Jumping off from the previous post, it seems to me that a good a priori statistic to know would be the following:
Suppose that two teams meet in a 7 game series. If you suppose that they are evenly matched, then you'd expect Team A to win the series 50% of the time. Okay. Now, suppose Team A loses the first game. A is down 0-1. What would have to be the odds of A to win each game such that Team A would have a 50% chance of winning the series? Does Team A have to be 55-45 good? 60-40? 70-30? What would bring the odds for winning the series back to parity? This, I think, would be a good set of figures to know to have a good intuitive grasp of how improbable a series win is for a team that falls behind.
Of course, these figures are just fixed mathematical calculations that you could apply in any sport. Unfortunately I don't know how to do the math to calculate them out, but maybe some bored genius will help us out.
Sunday, October 18, 2009
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In your scenario, Team A needs to win 4 out of the remaining 6 games to win the series. Let's use P to denote their prior probability of winning a game, which means that Team B's prior odds are (1-P). The question is: what value of P that would give Team A an even shot at the series?
The basic idea here is this: for each possible set of 4 games that Team A could win, calculate the probability that they win exactly that set of games, and sum all of these probabilities. It's useful to know that there are 15 ways to choose 4 out of the remaining 6 games. For example, Team A could win games 2,3,4,5; or they could win games 2,3,5,7; etc. (Note that you can google "6 choose 4", or any other such calculation.)
This would now be pretty straightforward if it weren't for the fact that the series doesn't always go to 7 games. Because the series ends when someone wins 4 games however, the only way I can think to do this is a bit more tedious:
How many ways are there for Team A to lose zero more games? Well, just 1, because they have to win four straight.
The probability of Team A winning four games is P*P*P*P, or more succinctly, P^4.
How many ways are there for them to lose one more game? Losing exactly one more game means that they win game six, so we need to choose one game out of games 2-5 for them to win, and "4 choose 1" = 4.
The probability of Team A winning four and losing one is (P^4)*(1-P).
How many ways are there for them to lose two more games? This means they win game seven, so we now need to choose two games out of games 2-6 for them to lose; "5 choose 2" = 10.
The probability of Team A winning four and losing two is (P^4)*((1-P)^2).
(As a sanity check, notice that we've covered all 15 possibilities, because 1+4+10 = 15.)
Finally, we want to solve for P in the following ugly mess:
1*P^4 + 4*(P^4)*(1-P) + 10*(P^4)*((1-P)^2) = 0.5
Fortunately, technology has come far enough that we can really just copy and paste that thing into WolframAlpha, and find out that Team A needs to have a 57.8593% prior probability of winning to have a 50% shot at winning the series.
Sorry, slight typo: when discussing how many ways they can lose exactly one more game, I should have said "we need to choose one game out of games 2-5 for them to lose".
Also, sorry if this was totally unenlightening; I really can't tell.
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