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Question:

In a lottery game, there is a card with $48$ numbers and you can choose $6$ numbers, or pay more and choose $7$ numbers. You win if you match all $6$ numbers drawn.

In which situation you have the most chance of winning the game?

$1$) buy a card of $7$ numbers

$2$) buy $7$ cards of $6$ numbers

Why? I can't understand this combination, but in the end I know the chance is the same.

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  • $\begingroup$ you would have to make sure that no two of your seven cards had the same six number combination $\endgroup$ – WW1 Mar 2 '18 at 0:16
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If you buy seven cards with six numbers each (assuming you don't have any two identical cards), there are seven combinations of draws that let you win. If you buy one card with seven numbers, there are again seven combinations that let you win-the combination made by leaving out each one of the seven numbers on the card. Either way you have seven chances to win, so the probability is the same.

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but in the end I know the chance is the same

Yes, the chance for both events occurring are equally likely.

Using this information, you can determine it is better to pay more to choose seven numbers, if that payment amount is less than the amount when paying $7$ times for a card with $6$ numbers.

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