There is a 1/48 chance for the first team to get the pick. The second team has a 1/47 chance, the third has 1/46, and the 15th team would have a 1/34 chance. However, using probably the event would be, "What is the probability that none of the first 15 teams pick player 34?"
Therefore P(event) = 47/48 x 46/47 x 45/46 ... x 33/34
If you multiply all 15 individual probabilities then you will see that P(event) = .6875. Which means that there is a 68.75% chance that player 34 (the player in question) will NOT be picked by the time team 16 makes his draft selection.
This would be assuming that all picks are made at random so 69% was basically right on. All those that say otherwise are wrong.
You can also do this simpler by just taking 33/48 as there are 15 teams picking before. Not all scenarios are you able to do the stats this simply, but this particular one you can.
Whoever said that there is a 1/48 then a 1/47 then a 1/46 for the first 3 teams is also correct. You would just have to multiply all 15 of those probabilities together and subtract from 1. This would also give you the same answer...
Last edited by Stauder at 6/19/2011 8:19:02 PM