Yeah, Spurs stole one in Miami. To me, since I think the teams are very evenly matched, it's all about probability at this point. If you assume each team has a 50% chance of winning the next game (which doesn't take HCA into account, but work with me here):
P(Spurs win it all) = P(Spurs in 4) + P(Spurs in 5) + P(Spurs in 6) + P(Spurs in 7)
P(Spurs in 4) = 0
P(Spurs in 5) = P(Spurs win next 2 games) = C(2,2) * 0.5^2 = 2! / (2!0!) * 0.25 = 0.25
P(Spurs in 6) = P(Spurs win exactly 1 of next 2) * P(Spurs win game 6) = C(2,1) * 0.5^2 * 0.5 = 2! / (1!1!) * 0.25 * 0.5 = 0.25
P(Spurs in 7) = P(Spurs win exactly 1 of next 3) * P(Spurs win game 7) = C(3,1) * 0.5^3 * 0.5 = 3! / (2!1!) * 0.125 * 0.5 = 0.1875
P(Spurs win it all) = 0 + 0.25 + 0.25 + 0.1875 = 0.6875
So as of right now, a very conservative estimate would be that the Spurs have a 68.75% chance of winning it all if the teams are evenly matched.
If we take HCA into account, so order matters (HCA usually confers about a 5% advantage in basketball, I think--some statshead back me up on this?):
P(Spurs in 4) = 0
P(Spurs in 5) = P(Spurs win next 2 games at home) = C(2, 2) * 0.55^2 = 0.3025
P(Spurs in 6) = P(Spurs win exactly 1 of 2 games at home) * P(Spurs win game 6 on the road) = C(2,1) * 0.55 * (1 - 0.55) * (1 - 0.55) = 0.22275
P(Spurs in 7) = (P(Spurs win exactly 1 of 2 games at home) * P(Spurs lose game 6 on the road) + P(Spurs lose next 2 games at home) * P(Spurs win game 6 on the road)) * P(Spurs win game 7 on the road) = (C(2,1) * 0.55 * (1 - 0.55) * 0.55 + C(2,0) * (1 - 0.55)^2 * (1 - 0.55)) * (1 - 0.55) = 0.16351875
P(Spurs win, taking HCA into account) = 0 + 0.3025 + 0.22275 + 0.16351875 = 0.68876875
So taking HCA into account the Spurs have basically the same chance of winning it all--around a 68.88% chance--as if we pretended HCA didn't exist. This makes sense--stealing one in Miami just evens the playing field.
(I feel dirty. Fuck advanced stats).
P(Spurs win it all) = P(Spurs in 4) + P(Spurs in 5) + P(Spurs in 6) + P(Spurs in 7)
P(Spurs in 4) = 0
P(Spurs in 5) = P(Spurs win next 2 games) = C(2,2) * 0.5^2 = 2! / (2!0!) * 0.25 = 0.25
P(Spurs in 6) = P(Spurs win exactly 1 of next 2) * P(Spurs win game 6) = C(2,1) * 0.5^2 * 0.5 = 2! / (1!1!) * 0.25 * 0.5 = 0.25
P(Spurs in 7) = P(Spurs win exactly 1 of next 3) * P(Spurs win game 7) = C(3,1) * 0.5^3 * 0.5 = 3! / (2!1!) * 0.125 * 0.5 = 0.1875
P(Spurs win it all) = 0 + 0.25 + 0.25 + 0.1875 = 0.6875
So as of right now, a very conservative estimate would be that the Spurs have a 68.75% chance of winning it all if the teams are evenly matched.
If we take HCA into account, so order matters (HCA usually confers about a 5% advantage in basketball, I think--some statshead back me up on this?):
P(Spurs in 4) = 0
P(Spurs in 5) = P(Spurs win next 2 games at home) = C(2, 2) * 0.55^2 = 0.3025
P(Spurs in 6) = P(Spurs win exactly 1 of 2 games at home) * P(Spurs win game 6 on the road) = C(2,1) * 0.55 * (1 - 0.55) * (1 - 0.55) = 0.22275
P(Spurs in 7) = (P(Spurs win exactly 1 of 2 games at home) * P(Spurs lose game 6 on the road) + P(Spurs lose next 2 games at home) * P(Spurs win game 6 on the road)) * P(Spurs win game 7 on the road) = (C(2,1) * 0.55 * (1 - 0.55) * 0.55 + C(2,0) * (1 - 0.55)^2 * (1 - 0.55)) * (1 - 0.55) = 0.16351875
P(Spurs win, taking HCA into account) = 0 + 0.3025 + 0.22275 + 0.16351875 = 0.68876875
So taking HCA into account the Spurs have basically the same chance of winning it all--around a 68.88% chance--as if we pretended HCA didn't exist. This makes sense--stealing one in Miami just evens the playing field.
(I feel dirty. Fuck advanced stats).