There is a lottery where is drawn six numbers out of 49 with two draws per day.

In this lottery is possible to select in one ticket more than six numbers - i.e. when you select 8 numbers then your tip contains all possible six-number combinations (in this case 28 combinations in total) from 8 numbers of your selection. It is possible to select up to 16 numbers in one ticket (this ticket would "generate" 8008 six-number combinations).

Let's presume that there is only main prize in this lottery - your tip must contain all 6 numbers drawn.

Scenario A: You will select 8 numbers on your ticket (28 six-number combinations) and this ticket you will bet to all draws in whole month (2 draws per day x 30 days = 60 draws in total).

Scenario B: You will select 9 numbers on your ticket (84 six-number combinations) and this ticket you will bet only to one draw per day in month (1 draw per day x 30 days = 30 draws in total).

Which scenario gives you better probability of winning the main prize in one month term -

A: more draws with less numbers (=more theoretical chances to win) or

B: less draws with more numbers (=better probability per draw)?

Or it doesn't matter?

  • $\begingroup$ In B, you have a 3x higher chance of winning, and playing half as often, so B is better by a factor of 3/2. In real life, I'm guessing more numbers cost more money, which should factor into your computation. $\endgroup$ – barrycarter Jul 15 '17 at 14:38
  • $\begingroup$ So if A will be selection of 15 numbers / 5005 combinations for 60 draws and B will be selection of 16 numbers / 8008 combinations for 30 draws then A will be better? $\endgroup$ – GilesNorthcott Jul 15 '17 at 14:57
  • $\begingroup$ Yes. Compute the total probability of winning. It's something like 1-(1-5005/C(49,6))^60 and 1-(1-8008/C(49,6))^30 where C(n,k) is the binomial coefficient. Do you see how I get those numbers? $\endgroup$ – barrycarter Jul 15 '17 at 15:04
  • $\begingroup$ Yes, I got it, thank you. $\endgroup$ – GilesNorthcott Jul 15 '17 at 16:05

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