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November 5

7-game series

The above question about the World Series got me wondering: Suppose the teams playing in it are Team H and Team A, with Team H having home field advantage. The World Series uses a 2-3-2 home-away-home format. If Team H's probability of winning any game against Team A is a constant x when Team H is at home and a constant y when Team H is away (with x > y due to home field advantage) (obviously this is simplified from reality, since chance of winning any particular game in the series is not going to be constant due to variables such as changing weather and injury conditions), then:

What is the formula for the probability that Team H wins the series?
If the series format was changed to 2-2-1-1-1 instead of 2-3-2, would the probability that Team H wins the series change or would it still be the same?

—SeekingAnswers (reply) 01:29, 5 November 2014 (UTC)[reply]

Somewhat counterintuitively, the order of home vs. away games does not matter, only the numbers of home and away games. To see this, note that if the series does no go to the seventh game, the remaining games could still be played without affecting the final outcome. So you might as well calculate the probabilities as if they were being played. Now the probability of winning the series is the probability of winning exactly 4 games plus the probability of winning 5 games plus the probability of winning 6 games plus the probability of winning 7 games. But none of these probabilities depends on the home/away order. That answers the second question. For the first question, I get x⁴+12x³y+18x²y²+4xy³-12x⁴y-48x³y²-24x²y³+30x⁴y²+40x³y³-20x⁴y³. More generally, if the probability of winning game i is x_i, then the probability of winning the series is e₄ - 4e₅ + 10e₆ - 20e₇, where e_i is the ith elementary symmetric polynomial in the x_i's. The fact that there are more home games than away in the format gives an advantage to the home team, but changing the order won't have any effect. This analysis may also apply to racket games such as tennis where having the serve gives an advantage. But the scoring of those games is more complicated so additional factors would have to be taken into account. --RDBury (talk) 13:08, 5 November 2014 (UTC)[reply]

Could you elaborate on how you arrived at x⁴+12x³y+18x²y²+4xy³-12x⁴y-48x³y²-24x²y³+30x⁴y²+40x³y³-20x⁴y³? Thanks. —SeekingAnswers (reply) 01:39, 6 November 2014 (UTC)[reply]

From the more general formula and evaluating e₄(x, x, x, x, y, y, y) = x⁴+12x³y+18x²y²+4xy³, etc. The more general formula can be derived as follows: Let f be the probability of winning the series. I know that f is a polynomial in the variable x_i because it can be written as the sum of the probabilities of all possible ways of winning, and each of these probabilities is a product of factors x_i or 1-x_i. In fact, if you look closely as these products you'll notice that none of the terms contains the square of one of the x_i's when expanded. In other words f is a polynomial of degree 1 in each variable. I also know, by the above, that the order of the games does not matter, so f is a symmetric function, therefore it must be a linear combination of the symmetric functions which are of degree 1 in each variable, namely the elementary symmetric functions. So f = c₀ + c₁e₁ + c₂e₂ + c₃e₃ + c₄e₄ + c₅e₅ + c₆e₇ + c₇e₇. I can also predict the outcome if the probabilities are 0 or 1, i.e. f(0, 0, 0, 0, 0, 0, 0) = 0, f(1, 0, 0, 0, 0, 0, 0) = 0, f(1, 1, 0, 0, 0, 0, 0) = 0, f(1, 1, 1, 0, 0, 0, 0) = 0, f(1, 1, 1, 1, 0, 0, 0) = 1, f(1, 1, 1, 1, 1, 0, 0) = 1, f(1, 1, 1, 1, 1, 1, 0) = 1, f(1, 1, 1, 1, 1, 1, 1) = 1, which gives the linear equations

c₀ = 0,

c₀ + c₁ = 0,

c₀ + 2c₁ + c₂ = 0,

c₀ + 3c₁ + 3c₂ + c₃ = 0,

c₀ + 4c₁ + 6c₂ + 4c₃ + c₄ = 1,

c₀ + 5c₁ + 10c₂ + 10c₃ + 5c₄ + c₅ = 1,

c₀ + 6c₁ + 15c₂ + 20c₃ + 15c₄ + 6c₅ + c₆ = 1,

c₀ + 7c₁ + 21c₂ + 35₃ + 35c₄ + 21c₅ + 7c₆ + c₇ = 1.

This system can be solved by back substitution to produce the coefficients c₀ = 0, c₁ = 0, c₂ = 0, c₃ = 0, c₄ = 1, c₅ = -4, c₆ = 10, c₇ = -20. Btw, the numbers 1, 4, 10, 20 appear in one of the diagonals in Pascal's triangle, and it's not too hard to state the more general case, the probability of winning at least m games out of n, in terms of binomial coefficients. --RDBury (talk) 09:06, 6 November 2014 (UTC)[reply]

PS. Poisson binomial distribution may be of some interest in this context. --RDBury (talk) 09:44, 6 November 2014 (UTC)[reply]

What are the equations in the Resurrection episode "Affliction"?

In the United States, on the ABC TV series Resurrection, Immigration Agent Bellamy is held in a government facility in the episode Affliction. This is probably available online though I haven't checked yet. Bellamy escaped from the room where he was being held and found his way to the office of the woman he is expected to report to in his job monitoring the "returned", people who have returned from the dead. On a whiteboard are some complicated equations. At a library I may be able to pinpoint exactly what time in the video these first appear, though they are onscreen a lot during the second half of the episode, before Bellamy is released.— Vchimpanzee • talk • contributions • 20:10, 5 November 2014 (UTC)[reply]

The video is here [1] (at least for the US)^{(see the note below)}. One good view of the whiteboard occurs at 18:09, though there are several views. It was a mess of things with no coherent theme. The sort of whiteboard you might write if you had no real understanding of math/physical sciences but wanted to be intimidating. It includes, Maxwell's Equations, the wave equation, the definition for a multidimensional normal distribution (and some manipulations thereof), E=mc^2, Schrodinger's equation, some differential equations I don't recognize, a diagram depicting the relationship between degrees / radians and the their sines / cosines, some form of a regression equation, a statement involving conditional probability, a couple graphs which aren't easily identified, and a variety of other things that I either can't make out or which don't make a ton of sense. Dragons flight (talk) 20:34, 5 November 2014 (UTC)[reply]

Thanks. I halfway suspected some of it might be meaningless when I saw E=mc^2.— Vchimpanzee • talk • contributions • 22:19, 5 November 2014 (UTC)[reply]

The note below: ;) Actually, only in US. I'm outsie US and got 'ERROR. En error occured while loading this video. Error code: 403-1; You appear to be outside the United States or its territories. Due to international rights agreements, we only offer this video to viewers located within the United States and its territories.' --CiaPan (talk) 06:10, 6 November 2014 (UTC)[reply]