Keno probability?

[Ohjay]

Sorry for the non-poker related topic but I need to figure this out.

My dad has been playing Keno for X amount of years now. Always using the exact same numbers, playing in the exact same game. And I need to figure out the probability of him hitting the "jackpot". Problem is I can't solve the math on my own, that's where you guys come in.

Dad has 10 constant numbers that he plays each time.
The Keno machine draws a total of 20 numbers out of 70 each time. They are numbered 1 through 70.
If dad's 10 numbers are drawn he hits the "jackpot". What is the probability that this will happen?

Thanks in advance

[the_hawk]

If only 10 numbers were drawn the odds would be:

(10/70)*(9/69)*.... (1/61)

There must be a C(x,y) "choose" expression for this but I can't be bothered to work it out off the top of my head.

Now, since 20 numbers are actually drawn the actual probability should be C(20,10) times this, shouldn't it?

Where's Bugs when you need him?

[the_hawk]

I think the probability you are looking for is given by:

EDIT: wrong.

[Fenris78]

I would do it like this: Count the number of combinations for 20 numbers that contain exactly the 10 numbers your father picks and divide it by all possible combinations.

You dad picks ten numbers. How many possible distribtions do exist which contain exactly those 10 numbers? Since 10 numbers are already fixed you just need to know how many combinations for the other 10 numbers of the 20 picked (that are irrelevant for your dad) exist? Those are simply C(60,10), because you look at the distribution of 10 numbers out of those 60 numbers that were not picked.

There are C(70,20) combinations of drawing 20 numbers out of 80, so the probability you are looking for should be:

C(60,10)/C(70,20) = 0.000047%

€: Fixed mistakes

[janeg]

C(70,10) = 1 in 396,704,524,216

I don't think you need to worry about the 20 numbers, all you care about is that the 10 numbers are chosen out of the 70 available numbers; if they are chosen from the 70 they are automatically a part of the 20, they are not re-drawn from the 20.

[the_hawk]

[janeg]

[ciaran]

The simplest solution to this problem is to teach him to play poker...

It appears, based on the link Jane provided, that her final answer is correct, and the odds of Papa Ohjay hitting are a bit more than 2.1M-1 against. I defer to Bugs' or Chillin's math if they show up later, though.

[Fenris78]

Of course Jane's final solution is correct since it coincides with mine

[the_hawk]

Jane, I'm now convinced your second formula is correct. As Fenris and Jane pointed out, the 10 balls drawn but not in Ohjay Sr's "hand" are really of little consequence.

I will also point out for clarity's sake that my "instinctive" solution in the first reply to this thread, namely:

(10/70) x (9/69) x .... x (1/61) x C(20,10) gives 4.65E-7 (i.e., Jane's answer).

So I was there first.

[Fenris78]

Just in case anybody doesn't know how to convert %

0,000047% = 4.7 E-7

So it seems that the 3 different ways lead to the same conclusion and that it will take Ohjays dad probably a couple more years to win that jackpot. Well, admittedly, Jane's Formula is the same as mine anyway since C(10,10)=1

[Ohjay]

Thanks a lot people. Definitely couldn't have done this without you
Time to gloat a bit the next time I see my dad

[dhwma]

In Mass, there are 80 #s.

I sometimes dabble in the 4 # play that pays $100 when you are 4 for 4.

[Bugsbunny]

People - for problems like this Google (as well as the Wizard of Odds site) are your friends.

For how to calculate See the wizard of odds site:
http://wizardofodds.com/keno
At the bottom is a section on how to calculate this. Also a pointer to this site:
http://www.mathproblems.info/gam470/games/keno/prob-keno.html

And what follows is a complete probability breakdown. One thing to remember is that (if I'm not mistaken) the jackpot payouts are capped, so even if it hasn't been hit in a while the max payout is still limited. I know this used to be true, not sure of it still is). Keno used to be the worst game in town. Nowadays that honor goes to most state lotteries, which have payouts in the 50% range on average.

From http://www.conjelco.com/faq/keno-odds.html

[Bugsbunny]

Also note that Keno is usually played with 80 spots, not 70 Don't know where he found a 70 spot game, but that changes odds quite a bit - assuming it wasn't a mistake.