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odds of winning a (lotto-like) picking 6 winning numbers from 1-51, plus a wild card number. formula please!!!

So please help me create an odds chart (for stupids like myself)
TICKET: Pick 6 numbers from 1 to 51.(actual value $1.75dlls)
First Prize draw , a machine "picks" 6 balls , each one contains a number.. then a show host picks an extra ball "wild-card" number. Today the first prize is about $ 40.0 million US DLLS - first prize is for the ticket(s) with the 6 numbers of the "machine balls", second prize (some 8.0 Million) is for any 5 "machine balls" plus the wild card.. -third prize is for 5 "m. balls" and fourth is 4 m.balls plus the wild card..and so on.. With the same original ticket ,A SECOND draw (1st prize $4.25M)goes on, now only six machine balls,no wild card.
What are the odds of winning(same numbers) the first prize on both draws?
FACT: last week draw 1 and 2 shared 4 numbers
What are the odds of winning any prize at all? the least prize is any 1 ball number plus the wildcard -ticket value back- Actual lottery in Mexico. No1st(on both) prize winners over 3 months

First prize:

1 in C(51,6) for ONE drawing
= 1 in 18,009,460

C(51,6) means the combination of 51 things taken 6 at a time. The value is C(51,6) = 51! / (45! * 6!)
where 6! (called six factorial) = 6*5*4*3*2*1, etc.
n! = n*(n-1)*(n-2)*(n-2)*...*3*2*1
There are C(51,6) possible combinations of numbers for a ticket. There is only one winning combination, so the probability is 1 in C(51,6)

The probability of winning on both draws is astronomical,
1 in 18,009,460 ^2 (^2 means squared), or
1 in 324,340,649,491,600

2nd prize on one draw:

There are C(6,5) = 6 ways to have five out of six of the winning numbers (each way is missing one of the six). Then the last number must be the wild card. So there are six winning combinations. The probability is then:

C(6,5) / C(51,6) = 6 / 18,009,460 or about
1 in 3,001,576.67

3rd prize: any 5 out of six:
Choose five out of the six winning numbers C(6,5) = 6, then there are 45 non-winning numbers left to fill the remaining spot. The probability is:

C(6,5) * 45 / C(51,6)
or about 1 in 66,701.7

that's enough for now. if you really need the rest of the odds email me.

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