In order to select draft position the NBA has a lottery system. The odds for each team winning the lottery are as follows:
Portland: 250
New York (To Chicago): 199
Charlotte: 138
Atlanta: 137
Toronto: 88
Minnesota: 53
Boston: 53
Houston: 23
Golden State: 22
Seattle: 11
Orlando: 8
New Orleans/Oklahoma City: 7
Philadelphia: 6
Utah: 5
What happens is that they randomly pick out a number or something to determine the draft order for the top three picks, then they go by order. Once a team is selected they are ineligible for any other picks but they are still in the lottery. In other words, they still have their names in the hat but if they are drawn it is ignored and another slip is drawn. What I want to calculate is the probability that New York (to Chicago) gets 1-5.
Its obvious that they have a 19.9% chance to get the first pick. Beyond that I am at a loss on how to calculate the odds of someone winning becuase it depends on who wins the lottery. For example, if Utah wins the first pick than NY (to Chi) still has basically a 19.9% chance of getting the second pick. On the other hand if Portland wins the first pick the probability that NY (to Chi) gets the second pick jumps up to 26.5%.
Besides that I can’t calculate anything easily becuase I don’t know how to get around the problem of the dependence on the odds on the previous results. Any ideas?
I think the dependences balance out in the wash. In other words, ignore all that. You have 199 balls out of the total, and 5 picks. The rest is irrelevant.
If you really want the exact probability, you can either make a big-ass tree diagram with all the possibilities, or you can make a big-ass Markov chain. Either way, it’s going to be a lot of work.
I don’t understand. 199/1000 is good enough for the first pick but its not good enough for anything else.
Thats kind of what I figured. I have been messing around in excel a little bit but haven’t figured out an easy way to get the probability for the 3rd, 4th and 5th picks. The chance of getting the second pick is by my calculations is 19.2%. If there is a computer program that does what I want. there is an off-chance I have access to it in a computer lab here at school.
Ach, I didn’t see that the picking stops after 3. Sheesh. I reread the OP a few times, I don’t understand the question. Is it the probability taht NY gets one of the top 5 picks? That seems like it’s 100%. Or is the probability of getting pick 1, then the p of pick 2, etc. This http://geocities.com/benschuarmer/draftlotto.html may help for that. It claims
If I’m following you correctly, the odds of getting any of the top 3 picks is 199/1000 (as Portland’s balls would still be in the hopper for any selection).
The odds of getting the #4 pick is dependant on the odds of “Portland plus any other two teams” landing in the top 3. The odds of getting the #5 pick is dependant on the odds of “Any three teams excepting Portland” landing in the top 3.
That’s as far as I can get without putting pen to paper
If the same number of balls are in the hopper for each of the first three picks, the above can’t be correct. Or is there some accounting for the chance of NY getting the first pick being subtracted from the odds of NY getting 2 or 3?
Yes, the odds for the second pick are more complicated, because they depend on what happened in the first pick. You have to go through all the possibilities for the first pick, and add up each of their probabilities for the second pick. Something like:
25% of the time Portland gets the first pick, so NY has 199/750 chance of getting #2.
19.9% NY gets the first pick, and has 0 chance of #2.
13.8% Charlotte, so NY has 199/862 of #2.
etc.
So NY’s total chance of getting #2 is: 0.25( 199/750 ) + 0 + 0.138(199/862) + …
