A Quantum Diaries Survivor

Ok, I got no feedback on what was the reference to the game suggested by Martin Gardner, so I decided I would investigate the matter by myself.

So today I wrote a simple fortran program that analyzes the possible configurations of the three wheels.

Of course, the wheels can be constructed in an infinity of ways. One can choose how many sectors to allow in each wheel, what is the maximum value of the score on each sector, whether it is a real or integer number, the minimum width of the sectors, and so on.

I decided to keep things simple:

• the wheels have three sectors each;
• to each sector is assigned a value, given by an integer smaller than V_max;
• the sectors have a angular extension that is a multiple of 360/N_max degrees, with N_max another small integer (this makes construction of real wheels easy)

I further assumed the following:

• all sectors are given different values

However, I allowed two sectors in the same wheel to have the same value, to examine "simple" configurations.

Since the program finds a very large number of solutions if N_max and V_max are not small, I decided to find a "best" configuration according to an estimator of how striking the different behavior of the wheels in 2-player and 3-player mode were. This estimator is the difference between three times the winning probability of the best wheel in 3-player mode, minus that wheel's winning probability with the other two in 2-player mode, minus the win-probability of the second-worse wheel in 2-player mode. This is just one of many possible choices for an estimator of "wheel weirdness", of course, but it selects interesting cases when maximized.  

Now, a few results.

First of all, my program finds no solutions (configurations for which 1>2, 1>3, 2>3 in 2-player mode, while 3>2>1 in 3-player mode) if N_max is smaller than 4, no matter how large V_max is. I believe a mathematical demonstration of this should not be hard, but I prefer to use my time otherwise.

Second, no solutions are possible if V_max is smaller than 6 (zero is allowed), no matter how large N_max is.

Third, there are some very nice cases for small values of both N_max and V_max. Here are a few.

For N_max=4, the solution with smallest V_max is for V_max=8, and it is very nice (below, for each wheel the extension of each sector and its value is given, and then the winning chances):

Wheel # 1 :
          0.250   5
          0.750   2
          Win chance in 3-player mode:  0.328
          Win chances in 2-player modes:
               against 2 :  0.562
               against 3 :  0.562
Wheel # 2 :
          0.250   4
          0.500   1
          0.250   6
          Win chance in 3-player mode:  0.328
          Win chances in 2-player modes:
               against 1 :  0.438
               against 3 :  0.625
Wheel # 3 :
          0.250   3
          0.500   0
          0.250   7
          Win chance in 3-player mode:  0.344
          Win chances in 2-player modes:
               against 1 :  0.438
               against 2 :  0.375

Wheel 1 is a big winner in 2-player mode, but it is wheel 3 the one that wins in 3-player mode, although by a small margin. The same solution above appears -with irrelevant differences in the value of sectors - if you increase V_max.

With N_max=5, things get more complicated. Here, the smallest V_max that yields solutions appears to be V_max=8. Here is a nice one:

Wheel # 1 :
          0.200   6
          0.400   4
          0.400   2
          Win chance in 3-player mode:  0.280
          Win chances in 2-player modes:
               against 2 :  0.520
               against 3 :  0.520
Wheel # 2 :
          0.600   5
          0.400   1
          Win chance in 3-player mode:  0.288
          Win chances in 2-player modes:
               against 1 :  0.480
               against 3 :  0.520
Wheel # 3 :
          0.400   0
          0.200   3
          0.400   7
          Win chance in 3-player mode:  0.432
          Win chances in 2-player modes:
               against 1 :  0.480
               against 2 :  0.480

Wheel 1 wins the 2-player game, wheel 3 wins hands down the 3-player one.

With N_max=6 (a tidy division of wheels into 60 degree sectors) there is a trivial solution with V_max=6:

Wheel # 1 :
          0.500   4
          0.333   1
          0.167   2
          Win chance in 3-player mode:  0.250
          Win chances in 2-player modes:
               against 2 :  0.500
               against 3 :  0.500
Wheel # 2 :
          1.000   3
          Win chance in 3-player mode:  0.250
          Win chances in 2-player modes:
               against 1 :  0.500
               against 3 :  0.500
Wheel # 3 :
          0.500   5
          0.500   0
          Win chance in 3-player mode:  0.500
          Win chances in 2-player modes:
               against 1 :  0.500
               against 2 :  0.500

Not too interesting… After all, the most interesting cases appear to be the ones where N_max is a prime number. Take 7, for instance. In that case, with just V_max=7 (the minimum in that case, by the way) we get a non-trivial configuration:

Wheel # 1 :
          0.571   4
          0.429   2
          Win chance in 3-player mode:  0.315
          Win chances in 2-player modes:
               against 2 :  0.551
               against 3 :  0.571
Wheel # 2 :
          0.143   1
          0.714   3
          0.143   6
          Win chance in 3-player mode:  0.318
          Win chances in 2-player modes:
               against 1 :  0.449
               against 3 :  0.633
Wheel # 3 :
          0.429   5
          0.571   0
          Win chance in 3-player mode:  0.367
          Win chances in 2-player modes:
               against 1 :  0.429
               against 2 :  0.367

————

Ok, that is basically it for three wheels. One question that springs to mind is: how can this riddle be generalized to more than three wheels ? After all, the whole point of the wheels game was to prove the fallacies of unwittingly usage of induction. What if we make four wheels: we can then imagine 2-player, 3-player, and 4-player games. I think the best enunciation, for N>3 wheels, is: find configurations such that in 2-player mode wheel 1 wins with 2…N, wheel 2 wins with 3…N, and so on; while in N-player mode N wins with N-1, second comes N-1, and so on, and 1 is the worst one.

I would be curious to know if the problem has been investigated…

Anyways, I have put the fortran source of the program I wrote to analyze the case of three wheels in http://www.pd.infn.it/~dorigo/wheels.f ,
in case you want to explore more complicated cases. Have fun!

I think I need to place an ad here…

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By looking at my page, I now realize it could take forever for anybody who occasionally visits this blog to notice that tab!

So that is the reason for this post: just plain good-old advertisement. That page contains links to a useful collection of posts!