In Keno, a player picks from 1 to 70 (at least in this version), 20 of these numbers are drawn, and the payouts are based on the number of matches. What I have tried to do is to check that the Swedish state owned gambling company Svenska spel is trustworthy when it claims that the chance of winning Keno $11$ - that is $11$ numbers match the $20$ picks - is $1$ in $6.2$ (a player wins if 5 or more numbers drawn match his/hers picks).
From the outset of the problem it is clear that this matches the description of hypergeometric distribution whose probability can be calculated using the following formulae: \begin{equation} P(X=k)=\frac{{m \choose k} {N-m \choose n-k}}{{N \choose n}} \end{equation} where $N$ is the total number of numbers, $m$ the amount of numbers we interested of, and $k$ is the amount of numbers that the player picks.
Now, even though I am fully aware that hypergeometric distribution is built into Mathematica, I went ahead and wrote down the definition as follows:
Urn[good_, bad_, draws_,
need_] := (Binomial[good, need]*Binomial[bad, draws - need])/
Binomial[good + bad, draws]
From this, I generated the following table: \begin{array}{|l|r|} \hline \text{Correct Picks} & \text{Rounded Odds (1 in …)} & \text{Odds (Percentage)} \\ \hline 0 & 29 & \text{0.0345} \\ \hline 1 & 7 & \text{0.1488} \\ \hline 2 & 4 & \text{0.2718} \\ \hline 3 & 4 & \text{0.2718} \\ \hline 4 & 6 & \text{0.1743} \\ \hline 5 & 14 & \text{0.0710} \\ \hline 6 & 53 & \text{0.0190} \\ \hline 7 & 300 & \text{0.0033} \\ \hline 8 & 2674 & \text{0.0003} \\ \hline 9 & 39445 & \text{2.5352 $\cdot 10^{-5}$} \\ \hline 10 & 1075780 & \text{$9.2955 \cdot 10^{-7}$} \\ \hline 11 & 72184861 & 1.3853 \cdot10^{-8} \\ \hline \end{array}
However, when I compared my calculations - using the following command
1/Sum[N[Urn[11, 70, 20, k]], {k, 5, 11}]
to Svenska spel's claim I discoverd that the two results did not match. According to my calculations, the chances of winning is $1$ in $10.7$, not $6.2$. Since I don't think Svenska spels odds are inaccurate, I suspect my calculations are faulty. Just to test this claim, I employed the Monte Carlo method:
n = 1000000;
nchoice = 11;
nball = 70;
draws = 20;
{#[[1]], N[ #[[2]]/n]} & /@ Sort@Tally@Table[
Length@Intersection[
RandomSample[Range[nball], nchoice],
RandomSample[Range[nball], draws]
], {n}]
which produced the following result:
0, 0.017371 0.0172627
1, 0.094738 0.0949448
2, 0.219235 0.219994
3, 0.283973 0.282849
4, 0.223563 0.223648
5, 0.113827 0.113857
6, 0.037821 0.0379524
7, 0.008278 0.00825053
8, 0.001096 0.00114103
9, 0.00009 0.0000950859
10, 8.*10^-6 4.26916*10^-6
11 --- 7.76212*10^-8
As you hopefully can make out, the probabilities from this test aren't close to those of my initial calculations, which fuels my suspicion that something went terribly wrong in my initial set up. The question I thus am left to ask is:
What exactly went wrong in my original calculations?
I get a different set of results for the distibution. Yours sum to $0.9948$ which seems too far from $1$ to be rounding. Using $P=\frac{{11 \choose wins}{59 \choose 20-wins}}{70 \choose 20}$I get $$\begin {array} {l r} Successes & Probability \\0&0.017262686\\1&0.09494477\\2&0.21999398\\3&0.282849403\\4&0.223648365\\5&0.11385735\\6&0.03795245\\7&0.008250533\\8&0.001141031\\9&9.50859E-05\\10&4.26916E-06\\11&7.76212E-08 \end {array}$$ and the sum for five or more is $0.161301$, which is duly $1$ in $6.2$