Backgammon Solutions

  1. To enter in the first turn, you must roll at least one 5. So the probability of not entering on your first turn is (5/6)^2 = 25/36, meaning that the probability of entering on the first turn is 1 - 25/36 = 11/36. The probability of not entering on any of the first three turns is then (25/36)^3 = 15625/46656. So the probability of entering during one of the first three turns is 1 - (15625/46656) = 31031/46656, or approximately .665. This type of probability is important to consider when using (or playing against) the blitz strategy described above.
  2. Intuitively, as this is a fair game (so the average amount of money you have at the end is the same as the amount with which you started), the answer should be 8/20 (why?). Let's see that this is correct.

    Call p the probability that you win. Then the probability that your opponent wins is 1-p (observe that we are supposing here that there is no possible tie; this is true and can be proved). If you win, you end up with 20 pennies; if you lose, you get 0. Thus the average amount of money you end up with is 20p + 0*(1-p) = 20p pennies. Since this is a fair game, this average amount has to be 8 pennies. Therefore, 20p = 8 so p = 8/20 = 0.4.

  3. The idea we will use is to assume that the threshold for offering a double is attained when the expected payoff you get is the same no matter if your opponent accepts or rejects the double. (This means, intuitively, that having more coins than the threshold implies that you would be better off if the opponent accepts the double, so you should try offering it).

    We will include doubling in this game in the following way: when a player gets all the coins, he wins 1 point, while the loser loses 1 point. If the doubling cube is at 2, he gets 2 points, and the loser loses 2, and so on.

    Let's call d the minimal amount of coins you should have before doubling. This threshold d should be such that the average payoff you get if your opponent rejects the doubling is the same as the payoff you get if she accepts it. We will suppose that the current bet is 1 (this does not make any difference). By the same argument used to solve problem 2 above, when you have d coins, the probability that you win is d/100, and your expected payoff is exactly:

    (d/100) - (100-d)/100 = (2d-100)/100
    (since here we are considering that upon winning you win 1 point, and upon losing you lose 1 point).

    Let's call q the probability that starting with d coins (where d > 50) you win before ever having 100-d coins. If this events happens, you win 1 point (forget the doubling for a while). If the opposite happens, then your expected return drops to (100-2d)/100. However, since you now have d coins, the current expected payoff is (2d-100)/100. Therefore we can calculate the expected value in two different ways. Setting these expressions equal yields:

    q + (1-q) * (100-2d)/100 = (2d-100)/100
    Solving this equation gives q = (2d-100)/d.

    Now, when you have the threshold value d of coins your expected payoff is 1, since that is what you would get if your opponent rejects the doubling. If he accepts the double you may win or lose the bet. The event of winning the bet before ever having 100-d coins has probability q, and that gives you a payoff of 2. If you reach 100-d coins, your opponent will redouble, so your expected payoff will be -2 regardless of if you accept or reject the redoubling. Therefore, we must solve the equation

    1 = 2q - 2(1-q) = 4q-2
    Replacing the expression for q in terms of d yields
    3/4 = (2d-100)/d
    3d = 8d - 400
    Solving gives d = 400/5, so you should have at least 80 coins before offering a double.

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