Probability of choosing objects same set of objects from a larger set?

Hello,

I am trying to figure out what formula I need to figure out this problem... I can't remember how to figure this out: if one person chooses three objects from a set of 50, what are the odds that a second person would pick the same three objects? I know it's probably not a lottery formula because the order of choice doesn't matter.... How do I figure the odds of this happening? Thank you very much!

I assume you mean that the three objects are returned before the new draw, and that all 50 objects are different from each other. It is a lottery formula. The number of ways to choose n objects (order not important) from a group of k objects is k!/(n!(k-n)!). In your case 50!/(3!47!) = 19600. So, the chances of choosing the same three objects again is 1/19600.

If order was important the formula would have been k!/(k-n)! in your case 50*49*48 (50 choices for the first item, 49 choices for the second, etc)

Dividing this by n! accounts for the n! orders they could have been chosen in. There are 3*2*1 orders in which the same three objects could have been chosen.

ETA: It doesn't matter about the objects being the same as a previous draw. The odds in a typical numbered PP ball lottery of getting the same result as yesterday are the same as getting 1,2,3,4,5,6 (or however many balls are drawn) are the same as getting whatever happens to be on your ticket.

Thank you!

Hello Counter!

Yes, you were exactly right, they got returned to the set after the first draw. Thank you so much, that was so, so helpful. I was trying to recall back to high school and college math. That was exactly the formula I needed, thank you so much!!! I appreciate your time in answering my post! Have a great weekend.