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Math Help - Lottery Probability 49 Choose 6

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    Lottery Probability 49 Choose 6

    Any input regarding this question would really be appreciated. I know that all possibilities of 49C6 is 13,983,816. So the odd of matching 6 numbers is 1/49C6, but let's suppose I choose 10 numbers and play all the combinations of the 10 ten numbers which is 10C6=210. Therefore if any of the 6 numbers that's is in my 10 number domain I am gauranteed to win the 1st prize along with other smaller prizes. My question is what are my probability of winning the lottery now if I play 10 numbers and of the 10 chosen numbers choose all the combinations of 6. I hope this make sense.. I thought the answer would be 10C6/49C6 or 210/13,983,816, but I know this is not correct. Please explain your answer. Thanks.
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    Quote Originally Posted by espn168 View Post
    Any input regarding this question would really be appreciated. I know that all possibilities of 49C6 is 13,983,816. So the odd of matching 6 numbers is 1/49C6, but let's suppose I choose 10 numbers and play all the combinations of the 10 ten numbers which is 10C6=210. Therefore if any of the 6 numbers that's is in my 10 number domain I am gauranteed to win the 1st prize along with other smaller prizes. My question is what are my probability of winning the lottery now if I play 10 numbers and of the 10 chosen numbers choose all the combinations of 6. I hope this make sense.. I thought the answer would be 10C6/49C6 or 210/13,983,816, but I know this is not correct. Please explain your answer. Thanks.
    10C6/49C6 is right.
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