# Probability of 1 of 7 golfers out of 130 placing in top 6

If I choose 7 golfers out of 130, and all golfers in the field have an equal chance of winning, how do I work out the probability of at least 1 golfer from my chosen 7 finishing in the top 6?

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Sorting the golfers from the winner to the loser (after the game), there are $\binom {130} 7$ ways to choose the golfers in general, and $\binom {124} 7$ ways to choose the golfers in a way that none of them end up in the top 6. So, the answer is
$$1 - \frac {\binom {124} 7} {\binom {130} 7}$$
Hint: The chance that none of your golfers winning is $\frac {123}{130}$. Given that none of yours win, the chance that none finish second is $\frac {122}{129}$. Keep going and multiply to get the chance than none are in the top six. Then subtract from $1$ for the chance that at least one is in the top six.