#Inline statbook binomial calculator how to

They hit in the 4th Q against a Pelicans teams that had gone about 8 or more games without a miss. Knicks played last night, they had missed the previous 2 games so I had them as a lock of sorts to hit, but also don't have much faith in those numbers yet. And so far no one has missed 3 games in a row. So far, probably have stats on 10 or so teams. I'm in the initial stages of getting that data and applying it. So far I fade a team that has gone over the average amount of games to get a miss and expect a team high on misses to get some hits.

#Inline statbook binomial calculator how to

Otherwise, so far, stats seems to be in agreement by team.Įventually I would like to figure out how to calculate regression to see specifically how hot or cold a team is running on the metric. So far 1 miss in 36 games with still about 9 games to look over. I was just going over the LA Lakers games and they seem to be an outlier. So far Q2-Q4 and miss numbers seem to follow expected probabilities (Q2 roughly 50%, Q3 roughly 25%, Q4 roughly 12.5%, and miss altogether 12.5%). I am compiling stats by team on how often the team hits, in which Q and how often they miss. I may be wrong with that but it's just the feeling. It just feels like 2 particular teams set the stage with whatever happens in Q1 and then they play it out from there, but say like if Q2 misses, then Q3 misses and then Q4 misses, whether it is the case or not, it feels more in line mathematically to let the next game's Q1 play out and to go from there instead of playing straight into Q1 from another game's Q4.

One thing I am considering is that 1 team may have a higher tendency to score odd or even than another and this of course would mean that one matchup in a game may be further from being 50-50 than another. But all things considered seems as if one quarter to another is roughly 50 or at least independent. How does that change things? I'm guessing it will be one of the other values on the calculator, probalby 0.219ġ) even or odd are in play depending on Q1.Ģ) Yes it is hard to say for sure if the probability is 50% I feel it is very close if not. P(X > x)' for 2 games according to the calculator, but I am considering odd or even in Q1 and a hit being the opposite hitting any time, Q2, Q3 or Q4. Looks like you found the 'Cumulative probability: If Q1 is even, I bet odd until odd hits for the rest of the game. If it's odd I bet against odd hitting in all of the remaining qtrs.

I don't know if this is an ideal strategy but I typically see what happens in Q1. Miss is just in relation to a team hitting a different O/E value from Q1. In this case a hit is a win because I am betting against all quarters being the same. If any quarter Q2-Q4 is opposite of Q1 then that's a hit. You could call it either way but for discussion any Q2-Q4 that is the same as Q1 (odd or even) is a miss. I hope somewhere in this response I answered your question. I would strongly think so but can’t be 100% certain.ģ.Ğven more unlikely is the possibility that the games are not independent or the probability is not constant, two conditions needed for the binomial to apply. Given Q1 is odd, is the probability each subsequent quarter has a 50% probability of being odd. The above assumes if the Q1 score is even then that game is not considered.Ģ. If the question is, for example, what is the probability of a hit in G1 or a hit in G2 or a hit in both, that is a different calculation, e.g., 2*0.125 – 0.125^2 = 0.2344 = 1 - 0.875^2 if both games meet the Q1 condition, whatever that is.ġ. Then you correctly used the binomial to find the probability of exactly one hit in N games. You gave that a (1/2)^3 = 12.5% probability. It seems like it is Q2-Q4 all odd given Q1 is odd. I’m not sure what you consider to be a miss (= success).