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I was always thinking about this questions when it comes to lottery. lets say you have a number range from 1 - 45. You need 5-6 numbers to win.

Let says you have 2 options Draw 3 sets of numbers with 10 numbers on each set. or 1 set of numbers with 19 numbers for that?

Which chance will have higher of winning?

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    While I was writing my post another answer was posted with a different interpretation of the question. Our answers are of course dependent on the question, which is not clear and hence both answers begin with a clarification of the question as we see it. Perhaps the asker could clarify, although he may well be interested in interpretations other than those he intended.2010-10-31

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We have $45$ numbers $1,2,\ldots,45$ of which $6$ are winners.

The choice is between (a) buying three cards, each of which contains $10$ distinct numbers chosen from the $45$ (the same number can appear on different cards) and (b) buying one card which contains $19$ distinct numbers chosen from the $45.$

To win we must have at least $5$ of the winning numbers on a card.

Now the probability of at least $5$ winning number on a card of size $10$ is

$$\frac{ \binom{6}{5} \binom{39}{5} + \binom{6}{6} \binom{39}{4} }{ \binom{45}{10} } = \frac{1}{902}.$$

And so the probability of $ \le 4 $ winners is $ 1 – 1/902 = 901/902.$ Hence the probability of at least one of our three cards of size $10$ winning is ($1$ minus the probability that they all lose)

$$ 1 - \left( \frac{901}{902} \right)^3 \approx 0.003322.$$

The probability of at least five winners on a card of size $19$ is

$$\frac{ \binom{6}{5} \binom{39}{14} + \binom{6}{6} \binom{39}{13} }{ \binom{45}{19} } = \frac{5491}{135751} \approx 0.040449.$$

Since $0.040449 > 0.003322$ you have a better chance of winning by buying one card of size $19.$

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    well lets say to make it more interesting. for the 2nd options you require 7 numbers in order to win. Will it still have a better probability?2010-10-31
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    If you are saying there are now 7 winning numbers amongst the 45 and, for option (a), we still need at least 5 of those on a card of size 10 to be a winner then the probability of winning is about 0.010370 and if all seven winning numbers are required on our card of size 19, in option (b), then the probability of winning in this case is about 0.001110. So (a) beats (b) in this case. I'd advise you to do the calculations along the lines I've given in my answer. It should help you formulate more clearly your question and understand the maths better. Good luck!2010-10-31
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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...?