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What are your chances of winning in the following scenario?

Consider a lottery game with 49 numbers and 6 are chosen as the winning numbers. You're allowed to select 10.

What is the probability that your selection contains all 6 winning numbers? (the remaining 4 can be anything else)

Please confirm if this is correct:

${43 \choose 4}$ / ${49 \choose 10}$

Reasoning:

From a selection of 10 numbers, there are ${43 \choose 4}$ combinations which contain the winning 6; i.e. ${6 \choose 6}$ * ${43 \choose 4}$

divided by the total number of combinations with 10 numbers: ${49 \choose 10}$

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    See your $3^{rd}$ para. Is it correct? – 2017-01-15
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    Thank you for the note, now fixed – 2017-01-15

3 Answers 3

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Going by what your question says, the remaining $4$ tickets don't matter, what is important is the winning $6$ tickets. And since you have all $6$ tickets, you are definitely going to win. So probability is $1$.

Hope this helps you.

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In this case your winning probability is 1 as you have all 6 winning numbers.

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lol 100% if you can predict lottery numbers using mathematics..... https://www.facebook.com/profile.php?id=100009190105045 top post on jan 10th prior to draw .... predicted ONLY 4 numbers... predicted 2 -that was drawn... predicted 17- that was drawn.... predicted 43 - that was drawn..... predicted 23 but as a comment stated "it could be 22 23 or 24" well 22 was drawn... that is 4 out of 4 ...