2
$\begingroup$

I found the following question for a quantitative research role on the website of a trading firm. I tried solving it and I think the answer is 0.5 for all three questions. I think so because both the expected absolute difference and expected squared difference from 0.5 is 0, with the expectation taken with respect to $U[0,1]$. However, I later heard back from the firm that I solved it incorrectly. I am curious about what is the correct answer to this question. I would be grateful if someone could explain where I went wrong.

Question Three players A, B, C play the following game. First, A picks a real number between 0 and 1 (both inclusive), then B picks a number in the same range (different from A’s choice) and finally C picks a number, also in the same range, (different from the two chosen numbers). We then pick a number in the range uniformly randomly. Whoever’s number is closest to this random number wins the game. Assume that A, B, and C all play optimally and their sole goal is to maximize their chances of winning. Also assume that if one of them has several optimal choices, then that player will randomly pick one of the optimal choices.

  1. If A chooses 0, then what is the best choice for B?
  2. What is the best choice for A?
  3. Can you write a program to figure out the best choice for the first player when the game is played among four players?
$\endgroup$
3
  • 1
    $\begingroup$ Interestingly, allowing negotiation changes the answers. If A chooses 0, C can say to B, "if you pick 1, I will pick 0.01, and we will each win with probability almost 1/2; otherwise, without any reduction my own chances of victory, I will pick so that your probability of winning is less than 0.34." $\endgroup$ – fblundun Dec 17 '20 at 13:09
  • 1
    $\begingroup$ And to see that 1/2 is suboptimal as the answer to question 2, note that if A picks 0.5, then B and C can pick 0.499 and 0.501 (or however close to 0.5 they are allowed to pick), leaving A with only a 0.001 probability of winning the game, and B and C each with a probability of about 0.5 of winning. $\endgroup$ – fblundun Dec 17 '20 at 22:52
  • $\begingroup$ Thank you for explaining $\endgroup$ – Durin Dec 18 '20 at 17:48
3
$\begingroup$

Part 1. B should pick 2/3, as then C will randomly pick a spot between A and B, giving B an average payout of 1/2. (If B picks b < 2/3, C picks b < c < 2/3, making c as small as possible; if B picks b > 2/3, C randomly picks c < 2/3. Both cases result in a worse expected payoff for B.)

Part 2. Without loss of generality, assume A's pick is at most 1/2. A should choose 1/4. By a similar argument to part 1, B then chooses 3/4 and C randomly chooses 1/4 < c < 3/4. A's expected payoff is 3/8. (B's is 3/8 and C's is 1/4, so neither B nor C has an incentive to choose a number < 1/4.) The intuition is that A wants to control the interval (0, a), so doesn't want to make a large enough to tempt the other players into trying to take over that interval.

If A chooses a < 1/4, then B chooses b = (2 + a)/3 and C randomly picks a number between a and b, so A's expected payoff is < 3/8.

If A chooses a > 1/4, then B chooses 1 - a + p for p as small as possible, so that C picks a - q for q as small as possible. Then A's expected payoff is (1 - 2a + p + q)/2 < 3/8. (In this case there is technically no "optimal" strategy for B - whatever value B takes for p, a better strategy would have been to take p/2. The same goes for C's choice of q.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.