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Another worked example

Posted By: Timothy Chow
Date: Sunday, 16 October 2016, at 10:48 p.m.

In Response To: The Backgammon Solution (Rick Janowski)

Let's ignore the fact that money comes in discrete multiples of 1 cent and suppose that the prior probability of the smaller prize being £P obeys an exponential distribution with mean P = 1000. Formally, this means that the probability density function is e^–P/1000/1000. Informally, this means that, before you open any envelopes, the prize money could in principle be arbitrarily large, but it will be exponentially unlikely to be that large, and on average, it will be £1000.

After you open the envelope and observe £Q, you're now down to two cases: Either P = Q or P = Q/2. In the former case, if you switch, you gain £Q, and in the latter case, if you switch, you lose £Q/2. The relative probabilities of the two cases are e^–Q/1000/1000 and e^–Q/2000/1000. So to find the breakeven point, you have to solve the equation

Qe^–Q/1000/1000 = (Q/2)e^–Q/2000/1000.

Numerically, one finds that the answer is about 1386.29. So you should switch if the amount you observe is less than this and you should not switch if the amount you observe is greater than this.

Of course there's no reason to believe this particular probability distribution; it's just an example. But presumably, most plausible probability distributions will have the general form of falling off smoothly as the prize amount gets larger. In such cases the rule will be to switch if the observed amount of money is below some threshold and to not switch if the observed amount of money is above some threshold. The actual value of the threshold is going to depend on just how unlikely you think it is that the prize money will be extremely high.

It would be more aesthetically pleasing if there were some plausible way to argue for always switching, and get rid of that annoying threshold, since its value depends on contingent features of the real world like how much disposable income people really have. However, I know of no way to construct such an argument. Any reasonable probability distribution that you postulate is going to have some threshold.

Messages In This Thread

• Tournament Prize Money Conundrum
  Rick Janowski -- Saturday, 15 October 2016, at 8:06 p.m.
  • Tournament Prize Money Conundrum
    KenB -- Saturday, 15 October 2016, at 8:30 p.m.
  • Tournament Prize Money Conundrum
    Tom Keith -- Saturday, 15 October 2016, at 8:46 p.m.
  • The Backgammon Solution *SPOILER ALERT*
    Fender -- Saturday, 15 October 2016, at 8:58 p.m.
    • The Backgammon Solution *SPOILER ALERT*
      Bob Koca -- Sunday, 16 October 2016, at 4:25 a.m.
      • The Backgammon Solution
        Rick Janowski -- Sunday, 16 October 2016, at 7:18 a.m.
        • The Backgammon Solution
          Casper Van der Tak -- Sunday, 16 October 2016, at 9:21 a.m.
          • The Backgammon Solution
            Rick Janowski -- Sunday, 16 October 2016, at 9:45 a.m.
            • The Backgammon Solution
              Casper Van der Tak -- Sunday, 16 October 2016, at 10:27 a.m.
              • The Backgammon Solution
                Casper Van der Tak -- Sunday, 16 October 2016, at 11:18 a.m.
                • The Backgammon Solution
                  Rick Janowski -- Sunday, 16 October 2016, at 11:40 a.m.
                  • The Backgammon Solution
                    Casper Van der Tak -- Sunday, 16 October 2016, at 12:15 p.m.
        • The Backgammon Solution
          Bob Koca -- Sunday, 16 October 2016, at 3:32 p.m.
          • The Backgammon Solution
            Rick Janowski -- Sunday, 16 October 2016, at 5:43 p.m.
            • The Backgammon Solution
              Casper Van der Tak -- Sunday, 16 October 2016, at 6:10 p.m.
              • The Backgammon Solution
                Rick Janowski -- Sunday, 16 October 2016, at 6:41 p.m.
            • The Backgammon Solution
              Timothy Chow -- Sunday, 16 October 2016, at 6:41 p.m.
              • The Backgammon Solution
                Rick Janowski -- Sunday, 16 October 2016, at 8:11 p.m.
                • The Backgammon Solution
                  Timothy Chow -- Sunday, 16 October 2016, at 10:01 p.m.
                • Another worked example
                  Timothy Chow -- Sunday, 16 October 2016, at 10:48 p.m.
                  • Another worked example
                    Bob Koca -- Sunday, 16 October 2016, at 11:42 p.m.
            • The Backgammon Solution
              Bob Koca -- Sunday, 16 October 2016, at 7:15 p.m.
              • The Backgammon Solution
                Rick Janowski -- Sunday, 16 October 2016, at 7:57 p.m.
                • The Backgammon Solution
                  Casper Van der Tak -- Monday, 17 October 2016, at 7:03 a.m.
  • Tournament Prize Money Conundrum
    Timothy Chow -- Saturday, 15 October 2016, at 9:09 p.m.
    • Tournament Prize Money Conundrum
      Rick Janowski -- Saturday, 15 October 2016, at 10:32 p.m.
      • Tournament Prize Money Conundrum
        Chuck Bower -- Sunday, 16 October 2016, at 12:52 a.m.
        • Tournament Prize Money Conundrum
          Rick Janowski -- Sunday, 16 October 2016, at 1:22 a.m.
      • Tournament Prize Money Conundrum
        Timothy Chow -- Sunday, 16 October 2016, at 5:07 p.m.
        • Tournament Prize Money Conundrum
          Rick Janowski -- Sunday, 16 October 2016, at 6:24 p.m.
    • Equity calculation question?
      Rick Janowski -- Sunday, 16 October 2016, at 9:30 a.m.
  • Tournament Prize Money Conundrum
    Casper Van der Tak -- Saturday, 15 October 2016, at 9:39 p.m.
  • Tournament Prize Money Conundrum
    Dmitriy Obukhov -- Monday, 17 October 2016, at 6:15 p.m.
  • Tournament Prize Money Conundrum
    Stein Kulseth -- Tuesday, 18 October 2016, at 8:45 p.m.