Hopefully I understand your question properly. To make sure, here's how I understand it:
There's a lottery, where 5 distinct random numbers will be chosen from 1-45 inclusive. If on your lottery card, you have all 5, you win.
Options:
- a) draw 3 cards with 10 random numbers per card
- b) draw 1 card with 19 random numbers
Which option is best?
a) The probability of winning if you drew one card with 10 numbers is 10/45 * 9/44 * 8/43 * 7/42 * 6/41 = 4/19393. Your probability of not winning is 1-4/19393. Hence your probability of winning with at least one card is 1-(1-4/19393)^3, which is approximately equal to 0.0006187.
b) On the other hand, your probability of winning with the 1 card option is 19 / 45 * 18/44 * 17/43 * 16/42 * 15/41 = 1292/135751, which is approximately equal to 0.009517.
So clearly the second option is the best.
A more intuitive (less numerical) explanation:
Your chances of winning with a card with 10 numbers is very low since half of the numbers on the card half to be winning numbers. Clearly with 19 numbers the chances are greater. If you crunch the numbers, it turns out that 3 10-number cards are not sufficient to overcome the advantage held by the 19-number card.
Perhaps you can extend the calculations above to show how many 10-number cards you'd need before that becomes the better option...?