so when we land the 14th pick again can we stop then know that stern didn't owe us a top 3 pick since he kept Gasol from us..?
Past events don't change future outcomes in this case, so no. If 2 years ago you said, "what's the chance we get a top 3 pick with 3 #14 finishes," then it would be 5.3% or whatever. But as of now, we have a 1.8% to finish in the top three.
"Rockets general manager Daryl Morey was not sure where he would be or if he would be where he could watch tonight's NBA draft lottery" What?
Is that even right, though? Because we can't get all top 3 picks. By my math, the highest odds we'll have to get one of the top 3 picks would be if the lotto went in order for the top 2 picks, so for pick #3 we'd have 5/551 chance, or 0.9% that may not be the right way to word it, or how probability works exactly, but seems logical in my head at the moment!
0.9% is the highest odds of getting the #3 pick. You left out the odds of each of the prior 2 picks. It is addition. Don't over-think it...think of it as three lotteries. Each draw of balls in the lottery has different odds based on who already won and can't win again...but they add up...since it is three, separate lottery drawings. Our worst odds of getting one of the picks is a little over 1.5%, and that occurs when the 12th, 13th and 14th picks all move into the top 3. Our best odds are somewhere around 2.07% in your scenario when the #1, #2 and #14 pick win. The range is 1.5 - 2.07%, and that's why they say our overall odds are somewhere in the middle at 1.8% covering all scenarios. Don't ask me to demonstrate that.
if Parsons lucky hand nets us a top 3 pick i say we all chip in 5 dollars each and get him the best hooker money can buy.
Kinda funny, if we get lucky and get either 2 or 3, he could be the good luck charm that helped us draft his replacement lol
I forget, who did we send last year? I remember we sent Brooks two years ago, because we were saying how funny it would be if we won the lottery for John Wall.
I always have problems conceptualizing why it's addition. Why aren't you also adding the denominator? In other words, if you ran all three scenarios together. Let's say the top two teams get the number 1 and two pick. So then what you have is 1,000 number options for the first draw, 750 for the second and 551 for the third. 2,301 number options. Of those, 15 would be assigned to the Rockets. Why isn't the probability then 15/2301 = 0.65% in this hypothetical? Not the biggest probability fan... though I do know Monty Hall's Dilemma!
You can't add fractions that have uncommon denominators. What is that game show about losing to 4 graders? 1/2 + 1/3 != 2/5 The math is: Code: 5 6.6666 9.074 20.74 ------ + --------- + -------- = ------- = 2.07% 1,000 1,000 1,000 1,000 5 of 750 balls equals 6.6666 of 1,000 5 of 551 balls equals 9.074 of 1,000 so, 2.07% is our highest odds when the #1 and #2 go in order.
