Problem No. 11, 2002


The problem

Our elfth problem is based on a mathematical problem that has appeared in many magazines lately, e.g. in Bridge Today and Scientific American.

The original problem goes something like this: You and two of your friends are taken into a large room where there is an infinite number of playing cards. Half of them are red, half of them black. The light is turned off, and the three of you pick up one card each at random, which you attach to your forehead with a string. The light is turned on, and all of you can see your friends' cards but not your own.
  Then you should try to guess which color your own card has (black or red), and all three have to do it simultaneously. Your choices are three: black, red or pass. If some of you guesses wrong, or all passes, you have lost. But if at least one of you makes a correct guess, without anybody guessing wrongly, you win.
  You may agree on a strategy on beforehand, like "A and B passes, while C guesses red." When you make your guess, you may not communicate with your friends in any way (like B looking at the floor if A has a black card). Now, the question is if there is a strategy which gives you a better chance to win than 50%. The answer to that question is found here.
  Now, when you know the best strategy for three players, the obvious question is if there is an optimal strategy for four or more players. What do you think: What should the strategy be then?


When I first heard of this problem, I eventually found a very beautiful solution, which aimed at winning two times out of three (how that is done, is explained here). That also happens to be the optimal strategy if all players have to guess in the same way, but that wasn't specified in the problem...
  The solution we had in mind was this: The best strategy is do exactly as you do when three players are involved, and let three chosen players, A, B and C, do what they would have done if only they had been present (look at each other's cards and ignore the rest), while the other players all pass. This means the chance for winning is 75% for any number of players above two.
  But this isn't the correct solution either... A few days after our solution had been made public, we got a letter from a Swedish mathematician, Per Hallberg of Stockholm, who showed that if there are at least seven participants, there is an even better strategy, one that succeeds 87.5% of the time. The solution is a real beauty, but to fully understand it you probably need to have studied mathematics. It's quite complicated, but if you simply have to know how it is done, just contact us, and we'll send it to you.

More problems:

[ 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | 43 ]

[ The competiton | Home ]


Copyright © 2013, Scania Bridgekonsult