# 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?

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

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Ah, yes, that's a bit more straightforward than what I did :-) – joriki Jan 3 '12 at 11:11
This is for even $m$ :) I am confident you won’t have problem to modify it for odd $m$ – Elvis Jan 3 '12 at 15:06
Actually, both of our answers are related by the well-known identity $$\begin{pmatrix} n \\ k \end{pmatrix} + \begin{pmatrix} n \\ k +1\end{pmatrix} = \begin{pmatrix} n+1 \\ k+1 \end{pmatrix}.$$ Send one term to the other side and your terms in your sum becomes our terms in our sum. – Patrick Da Silva Jan 3 '12 at 15:10
@PatrickDaSilva This is the "recursive formula" I was alluding to. Besides that, no offense, don’t worry! – Elvis Jan 3 '12 at 15:12
Didn't understand what recursive formula you were referring to... lol. Let's stop talking here. I hate it when they say that "move this discussion to chat" thing. – Patrick Da Silva Jan 3 '12 at 15:13

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

$$\begin{eqnarray} \sum_{k=0}^{5}\frac{\binom{21-(2k+1)}{10}-\binom{20-(2k+1)}{10}}{\binom{20}{10}} &=& \sum_{n=0}^{10}(-1)^n\frac{\binom{20-n}{10}}{\binom{20}{10}} \\ &=& \frac{122464}{184756}\\ &\approx& \frac23\;. \end{eqnarray}$$

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You might be interested by this question I asked today, related to this problem... – Elvis Jan 6 '12 at 13:10

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,

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I realized I had put the numbers wrong in there ; I was editing my answer too much along the way and forgot to switch some stuff. Now I hope it's correct. – Patrick Da Silva Jan 3 '12 at 14:01