None of you guys took/remember statistics. If you have a 50% chance to get heads, does that mean if you flip a coin twice, you have a 100% chance of getting it once? The answer is no. What you do is you have a 50% chance of not getting heads once and a 50% chance of not getting heads again. You multiply the odds of it not happening and then subtract this number from 1. Odds = 1-((1-.5)*(1-.5))=.75 or 75%. So based on those numbers NJ odds are 1-((1-0.053)*(1-0.06)*(1-0.073))=17.48% And for us it's 82.52% of keeping their pick.
This is why I asked the question, but I'm not sure that you're correct. In this instance, there's only one "flip". Given NJ's current position, it has a 5.3% chance of being the #1 pick, a 6% chance of being the #2, and a 7.3% chance of being the #3 pick. It seems to me that addition of those 3 numbers is the correct way to proceed here, but I'm more than willing to be shown otherwise.
Agree. No way its rigged, especially now days. Way to much at stake if they got caught. It'd be like the Saints situation in the NFL but 100 times worse.
e_blazer, I am taking a statistics refresher course April 30 if you can wait that long. My recollection (damn, college was a long time ago) is that westnob is correct. The question is whether these are 3 independent events like coin toss or 3 related events where one outcome precludes any others.
Or... we could just say that there is a 50/50 chance. We either will or we wont get the pick this year.
As of today, I think any discussion of probabilities is a waste of time. Looking at the standings, there are 4 teams with 43 losses. NOH has 44 losses. If everybody loses out the rest of their remaining games (yeah, tankers) then NOH goes into the lottery with the 3rd spot, Sacramento & Cleveland would flip for the 4 & 5 spots, and NJ and Toronto would flip for 6 & 7. All it would take is a win by one or two of these teams to throw everything up in the air. Washington, currently with the 2nd pick, is for some insane reason on a 4 game winning streak. Their management couldn't be so inept as to allow them to win any more could they? Crazy.
they do play the cavs next we should be happy NJ decided to win a few games when they did 2 weeks ago. they could easily be in 2nd right now.
Yeah, I didn't consider e_blazer's point, but I think crandc is right. Each time you draw a ping pong ball, it's another instance. And technically each time someone else gets drawn the chances change slightly.
When the Blazers won the 2007 lotto they had a winning% of .390; If the Blazers lose the last game against Utah they'll end the season with a winning% of .420. We need this!
