More Keno Info...


Q:K3 How do I calculate the odds of winning at keno? 
A:K3 [John Hallyburton] 

It's straightforward but takes a little math. For reference, a full table of keno odds is at
http://www.conjelco.com/faq/keno-odds.html . 

Occasionally I get email requests for "the keno formula" from people who want
something to plug into a spreadsheet. There is no simple formula, though if you slog
through this section you can come up with a complicated formula. 

The proper buzzword for keno odds is "hypergeometric distribution". But as usual,
understanding the math is far less important than understanding how to apply it properly.
First, let's do the basics: if you mark N spots, the probability of hitting exactly K of them
is given by the formula: 

               C(N,K)  *  C(80-N,20-K)
p(N,K) =   -------------------------------
                     C(80,20)

The expression C(X,Y) represents the number of possible ways to select Y items from a
larger collection of X items, where order of selection is unimportant. Many calculators,
spreadsheets and math libraries have a built-in facility for calculating this function. Both
Lotus 1-2-3 ™ and Excel ™ name this funcion COMBIN(n,r); it is also known as the
"binomial coefficient" function. (Caution: even if defined by your spreadsheet you may
find the numbers involved too large to be handled by your spreadsheet program). Direct
evaluation comes from the following formula: 

                  X!
C(X,Y) =   -----------------
              Y! x (X-Y)!

... where "X!", pronounced "X factorial", is the product of all whole numbers from 1 to
X. Thus 4! = 1 x 2 x 3 x 4 = 24. As a degenerate case, 0! = 1. So C(5,3) = 5!/(3!x2!) =
120/12 = 10. There are 10 ways to select 3 items from a bag of 5 items. Again, order of
selection is unimportant. Note that N! = N x (N-1)!, for N>0. This can be useful in
simplifying calculation. 

Here's a sample of how to calculate C(80,6) by hand: 

              80!       80 x 79 x 78 x 77 x 76 x 75 x 74!
C(80,6) = ---------- = -----------------------------------
            6! x 74!     6 x 5 x 4 x 3 x 2 x 1 x 74!

                         80 x 79 x 78 x 77 x 76 x 75
                     = -------------------------------
                          6 x 5 x 4 x 3 x 2


Now we start canceling: 6 into 78 13 times, 5 into 80 16 times: 

                         16 x 79 x 13 x 77 x 76 x 75
                     = -------------------------------
                                   4 x 3 x 2

Cancel some more: 4 into 16 4 times, 3 into 75 25 times: 

                         4 x 79 x 13 x 77 x 76 x 25
                     =  ---------------------------
                                    2

Finally, 2 into 4 2 times. You can always divide all numbers in the bottom into the
numbers on top. We are left with: 

                     = 2 x 79 x 13 x 77 x 76 x 25

                     = 300500200

the number of ways to select 6 items from 80. 

Notice if you select 6 items from a group of 80, you are "leaving" 74 items unselected.
They form a group of their own! That means C(80,74) = C(80,6). Whenever you select a
group, you actually select two groups, the "ins" and "outs". In other words: C(X,Y) =
C(X,X-Y). This amounts to switching the order of the multiplication of the bottom half of
the fraction in the definition of C(X,Y). 

For reference, C(80,20) = 3.535316142 x 10^18 or about 3 1/2 quintillion ways for the
house to draw 20 balls. That number is so huge it is unlikely any random keno draw has
ever happened twice in all of history. Well, maybe one repetition somewhere, if you're
generous enough estimating how many games have taken place over time. 

Let's use the formula to calculate our chances for hitting a 6-spot: 

p(6,6) =  C(6,6) * C(74,14) / C(80,20) ~= 0.00013 or 1 in 7753.

Rarely do we hit all 6. Let's calculate the whole 6-spot table from the above formula.
Note these numbers are independent of the house payoff as they are merely the
probability of an event happening, regardless of whether any money is wagered. 

p(6,6) = C(6,6) * C(74,14) / C(80,20) ~= 0.00013     Catch 6
p(6,5) = C(6,5) * C(74,15) / C(80,20) ~= 0.00310     Catch 5
p(6,4) = C(6,4) * C(74,16) / C(80,20) ~= 0.02854     Catch 4
p(6,3) = C(6,3) * C(74,17) / C(80,20) ~= 0.12982     Catch 3
p(6,2) = C(6,2) * C(74,18) / C(80,20) ~= 0.30832     Catch 2
p(6,1) = C(6,1) * C(74,19) / C(80,20) ~= 0.36349     Catch 1
p(6,0) = C(6,0) * C(74,20) / C(80,20) ~= 0.16602     Catch 0
                              Total      0.99932

The total should always be 1, or very close due to rounding, since one of the above
outcomes will happen. 

To find the probability of one of several outcomes, you add the numbers for each entry.
In the above 6-spot table the chances of catching "exactly 0 OR exactly 1" are 0.52951.
Meaning more than half the time you'll catch at most one number on your 6-spot ticket.
Similarly, roughly 1 in 6 tickets will be a winner according to the payoff table presented
above. (Since Catch3+Catch4+Catch5+Catch6 probabilities total 0.16159 or 1/6.19). 

Of course the chance of having a winning ticket is not as interesting as knowing what the
expected return is. This is calculated by adding together the expected return for every
possible outcome. Now for any possible outcome you can calculate the expected return
for that outcome by multiplying the payoff for that outcome by the probability of that
outcome. This leads directly to the following set of calculations, using the 6-spot payoff
and odds already presented:

     Catch  chance  payoff   expected
                   ($1 bet)   return

       6   0.00013  1500      0.195
       5   0.00310    50      0.155
       4   0.02854     8      0.228
       3   0.12982     1      0.130
       2   0.30832     0      0
       1   0.36349     0      0
       0   0.16602     0      0
                              -----
    Total                     0.708

Contract Bridge - Possible auctions - The number of possible auctions with North as dealer is 
128,745,650,347,030,683,120,231,926,111,609,371,363,122,697,557.

one hundred and twenty eight quindecillion.