Here's a quick derivation which gives the same solution as that provided by gar, although a slightly different recursion.
A winning $k$ ball combination consists of integer sequences $0<x_1<\cdots<x_k$ and $0<y_1<\cdots<y_k$ so that $|x_i-y_i|\le1$.
Let $A_k(n)$ be the number of ways to pick winning $k$ ball combinations so that $x_k,y_k\le n$. This is the same as $T(n+1,k)$ defined by gar.
Similarly, let $B_k(n)$ be the number of ways to pick $k$ ball combinations so that $x_k\le n$ and $y_k\le n+1$. We'd get the same number of instead we require $x_k\le n+1$ and $y_k\le n$ due to symmetry.
Obviously, $A_k(n)=B_k(n)=0$ for $n<k$. Otherwise, $A_0(n)=B_0(n)=1$.
Let's first consider $A_k(n)$. If $(x_k,y_k)=(n,n)$, what remains is picking the remaining $k-1$ balls with $x_{k-1},y_{k-1}\le n-1$, which can be done in $A_{k-1}(n-1)$ ways. If $(x_k,y_k)=(n,n-1)$, the remaining $k-1$ balls should have $x_{k-1}\le n-1$ and $y_{k-1}\le n-2$, which can be done in $B_{k-1}(n-2)$ ways; the same applies to $(x_k,y_k)=(n-1,n)$. Otherwise, $x_k,y_k\le n-1$, which leaves $A_k(n-1)$ alternatives. Thus,
$$
A_k(n)=A_{k-1}(n-1)+2B_{k-1}(n-2)+A_k(n-1).
$$
We derive a recursion for $B_k(n)$ in a similar way. If $(x_k,y_k)=(n,n+1)$, the remaining $k-1$ balls require $x_{k-1}\le n-1$, $y_{k-1}\le n$, leaving $B_{k-1}(n-1)$ alternatives. Otherwise, $x_k,y_k\le n$, leaving $A_k(n)$ alternatives. Thus,
$$
B_k(n)=B_{k-1}(n-1)+A_k(n).
$$
Plugging in $k=6$, $n=50$ gives $A_6(50)=8000567708$. Since each of the two integer sequences are drawn randomly with probability $1/\binom{n}{k}$, this gives the probability of winning
$$
\frac{A_6(50)}{\binom{50}{6}^2}
=\frac{8000567708}{15890700^2}\approx 0.00003168361647.
$$
The first thing to do is determine the number of possible tickets, in this case ${59 \choose 6}=45057474$, and the number of ways of winning each of the prizes:
- Matching 3: ${6 \choose 3} \times {53 \choose 3} =20\times 23426 =468520$ (The number of ways of choosing 3 from the winning set and 3 from the loosing set.)
- Matching 4: ${6 \choose 4} \times {53 \choose 2} =15\times 1378=20670$
- Matching 5: ${6 \choose 5} \times {53 \choose 1} = 6\times 53 =318$
- Matching 6: ${6 \choose 6} \times {53 \choose 0} = 1$
The probability of winning in each of these ways is determined by dividing by ${59 \choose 6}$. (The probability of winning anything at all is approximately $0.01$.) The expected value of a single random ticket depends on what the actual prizes are. As a function of these prizes we get:
$$\frac{234260}{22528737}E_3 + \frac{3445}{7509579}E_4 + \frac{53}{7509579}E_5 + \frac{1}{45057474}E_6$$
Let's assume that the prizes are $E_3=\$5$, $E_4=\$100$, $E_5=\$10000$, $E_6=\$1000000$. Then we get an expected value of about $\$0.19$.
If you do this twice, where the two tickets are independent of each other, you simply get twice the expected value, about $\$0.38$.
Now, if the second ticket has distinct numbers in it, the first ticket proceeds the same way but for the second, we get a minor plague of cases. First, there are ${53 \choose 6}$ ways of choosing the second ticket in each of the cases. Then, there are seven cases (each with subcases) depending on how many of the winning digits were matched by the first ticket:
- The first ticket matched 0 numbers. $P={53 \choose 6}/{59 \choose 6}$
Then the second ticket could match any number of winning numbers. There are 6 unpicked winning numbers and 47 unpicked losing numbers.
Number of ways of matching 3: ${6 \choose 3}\times {47 \choose 3}$
Number of ways of matching 4: ${6 \choose 4}\times {47 \choose 2}$
Number of ways of matching 5: ${6 \choose 5}\times {47 \choose 1}$
Number of ways of matching 6: ${6 \choose 6}\times {47 \choose 0}$
- The first ticket matched 1 number. $P={6 \choose 1}{53\choose 5}/{59 \choose 6}$
Then the second ticket could match any number of winning numbers except all of them. There are 5 unpicked winning numbers and 48 unpicked losing numbers.
Number of ways of matching 3: ${5 \choose 3}\times {48 \choose 3}$
Number of ways of matching 4: ${5 \choose 4}\times {48 \choose 2}$
Number of ways of matching 5: ${5 \choose 5}\times {48 \choose 1}$
- The first ticket matched 2 numbers. $P={6 \choose 2}{53\choose 4}/{59 \choose 6}$
Then the second ticket could match up to 4 winning numbers. There are 4 unpicked winning numbers and 49 unpicked losing numbers.
Number of ways of matching 3: ${4 \choose 3}\times {49 \choose 3}$
Number of ways of matching 4: ${4 \choose 4}\times {49 \choose 2}$
- The first ticket matched 3 numbers. $P={6 \choose 3}{53\choose 3}/{59 \choose 6}$
Then the second ticket could match up to 3 winning numbers. There are 3 unpicked winning numbers and 50 unpicked losing numbers.
Note that, in this case, we have a win even if the second ticket doesn't win so this is really two cases: One where you win twice, one where you only win once.
Number of ways of matching 3: ${3 \choose 3}\times {50 \choose 3}$
- The first ticket matched 4 numbers.
- The first ticket matched 5 numbers.
- The first ticket matched 6 numbers.
The second ticket cannot win in these cases.
That's 14 cases where you win something. The expected value of playing this way is then
$P(T_1=0,T_2=3)(0+E_3)+P(T_1=0,T_2=4)(0+E_4)+P(T_1=0,T_2=5)(0+E_5)+P(T_1=0,T_2=6)(0+E_6)
+P(T_1=1,T_2=3)(0+E_3)+P(T_1=1,T_2=4)(0+E_4)+P(T_1=1,T_2=5)(0+E_5)
+P(T_1=2,T_2=3)(0+E_3)+P(T_1=2,T_2=4)(0+E_4)
+P(T_1=3,T_2=3)(E_3+E_3)+P(T_1=3,T_2<3)(E_3+0)
+P(T_1=4)E_4
+P(T_1=5)E_5
+P(T_1=6)E_6$
I made Sage do the algebra on this to find the expected value as a function of the prizes:
$$\frac{268671400163}{25860151455138}E_3 + \frac{6890}{7509579}E_4 + \frac{106}{7509579}E_5 + \frac{1}{22528737}E_6$$
This is $\frac{1573862727}{151228955878}E_3$ less than the expected value for two independent tickets. For the prize amounts I gave above, the expected value is about $\$0.33$ or about five cents less than the expected value for independent tickets.
Best Answer
You need $500$ distinct numbers to have a $50\%$ chance of winning. This is the coupon collector's problem except that you do not need to collect all the coupons. We can compute the expected number to get $500$ distinct numbers. The first coupon is guaranteed to get a new number. To get a second new number takes on average $\frac {1000}{999}$. Once you have two, the third takes $\frac {1000}{998}$ and so on. We can compute the expected number to get $500$ distinct numbers as as $$\begin {align} 1+\frac {1000}{999}+\frac {1000}{998}+\ldots \frac {1000}{501}&=1000(H_{1000}-H_{500})\\ &\approx 1000(\log(1000)-\log (500))\\&=1000\log (2)\\ &\approx 693 \end {align}$$ This gives the expected number of days to get $500$ different tickets. It is close to, but not guaranteed to be the same as, the number of days to get your chance of $500$ distinct tickets over $50\%$. I don't have a good way to calculate the second of these. I suspect the $693$ is an overestimate of the number of days to have a $50\%$ chance to have $500$ different because there will be a long tail.