Probability of an odd number in 10/20 lotto Say you have a lotto game 10/20, which means that 10 balls are drawn from 20.
How can I calculate what are the odds that the lowest drawn number is odd (and also how can I calculate the odds if it's even)?
So a detailed explanation:
we have numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19 and 20
and the drawn numbers were for example 3, 5, 8, 11, 12, 13, 14, 15, 18 and 19
so, we see now that lowest number is 3 and he is an odd number. 
So, as stated above, can you help me in finding out how to calculate such probability?
 A: The probability that the lowest number is at least $n$ is
$$\frac{\binom{21-n}{10}}{\binom{20}{10}}\;.$$
Thus the probability that the lowest number is exactly $n$ is
$$\frac{\binom{21-n}{10}-\binom{20-n}{10}}{\binom{20}{10}}\;.$$
Thus the probability that the lowest number is odd is
$$
\begin{eqnarray}
\sum_{k=0}^{5}\frac{\binom{21-(2k+1)}{10}-\binom{20-(2k+1)}{10}}{\binom{20}{10}}
&=&
\sum_{n=0}^{10}(-1)^n\frac{\binom{20-n}{10}}{\binom{20}{10}}
\\
&=&
\frac{122464}{184756}\\
&\approx&
\frac23\;.

\end{eqnarray}

$$
A: It would help if you could compute the probability that $\mathbb P(X = i)$, where $X$ would be your minimal number out of the $10$. Here $X$ would stand for for the minimum of the $10$ variables $X_1, X_2, \dots, X_{10}$, where $X_i$ is the value of the $i^{\text{th}}$ ball. 
An easier way to do this is to compute $\mathbb P(X \ge i)$ and then compute $\mathbb P(X = i)$ by computing $\mathbb P(X \ge i) - \mathbb P(X \ge i+1)$. Why? Because in this way computing leads to cleaner formulas. The idea is that it is usually more easy to compute $\mathbb P(X \ge i)$ when $X$ is a minimum of i.i.d. than to compute $\mathbb P(X = i)$, and the same goes for a maximum of i.i.d : it is usually more easy to compute $\mathbb P(X \le i)$ than $\mathbb P(X=i)$. Notice that I said "usually" though, I wouldn't say this is a general method, even though it works most of the time.
We will compute $\mathbb P(X \ge i)$ just by counting the ratio of number of possible cases over the number of total cases. $X \ge i$ if and only if all balls have a number on it greater than or equal to $i$, so there is $\begin{pmatrix} 21-i \\ 10 \end{pmatrix}$ possibilities that those $10$ balls satisfy this property (this is the number of ways to pick 10 balls in the ,last balls $i$, $i+1$, $\dots$, $20$). Note that for $i > 11$ this is zero, in this notation, so you can either look at it that way or assume $i \le 11$. (Try to understand why it must clearly be zero for $i \ge 12$.)
Now the number of ways to pick $10$ balls amongst $20$ is $\begin{pmatrix} 20 \\ 10 \end{pmatrix}$, hence
$$
\mathbb P(X = i) = \mathbb P(X \ge i) - \mathbb P(X \ge i+1) = \frac{ \begin{pmatrix} 21-i \\ 10 \end{pmatrix} - \begin{pmatrix} 20-i \\ 10 \end{pmatrix} }{\begin{pmatrix} 20 \\ 10 \end{pmatrix} }.
$$
If you want to know the probability that $X$ is even, just sum over the even possible values of $X$. Same goes if you want to know the probability that $X$ is odd ; sum over the odd possible values of $X$.
For instance, the probability that $X$ is even is
$$
\sum_{i=0}^5 \frac{ \begin{pmatrix} 21-2i \\ 10 \end{pmatrix} - \begin{pmatrix} 20-2i \\ 10 \end{pmatrix} }{\begin{pmatrix} 20 \\ 10 \end{pmatrix} } 
= \frac{ \begin{pmatrix} 20 \\ 10 \end{pmatrix} - \begin{pmatrix} 19 \\ 10 \end{pmatrix} + \begin{pmatrix} 18 \\ 10 \end{pmatrix} - \begin{pmatrix} 17 \\ 10 \end{pmatrix} + \dots + \begin{pmatrix} 10 \\ 10 \end{pmatrix}}{\begin{pmatrix} 20 \\ 10 \end{pmatrix} }
$$
Hope that helps,
A: The total number of outcomes is ${20 \choose 10}$. Now count the total number of favorable outcomes:


*

*outcomes with lowest element 1 : ${19 \choose 9}$ ;

*outcomes with lowest element 3 : ${17 \choose 9}$ ;

*outcomes with lowest element 5 : ${15 \choose 9}$ ;

*outcomes with lowest element 7 : ${13 \choose 9}$ ;

*outcomes with lowest element 9 : ${11 \choose 9}$ ;

*outcomes with lowest element 11 : ${9 \choose 9} = 1$ ;


So the probability is $$\sum_{k\in \{9, 11, 13, 15, 17, 19 \}} { {k \choose 9} \over {20 \choose 10}} = {30616 \over 46189} \simeq 0.662842.$$
