a long story short, one particular number was finally selected on the 34th consecutive game. I concluded the odds against it failing to be selected were astronomical, and this morning attempted to determine the probability. Unfortunately, it is beyond me. I have spread sheeted your formula, but am not sure how to reduce the numbers to return to the probability..."

My answer:

The probability for one Keno number to come out is p=20/80 = 1/4. The probability of one number NOT to come out is q=3/4. That's the probability to skip one drawing. The probability to skip N consecutive drawings is (3/4) to the power of N. For example, to skip 3 drawings in a row is: 3/4= 0.75 to the power of 3 = 0.422 (42%). The probability to skip 34 drawings in a row is: 0.75 to the power of 34 = 0.0000566 = 5.66 times in 100,000 drawings. That's why it's better to select numbers below the median (with 2 skips or less).

Read more on the “Winning Strategy” page. The same methods apply, as for lotto-6. For casino Keno, everything is more difficult, because you cannot use a computer to track the drawings and do the WS reports. You can only focus on numbers that came out below the median for p=1/4.

Good luck!

Ion Saliu

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