2009
2009-01-28 12:37:01
rating 2.9
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2009-01-28 12:37:01
23 votes, rating 2.9
23 votes, rating 2.9
Probability after a hail mary pass to scatter in a zone with diving catch
Perhaps someone has calculated already the probability for following:
Player throws successfull a hail mary pass.
Ball scatters 3 squares.
Probability for the ball to scatter in a field the player with diving catch can reach.
It is 50%?
Player throws successfull a hail mary pass.
Ball scatters 3 squares.
Probability for the ball to scatter in a field the player with diving catch can reach.
It is 50%?
Comments
Posted by UberEvil on 2009-01-28 13:40:35
Since I found it amusing I did a quick check using a Markov Chains and these are the probabilities:
Scatter back to catcher: 3/64
Scatter 1 square away from catcher (withing Diving Catch reach): 27/64
Scatter 2 squares away from catcher: 21/64
Scatter 3 squares away from catcher: 13/64
Over and out!
Scatter back to catcher: 3/64
Scatter 1 square away from catcher (withing Diving Catch reach): 27/64
Scatter 2 squares away from catcher: 21/64
Scatter 3 squares away from catcher: 13/64
Over and out!
Posted by wackyone on 2009-01-28 13:44:49
Slightly less, I think.
Between 47,9% and 49,5% (beware - calculations made with a get-results-fast model... may or may not have an actual basis in reality).
Between 47,9% and 49,5% (beware - calculations made with a get-results-fast model... may or may not have an actual basis in reality).
Posted by UberEvil on 2009-01-28 13:53:07
You were slightly off wackyone: (3+27)/64 = 46,875% ;)
Posted by wackyone on 2009-01-28 17:25:58
oh... then the get-results-quick model was only almost based in reality XD
could have been worse, I guess...
could have been worse, I guess...
Posted by UberEvil on 2009-01-28 21:49:37
Could be way worse. I'm a bit interested in how you obtained your numbers though. Simulated?
Posted by wackyone on 2009-01-28 23:36:59
Manually, actually, just with a little... imperfection ;)
I took the chance it was in range in one dimension, and then just squared it.
In one dimension 1,2,3 = +1; 4,5 = 0 and 6,7,8 = -1... and now I can treat it like a permutation problem (with 3 values - I can evaluate all the combinations easily).
It's not correct, it adds a "central" scatter (a chance that the ball actually doesn't scatter) - that's what the disclamer was for!
I think that, to get the 2 percentages, once I considered all the permutations, and once I excluded the "0-0-0" cases... I would have thought that excluding those I'd get a lower-than-real percentage, to get semi-decent boundaries!
I took the chance it was in range in one dimension, and then just squared it.
In one dimension 1,2,3 = +1; 4,5 = 0 and 6,7,8 = -1... and now I can treat it like a permutation problem (with 3 values - I can evaluate all the combinations easily).
It's not correct, it adds a "central" scatter (a chance that the ball actually doesn't scatter) - that's what the disclamer was for!
I think that, to get the 2 percentages, once I considered all the permutations, and once I excluded the "0-0-0" cases... I would have thought that excluding those I'd get a lower-than-real percentage, to get semi-decent boundaries!