The first one would have two shots in 48 to get one of his selections, because the op listed two players he is looking at which would be 2 chances in 48 or a ratio of 23/24. That means the first pick has a 95.8% chance of not selecting one of the two guys the op is looking to select.
so you are believing seriously that he would draft two players, before the third team get his first draft pick? I was pretty sure, he will draft hust his number 1, which makes it to a 1/48 chanche.
That means the first pick has a 95.8% chance of not selecting one of the two guys the op is looking to select. That has to happen first, then the team drafting two would need to select the other draftee at a 1 chance in 47 so there is a 97.9% chance of that not happening
now it becomes more curious, you say that the first draft will get the first draft of the thread opener in 4,2% of the cases, but when the third team draft his number one pick would be drafted just in 2,1% of the cases. Did he come back after he was drafted?
So the 1/47 is in reality a negativ value, which free the player in 50% of the cases from team one :)
So Crazyeye, before you go correcting someone, make sure they are wrong :)
i am, you are?