I can work out one thing but not another. I know that if you played 20 lines twice a week for a year, the probability you'd win the jackpot at least once during that year is

1 - (1 - 20/10737573)^104 = 0.00019369372287 = odds of 1 in 5163.

The problem I'm having trouble working out is the probability you'd win the second place prize at least once during that year.

The probability of getting 5 out of 6 balls on a single ticket is (6 choose 5)*(41 choose 1)/(47 choose 6) = 246/10737573. So if you've got 20 tickets, no two of which have the same set of 5 numbers, then your chance of winning the second place prize in a single drawing is 20*246/10737573 = 4920/10737573.

To solve your problem, you just put that new fraction in the place of where you have "20/10737573" and keep the rest of the equation the same. The whole thing works out to about a 1 in 21.48 chance of hitting the second prize at least once in a year's time.

You can do it for the third and fourth prizes similarly:

4/6: 20*(6 choose 4)*(41 choose 2)/(47 choose 6) = 246000/10737573. Probability of hitting at least once = 91.02%

3/6: 20*(6 choose 3)*(41 choose 3)/(47 choose 6) = 4264000/10737573. Probability of hitting at least once = 99.999999999999999999998604455426%

(In fact, if you just bought *one* line for each drawing, your chances of hitting the 3/6 prize at least once would be 87.58%, and if you played *two* lines each drawing your chances would be 98.52%.)

May 21, 2020, 2:58 am

In response to cottoneyedjoe

**Quote:** Originally posted by cottoneyedjoe on May 20, 2020

The probability of getting 5 out of 6 balls on a single ticket is (6 choose 5)*(41 choose 1)/(47 choose 6) = 246/10737573. So if you've got 20 tickets, no two of which have the same set of 5 numbers, then your chance of winning the second place prize in a single drawing is 20*246/10737573 = 4920/10737573.

To solve your problem, you just put that new fraction in the place of where you have "20/10737573" and keep the rest of the equation the same. The whole thing works out to about a 1 in 21.48 chance of hitting the second prize at least once in a year's time.

You can do it for the third and fourth prizes similarly:

4/6: 20*(6 choose 4)*(41 choose 2)/(47 choose 6) = 246000/10737573. Probability of hitting at least once = 91.02%

3/6: 20*(6 choose 3)*(41 choose 3)/(47 choose 6) = 4264000/10737573. Probability of hitting at least once = 99.999999999999999999998604455426%

(In fact, if you just bought *one* line for each drawing, your chances of hitting the 3/6 prize at least once would be 87.58%, and if you played *two* lines each drawing your chances would be 98.52%.)

Question?... Now, does these equations take into consideration the studying of the Stats of a Game?, or is it based on just random, trust ya rabbit foot playing?

In response to Stat$talker

**Quote:** Originally posted by Stat$talker on May 21, 2020

Question?... Now, does these equations take into consideration the studying of the Stats of a Game?, or is it based on just random, trust ya rabbit foot playing?

Hi Stats Talker, I will take your question as a mathematical inquiry made in good faith. These calculations are based on the assumption that each of the 10,737,573 combinations is equally likely to be drawn and recorded as the official result.

If some combinations are more likely to be drawn than others, and a player knows which ones they are and selects those combinations, then the probabilities would be higher.

May 21, 2020, 3:46 pm

In response to cottoneyedjoe

**Quote:** Originally posted by cottoneyedjoe on May 21, 2020

Hi Stats Talker, I will take your question as a mathematical inquiry made in good faith. These calculations are based on the assumption that each of the 10,737,573 combinations is equally likely to be drawn and recorded as the official result.

If some combinations are more likely to be drawn than others, and a player knows which ones they are and selects those combinations, then the probabilities would be higher.

Yes, my question was in good faith.. so, it may be a good idea to express that to most folks that ask a probability question, for those kinda specs, as you have so honestly and academically admitted, does affect the probability outcome... I was just curious, no pun intended this time..

BTW.. my avatar is 1 word... as in Stalker (someone who FOLLOWS/WATCHES)..."Stat$talker"... I just use the $ as an S...

In response to cottoneyedjoe

**Quote:** Originally posted by cottoneyedjoe on May 20, 2020

The probability of getting 5 out of 6 balls on a single ticket is (6 choose 5)*(41 choose 1)/(47 choose 6) = 246/10737573. So if you've got 20 tickets, no two of which have the same set of 5 numbers, then your chance of winning the second place prize in a single drawing is 20*246/10737573 = 4920/10737573.

To solve your problem, you just put that new fraction in the place of where you have "20/10737573" and keep the rest of the equation the same. The whole thing works out to about a 1 in 21.48 chance of hitting the second prize at least once in a year's time.

You can do it for the third and fourth prizes similarly:

4/6: 20*(6 choose 4)*(41 choose 2)/(47 choose 6) = 246000/10737573. Probability of hitting at least once = 91.02%

3/6: 20*(6 choose 3)*(41 choose 3)/(47 choose 6) = 4264000/10737573. Probability of hitting at least once = 99.999999999999999999998604455426%

(In fact, if you just bought *one* line for each drawing, your chances of hitting the 3/6 prize at least once would be 87.58%, and if you played *two* lines each drawing your chances would be 98.52%.)

OK, thanks. I was curious about the odds of nobody in particular hitting the big one, or almost hitting the big one, and treating us all to an epic thread.

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