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Basketball Playoffs Probability Question

Posted By: Kevin Whyte
Date: Monday, 29 April 2013, at 7:19 p.m.

In Response To: Basketball Playoffs Probability Question (Bob Koca)

That was my intuition as well, but it is misleading. One thing to realize is that if everyone wins their home games then the schedule is irrelevant - the higher seed will win the series, at home, whenever their 4th home game occurs.

If the probability of a road win is really low, then it's easy to see what's happening. In that case we need only look at the scenarios with exactly one home loss. With the 3-4 schedule, if that occurs in the first three games the higher seed still wins (and at home), while if it occurs in the last 4 the lower seed wins (on the road). Thus the probability of the winning team doing so at home is (asymptotically as p->0) 1-4p where p is the chance of a road win.

With the schedule of 1-4-2 (there was a typo in my original as I outlined an 8 game schedule, not a 7 game one) then any scenario with only one road win has the clinching game at home. Thus the probability is 1- O(p^2) as p>0, which is definitely higher.

As to why we should maximize this, I'm not sure. It certainly doesn't maximize revenue. It is better for the fans at the game, but maybe irrelevant for the TV viewers. I admit to being more a fan of hockey than basketball, and that is more gate driven. Mainly I mentioned it because it seemed like an amusing probability question. I still don't see a slick argument that works for all p.

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