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Coin....

That coin flip lowered our chance to fall to 7th noticeably. Nice.

We have an 81% chance of landing a top-5 pick.

What were our odds of a top-5 pick before we fell into a tie with Boston?


Edit: Our odds of landing outside the top 6 were 17.2% before, meaning our win at the end of the season only added a 1.8% risk of falling out of the top 5. That's what I was worried about.
 
My calculations seem to be a bit different than the couple I've seen online.

For pick probabilities, I have:

1: 10.400%
2: 11.201%
3: 12.092%
4: 9.855%
5: 37.291%
6: 17.747%
7: 1.415%

nice, i copied mine from a news article (who knows how they got to it). we have decent shot at something good this year
 
nice, i copied mine from a news article (who knows how they got to it). we have decent shot at something good this year
I wrote some code that allows me to adjust the number of balls each team gets. If you're curious what the probability of any particular lottery outcome is, I've got that info (although calculating this is pretty straightforward).
 
I wrote some code that allows me to adjust the number of balls each team gets. If you're curious what the probability of any particular lottery outcome is, I've got that info.

that's pretty slick. What languages do you code in? I've been learning on the job some different languages, (python and javascript/html/css are really the only things i'm knowledgeable with at this point), about to start learning java though.
 
Thought I'd post this here.

@Mitch_Lawrence: #Utah Jazz ready to pull the plug on Tyrone Corbin and go for a new coach, according to #Jazz exec.
 
that's pretty slick. What languages do you code in? I've been learning on the job some different languages, (python and javascript/html/css are really the only things i'm knowledgeable with at this point), about to start learning java though.
It's in Matlab (which I think is pretty similar to Python). Not a coder, really...I use STATA more than anything else.

This particular code is pretty ****ing clumsy, but I'm pretty sure it's accurate (this is the code for this year's lottery, with the Jazz and Celtics splitting the balls at 4th and 5th):

format long

%The default Balls vector (no ties) is: [250 199 156 119 88 63 43 28 17 11 8 7 6 5]
Balls = [250 199 156 104 103 63 43 28 17 11 8 7 6 5]';


%Creating probability matrix of possible outcomes
ProbMat = zeros(14,14,14);
one = 1;
while one < 15
two = 1;
while two < 15
three = 1;
while three < 15
ProbMat(one,two,three) = (Balls(one)/sum(Balls))*(Balls(two)/(sum(Balls)-Balls(one)))*(Balls(three)/(sum(Balls)-Balls(one)-Balls(two)));
if one == two
ProbMat(one,two,three) = 0;
end
if one == three
ProbMat(one,two,three) = 0;
end
if two == three
ProbMat(one,two,three) = 0;
end
three = three + 1;
end
two = two + 1;
end
one = one + 1;
end


%Matrix of Pick Probabilities
ProbPick = zeros(14,14);


%Calculating Prob of Top 3 Pick
l = 1;
while l < 15
n = 1;
while n < 15
t = 1;
while t < 15
ProbPick(l,1) = ProbPick(l,1) + ProbMat(l,n,t);
ProbPick(l,2) = ProbPick(l,2) + ProbMat(n,l,t);
ProbPick(l,3) = ProbPick(l,3) + ProbMat(n,t,l);
t = t + 1;
end
n = n + 1;
end
l = l + 1;
end


%Calculating Prob of keeping pick (for teams in the 4 - 14 slots)
n = 4;
while n < 15
one = 1;
while one < n
two = 1;
while two < n
three = 1;
while three < n
ProbPick(n,n) = ProbPick(n,n) + ProbMat(one,two,three);
three = three + 1;
end
two = two + 1;
end
one = one + 1;
end
n = n + 1;
end


%Calculating Prob of sliding three slots (for teams in the 1 - 11 slots)
n = 1;
while n < 12
r = n + 1;
while r < 15
s = n + 1;
while s < 15
t = n + 1;
while t < 15
ProbPick(n,n+3) = ProbPick(n,n+3) + ProbMat(r,s,t);
t = t + 1;
end
s = s + 1;
end
r = r + 1;
end
n = n + 1;
end


%Calculating prob of sliding two slots (for teams in the 2 - 12 slots)
n = 2;
while n < 13
r = 1;
while r < n
s = n + 1;
while s < 15
t = n + 1;
while t < 15
ProbPick(n,n+2) = ProbPick(n,n+2) + ProbMat(r,s,t) + ProbMat(s,r,t) + ProbMat(s,t,r);
t = t + 1;
end
s = s + 1;
end
r = r + 1;
end
n = n + 1;
end


%Calculating prob of sliding one slot (for teams in the 3 - 13 slots)
n = 3;
while n < 14
r = 1;
while r < n
s = 1;
while s < n
t = n + 1;
while t < 15
ProbPick(n,n+1) = ProbPick(n,n+1) + ProbMat(r,s,t) + ProbMat(r,t,s) + ProbMat(t,r,s);
t = t + 1;
end
s = s + 1;
end
r = r + 1;
end
n = n + 1;
end
ProbPickT = ProbPick'
 
There's only about a 10% chance that the worst 3 teams all end up picking in the top 3 (in any order). There's roughly a 34% chance the Jazz end up in the top 3. That leaves a 56% chance that at least one non-Jazz team outside the top 3 ends up picking in the top 3 and the Jazz don't pick in the top 3, which is about 5.5x more likely than the top 3 picking in the top 3, which is what's required for the Jazz to end up picking 4th. That's why 5th is the most likely outcome.

6th is more likely than 4th because it's more likely 2 non-bottom-3 teams end up picking in the top 3 than all 3 bottom-3 teams picking in the top 3 (again, 10% chance).

Nice, thanks. That oddly makes sense.
 
It's in Matlab (which I think is pretty similar to Python). Not a coder, really...I use STATA more than anything else.

This particular code is pretty ****ing clumsy, but I'm pretty sure it's accurate:

nice, well done!

yeah the syntax looks pretty similar to python. My prof's recommended i learn to use matlab, i defiantly never touched it, and regret it now looking back.
 
How do you correctly calculate the possibility of neither Boston nor Utah getting into the top 3?

Chance of Utah landing top 3 = 23.7%
Chance of Boston landing top 3 = 23.4%
Chance of both Utah and Boston landing top 3 = (23.7%) * (23.4%) = 5.55%
Chance of neither Boston nor Utah landing in the top 3 = (76.3%) * (76.6%) = 58.45%

Is this correct?

Then the chance of Utah landing top 3, while Boston does not land top 3 = (23.7% * 76.6%) = 18.15%

So then there is a greater than 76% chance that the Jazz will, in fact, pick ahead of Boston (i.e., Utah lands top 3 while Boston does not + neither Utah nor Boston land top 3).

Hence, we've beaten out Boston in over 76% of scenarios.

The coin flip matters.
 
How do you correctly calculate the possibility of neither Boston nor Utah getting into the top 3?

Chance of Utah landing top 3 = 23.7%
Chance of Boston landing top 3 = 23.4%
Chance of both Utah and Boston landing top 3 = (23.7%) * (23.4%) = 5.55%
Chance of neither Boston nor Utah landing in the top 3 = (76.3%) * (76.6%) = 58.45%

Is this correct?

if i remember statistics, 'AND' sequences (condition1 AND condition2) get treated with multiplication yes, but as GVC pointed out, we have about a ~33.69% chance at the top 3, not a 23.7%.
 
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