What a shot!
A young man stands on the top tier in a football stadium
(no, not soccer). He throws a basketball to the football field
below (let's assume it's 500 feet horizontal and 50 foot lower)
to a bastetball goal, and makes it. What was the probability of
making this basket?
Assuming that he has tried many, many times, and 85%
of the time he can come within a 20 foot radius of the
center of goal.
You can see this video at http://RagKnot.blogspot.com
(Texas A&M University is where I went to school.)
To save time: diameter of goal 17 inches. diameter of ball 9 inches.
The shot was what we call a bank shot, not a "net only".
(An only net would be harder to make)
(no, not soccer). He throws a basketball to the football field
below (let's assume it's 500 feet horizontal and 50 foot lower)
to a bastetball goal, and makes it. What was the probability of
making this basket?
Assuming that he has tried many, many times, and 85%
of the time he can come within a 20 foot radius of the
center of goal.
You can see this video at http://RagKnot.blogspot.com
(Texas A&M University is where I went to school.)
To save time: diameter of goal 17 inches. diameter of ball 9 inches.
The shot was what we call a bank shot, not a "net only".
(An only net would be harder to make)
Labels: Probability
posted by Ragknot at
3:53 PM
18 Comments:
Pretty good video. Have you seen:
http://www.youtube.com/watch?v=NHH-6ZQktRQ
I will make a wild guess of about 5%.
Chris,
From my rough computations, your 5% is way off.
How do you kick a comment thru?
send a 2nd comment?
Then do you send a 3rd to get the 2nd thru?
not sure but...
if its a bank shot-you have to rule out the chances of a net only shot
i got about 0.00085 ?
Make is easy.
a circle of 20 radius has x square inches inches
Now, assume the ball must passes thru a 13 inch diameter circle for a "nothing but net". Area = y square inches.
If you figure the ratio of x to y, then apply 85% to 20' radius, what is the probability of the 13 diameter?
would chances for a "bank shot" be-.063014% ?
Hi Ragknot. I stayed away from this problem as I haven't got the a normal probability distribution table to hand. I'm also aware that this should be treated as a 2-dimensional problem in view of the way you expressed your information (that also rules out 3D). So I really need a suitable circular type normal distribution table, then I'll need to know how to choose the standard deviation from the 85%, 20 foot info. I'm very rusty on doing that sort of thing. I'll try later tonight. But I've got to fix a computer too - that takes priority.
But I'm sure that the probability is very low. Less than 0.01% doesn't seem unreasonable (nor does 0.001%) - but they're very wild guesses. Greetz.
I would appreciate any critism. But this was my calculations.
A 20 foot radius circle is 180,956 sq inches
The radius of the goal is 8.5 inches. The radius of the ball is 4.5 inches.
This means the center of the ball must fit into the center of a 4 inch radius circle which gives an area of 50.27 square inches. (for a "nothing but net shot)
180,956 / 85% = 212,889
50.27 / 212,889 = 0.0236%
That would mean, 1 shot in 4,235 shots
or...
The area of the 20 foot radius circle is 3600 times the size of 4 inch radius area. And 3600/.085 = 4235
The area on the back board for making a bank shot would be roughly the same size, it simply allows for a flatter trajectory to still make the basket.
If you were to allow either a net or a backboard, I'd estimate an approximately a 1 in 2000 probability,
not this 3600/.085 = 4235
but 3600/0.85 = 4235
This post has been removed by the author.
This post has been removed by the author.
Hi Ragknoot. It looks as though one can get fairly good estimates by assuming a uniform(rather than a normal) distribution. However I get almost 2% succes rate if I assume a 1-dimensional ND problem. Quick check, 4/240 1.67%. I get a larger value as the norm dist peaks in the middle. Ignore the extra difficulties of going to a 2-dim ND. It would seem reasonable to do it by the ratio of areas as you have done. But I would have simply gone for 50/180000 = 0.028%. But I'd expect that after correcting my major approximations and not fully appreciating the backboard, to get thrice that value. Say 0.1%. i.e. I don't have a real issue with your calculations. Greetz.
Oooops. Sorry about spelling your name wrong :)
Thanks for the attempt, but I can hardly buy the almost 2%. I don't think a normal distributation would fit unless the shooter has had a lot of practice. I assumed that anywhere in the 20 foot radius was just as probable as anywhere else within the 20 foot radius.
The 50/180000 is the same as the the 1/3600 I had... the small area / large area.
But I said the probility was 85% of a hit within the 20 foot radius and I did see how you used that.
I'm not very good with probabalities. Maybe I'm close 50% of the time. LOL.
Thanks for reply.
Hi Ragknot. It's a fairly awkward problem. The 4/240 hasn't really used the 85% info. But the correct application of a 1-dim ND gives, very close to 2% - that seems reasonable to me. As for 2D, that's harder. I have stumbled across a problem elsewhere, which is about throwing darts at a dartboard with an assumed uniform probability/unit area. The mean radius is 2/3 of the actual radius (if my memory is right). i.e. you are just as likely to hit between the centre and 2/3 radius as between 2/3 radius and outer ring - that result is not "obvious" BTW. and you'd think The 2-dim ND stuff will make the mean radius even closer to the centre. Of course, I really beileve that it's a 3-dim problem, and not trivially so either because of the backboard.
Whatever, the film showed someone doing vastly better than I would.
Did you check the link I posted? My son (who's nearly 14 and already far smarter than me)suspects that it's faked - if so, what's the point of that?
Is there a basketball goal in a football field???
I DON"T THINK SO
Ahmed, listen to the commentary; the video was taken at a football stadium (in Texas).
Post a Comment
Links to this post:
Create a Link
<< Home