Another math puzzle

If the probability of my winning a service game in tennis against a particular opponent is equal to my probability of winning the game once the score has reached “my ad” (or 40-30), what is this probability?

This puzzle has a mysterious connection to the previous puzzle. If you can solve one exactly, you can solve the other exactly.

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2 Responses to Another math puzzle

  1. Let p be the probability of winning a service point. Once the game has reached Deuce, the probability of winning the game is the probability of being the next player to win 2 points in a row from deuce, which is (p^2)/(p^2 + (1-p)^2). Now you should be able to write a polynomial equation which sets equal the probabilities of winning from 40-30 and from 0-0 (the binomial theorem is helpful). It’s not cheating to use online equation solvers to factor and find roots of the polynomial; but it’s more fun to solve exactly by radicals when this is possible. In both this puzzle and the other one, you get very simple equations in the end, but they still have amusingly complicated exact solutions. I’ll post the answers in 2 or 3 days.

  2. No takers?

    Using the binomial theorem, the probability of winning at least 4 points out of 6 (since you can assume the players play 6 points out even if the game is decided earlier) is

    p^6 + 6p^5*(1-p) + 15p^4*(1-p)^2

    and the probability of winning exactly 3 out of 6 points (Deuce) is 20p^3*(1-p)^3

    Therefore the overall probability of winning the game is

    p^6 + 6p^5*(1-p) + 15p^4*(1-p)^2 + (20p^3*(1-p)^3)*(p^2)/(p^2 + (1-p)^2)

    which is to be set equal to the probability of winning from 40-30. That latter probability is just

    p + (1-p)*(p^2)/(p^2 + (1-p)^2)

    Multiplying out and equating the polynomials after clearing the (p^2 + (1-p)^2) denominator, we get

    8p^7 – 28p^6 + 34p^5 – 15p^4 -p^3 + p^2 – p = 0.

    This is not so hard to solve, because we know p=1 is one solution to equate the probabilities as is p=0 so divide through by (p-1) and p to factor the polynomial as

    (p-1)^3 * p * (8p^3 – 4p^2 – 2p -1).

    We are seeking the solution to the cubic factor — obviously we can substitute x=2p to get

    x^3 = x^2 + x + 1

    and then divide by 2 after solving for x.

    This is a simple-looking cubic equation, just like in the earlier “draw poker” problem, and it is easy to get the numerical answer x=1.839286, giving p=0.919643. But the EXACT answer is messier:

    p = (1/6) * (1 + cbrt(19-sqrt(297)) + cbrt(19+sqrt(297))

    where sqrt is the square root symbol and cbrt is the cube root symbol.

    How do you solve cubic equations anyway? Well, any cubic equation can be reduced by a linear substitution (of the form y = ax + b) to one of the forms

    x^3 + x = c
    or
    x^3 – x = c.

    (How? First shift [add a constant to the variable] to eliminate the quadratic term, then scale [multiply the variable by a constant] to make the linear coefficient 1.)

    The answer in the first case is a unique real root:

    cbrt(c/2 + sqrt((c/2)^2 + 1/27) + cbrt(c/2 – sqrt((c/2)^2 + 1/27))

    and in the second case you can substitute x’=-ix and c’=ic to get

    i * (cbrt(ic/2 + sqrt((ic/2)^2 + 1/27) + cbrt(ic/2 – sqrt((ic/2)^2 + 1/27))

    where you are now taking square and cube roots of complex numbers (which geometrically comes down to expressing them in polar coordinates, taking the real square or cube root of the radius, and bisecting or trisecting the angle). Here there are THREE real roots instead of 1, but the only way to get exact real expressions for them is by going through complex numbers, trisecting angles, using trigonometric formulas, etc.

    And don’t forget to substitute back at the end since you started with a more complicated equation that you reduced to one of the two simple forms by a linear substitution.

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