Re: Beauty and the Geek Game Theory: Answering the Freakonomics Challenge

Niels_Kjaer — Thu, 28 May 2009 05:05:19 -0000

Dear Peter, beauties, and geeks,

I looked at your "baby Monte Carlo" solution for this game problem and decided also to approach it in my own way. I find this problem very treatable since it is finite and there are is no hidden information. The strategies of the other teams are "fully" described by being optimal(rational).

Even though this problem is analytically solvable, I find "baby Monte Carlo" approaches extremely useful. It facilitates debugging and makes it easy to examine variations of the game and its strategies. The latter is very useful for getting robust results in a world of freakonomics.

I used your program to teach myself a bit of Python, and I did not "manage" to find any problems with your numerical results. However, I did find that this games contains features not fully covered as far as I can see. I concentrated on extending the strategies, like weak "w" and strong "s" to optimal "o" ones depending on the specific status of the game. For optimal strategies I need to assume that the strategies of the other teams are known and no cartels exist. This eliminates the problem of non-credible threats. I will be looking for fix points, Nash equilibria or N-E if you like, where no single team improves by changing strategy.

I have made three technical observations about this game:

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1) There is one and only one series of games to be played by the teams. This means that optimal strategies will depend significantly on the actual situation in that series. Assume as an example a situation where there are 4 teams left: 3 strong "S" and one weak "W". One of the strong teams "T"="C1"="C2" wins both contests and will pick the two teams "E1","E2" for the elimination round. In this case it is advantageous for "T"
to select two other strong teams. In a limit of "weak W strength", W->0, the probability for "T" to become the final winner is p("T")=1/2 if "E1","E2"
are both "S", otherwise p("T")=1/3.

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2) In your approximation to Nir's version of the game the "safe" feature has been removed. In your version (and code) both "E1"="C2" and "E2"="C1" are allowed. This feature gives rise to some peculiar situations. Consider a situation where we are left with 3 teams "T1","T2","T3", "C1"="T1", and "C2"="T2". (The strength of the teams are irrelevant). In this case your version forces the strategy of "T1" in the following way: If "E1"="T3",
"T2" will be forced (independently of its strategy) to select "E2"="T1". Hence, "E1"="T2" by rule of exclusion. When "E1"="T2", "T2" will always face two elimination rounds with "T1" and "T3" and p("T2") will hence not depend on which choice "T2" makes for "E2". This makes the final stage(s) of this game quite awkward.

(-:"In the "real" game "T2" would probably select the weakest of "T1" and "T3" in order to maximize "T2"'s appearance on the show. In a "real theory" game "T2" might select "E2"="T1" in order to take revenge, not on "T1", but on the "producer" (Norvig, Nir, or Niels(me)) of the "real theory" game show. If "E1"="C2" and "E2"="C1" is presented as optimal in a "real theory" game show, a simple variation of the (unwritten) game rules shows that
the game is not robust: After "T2" did not win the first challenge "T2" should try every trick to force "T3" to become the winner of the second challenge.":-)

It is natural to remove the "safe" aspect of Nir's version. Nir's version is not consistent when there are 3 teams and two different teams win the challenges, since they cannot be not allowed to pick the same team for the elimination round. Since the 'safe' feature is important for the game
strategies I decided to investigate both, changing the original slightly, so my versions become less inconsistent:

My version of [NORVIG]: When a team has several optimal strategies it will make a random selection in the same way that it makes a random selection between pairs with the same strengths.

My version of [NIR]: There will be only one challenge in the penultimate round, which has 3 teams. The winner automatically selects both other teams for the elimination round.

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3) There is another interesting feature related to "choice between identical teams". (-:"Most "geeks" would quickly realise that this implies teams obeying Bose-Einstein statistics and can lead to strange effects like first order phase transitions (condensations) in the game. "Geekier geeks" would probably realise that such effects only occur in "real theory" games where pure chance is involved.":-) In order to enhance some of these funny effects, I decided to make my own simplified version of the game called [NIELS]: In each round there is only one challenge, and the winner selects a single team which is eliminated.
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I implemented the optimal "o" strategy by simply looking on all the possible selections "E1","E2" given "C1","C2". If both "C1" and "C2" are using the "o" strategy I let "C2" maximize p("C2") over "E2" for all allowed values of "E1", and I then let "C1" maximize p("C1") over "E1". The strategies are constructed using reverse recursion. For each configuration consisting of NT teams I keep the probabilities that any team wins which is needed for determining the optimal strategy for the configurations with NT+1 teams. Since I allow for more equal solutions to the maximization the code becomes less elegant than Peter's. I will show a few selected optimization features below.

Results from the simulation and analysis of the game [NIELS]:

A) "Democracy and vulnerability"

Each team "TN" is assigned a strength N+epsilon. Introducing a small number epsilon avoids the few situations where two selections are equally optimal for the selector. The first table show the probability (all probabilities are shown in percent) p("TN") for a team to win as a function of N and the total number of teams. In parentheses are which other team will be selected by the team in question if the team in question wins the first challenge.
Take for example a game with only 3 teams: "T1","T2","T3". The winner of the first challengewill always adopt a strong "s" strategy. "T1" wins 1/6 eliminating "T3", "T2" wins 2/6 eliminating "T3",and "T3" wins 3/6 eliminating "T2". This leads to 1/2 of the final pair being "T1","T2" and 1/2 being "T1","T3". This, gives p("T1")=7/24,p("T2")=1/3,p("T3")=3/8 as final probabilities.

"T1" "T2" "T3" "T4" "T5" "T6" "T7" "T8" "T9" "T10"
100.0(X)
33.3(2) 66.6(1)
29.2(3) 33.3(3) 37.5(2)
20.0(4) 22.2(1) 21.7(2) 36.2(3)
17.6(4) 11.8(4) 22.9(2) 21.9(5) 25.6(2)
15.1(6) 11.6(4) 14.7(5) 19.3(2) 21.2(6) 18.0(3)
13.6(6) 10.1(1) 16.6(6) 16.1(7) 13.4(7) 18.1(2) 12.0(5)
9.1(5) 9.1(1) 13.0(2) 11.5(8) 13.8(1) 15.8(4) 13.4(2) 14.0(7)
7.4(5) 8.9(1) 12.1(8) 12.1(8) 12.3(7) 12.0(8) 9.6(2) 10.8(6) 14.7(7)
8.2(5) 10.3(1) 10.3(10)10.1(10) 9.0(1) 11.0(5) 10.3(8) 9.8(9) 9.8(8) 11.2(9)

A clear picture of equality emerges, where in the end all teams will have nearly equal probability to win. In order to understand this, I can follow the probabilities for a specific setup. If I start with 10 players with strength ranging from 1(+epsilon) to 10(+epsilon) I can for each round of the game compute the probability that each team remains which is shown in the next table:

Round "T1" "T2" "T3" "T4" "T5" "T6" "T7" "T8" "T9" "T10"
1 100.0 100.0 100.0 100.0 100.0 100.0 100.0 100.0 100.0 100.0
2 87.3 100.0 100.0 100.0 87.3 100.0 100.0 70.9 67.3 87.3
3 79.9 94.7 90.9 97.6 72.8 86.3 79.3 59.1 62.8 76.4
4 73.0 84.3 81.8 82.9 65.6 74.2 65.5 52.1 54.1 66.6
5 65.7 70.0 71.8 66.2 55.4 62.3 56.5 47.0 47.2 57.9
6 57.9 58.2 56.1 51.6 46.1 53.0 48.1 41.5 40.2 47.2
7 55.0 49.3 41.5 39.3 35.9 40.9 37.2 34.2 31.9 34.9
8 47.7 43.0 35.2 29.9 24.8 27.7 24.5 22.1 21.5 23.7
9 47.7 31.5 21.1 17.2 14.3 15.8 14.0 12.6 12.3 13.5
10 8.2 10.3 10.3 10.1 9.0 11.0 10.3 9.8 9.8 11.2

In the first rounds, (typically strong) teams winning the challenges will select other strong teams to be eliminated. "T1" will on average survive the longest, but only to be eliminated with a high probability in the ultimate round. One could think it is best to get rid of the strongest team "T10" first, but there is normally plenty of time to get rid of "T10" in rounds 5-8. Therefore, most teams using optimal strategy aim at removing "T8" and "T9" early on. This illustrates the conflicting interest of becoming the strongest effectively without the other teams knowing about it and taking advantage of that knowledge.

B) "Hiding and self-organization"

Intuitively, it is an advantage to hide in a group of indistinguishable peers if your team risks elimination in the next round. This effect is most easily examined when there are only two groups of "W" and "S" teams. When one of the two groups have large probabilities to win challenges they will dominate the elimination in the next round(s) and thereby bias the evolution of the game. This effect creates positive feedback such that a slightly more
powerful becomes even more powerful. This effect competes with the previous effect in the following way: In a powerful group each team has larger than 1/(rounds left) probability to win, but if the non-powerful group becomes much smaller this bias will decrease and eventually disappear when the non-powerful group is totally eliminated.

As an example I created 2 groups consisting of NW teams with strength=1.0 and NS teams with strength=1+epsilon. The results are simple fractions in the limit epsilon -> 0, I first give the full results in the table below for games with 1-6 teams. The first 2 columns give the number of teams in each group, the next two are the probabilities that a given team, i, in the group ends up victorious in the final. The probabilities are given
like Peter gives them so that NW*PWi+NS*PSi=1. The final two columns show which type of team will be eliminated if a "W" team or "S" team wins the challenge.

NW NS PWi PSi WE SE
----------------------------
0 1 0 1 X X
1 0 1 0 X X

0 2 0 1/2 X S
1 1 1/2 1/2 S W
2 0 1/2 0 W X

0 3 0 1/3 X S
1 2 2/4 1/4 S S
2 1 5/12 2/12 S W
3 0 1/3 0 W X

0 4 0 1/4 X S
1 3 3/24 7/24 S W
2 2 1/4 1/4 W W
3 1 5/18 3/18 W W
4 0 1/4 0 W X

0 5 0 1/5 X S
1 4 4/32 7/32 S S
2 3 33/240 58/240 S W
3 2 14/60 9/60 S W
4 1 5/24 4/24 W W
5 0 1/5 0 W X

0 6 0 1/6 X S
1 5 5/40 7/40 S S
2 4 22/160 29/160 S S
3 3 39/240 41/240 S W
4 2 67/360 46/360 W S
5 1 1/6 1/6 W W
6 0 1/6 0 W X

This process continues for games with a large number of teams. Having simulated games with more than NW+NS=10,000 teams I observe a slow and seemingly chaotic convergence of these quantities. It might be interesting to note that the bias seems to become maximal when PWi/PSi=NW/NS and this values seems to be consistent with phi=(1+sqrt(5))/2, which might not be a huge surprise in such a self-organizing environment.

When the strength of the two groups become different the democratization effects are added on top. If for example the strength of "S" becomes 4.0 as in the original proposed game, the maximal bias PSi/(rounds left)=1.32 compared with 1.19 when the strengths are equal. When there are many more "W" teams than "S", also the "W" teams can dominate, but in the case (d=4.0) the maximal dominance becomes PWi/(rounds left)=1.09.

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Results from simulation and analysis of the [NIR] game:

Compared to the [NIELS] strong players have an advantage in [NIR] and [NORVIG] games because they cannot be eliminated directly, but can use their strength also in the elimination round. A direct comparison shows for a game with 10 different players:

Game "T1" "T2" "T3" "T4" "T5" "T6" "T7" "T8" "T9" "T10"
[NIR] 1.6 3.9 6.6 8.0 9.7 12.4 13.8 13.2 14.7 16.0
[NIELS] 8.2 10.3 10.3 10.1 9.0 11.0 10.3 9.8 9.8 11.2

But this only means that democracy requires more players to to play a dominant role in [NIR] (and [NORVIG]) compared to [NIELS] games. The self-organization effect in [NIR] (and [NORVIG]) works basically in the same way as for the [NIELS] games.

It is more interesting to study the exact team strategies, because each round contains 2 challenges and the optimal strategies might interfere. When the winners of the two challenges "C1,"C2" are different it is necessary to include explicitly the strategy of "C2" to find the optimal strategy for "C1". For each pair of possible teams for elimination, both "C1" and "C2" can separately compute the probability to become the final winner. For each
choice of "E1", "C2" will be able to find the maximum probability. This gives a reduced set of probabilities from which "C1" will choose the maximum. But the choice of "E1" requires that "C1" that "C1" can trust "C2" also to make an optimal choice.

As an example of beliefs and threats: Assume there are 6 teams left: "T1","T5","T6","T7","T8","T10". The two winners of the challenges are: "C1"="T10","C2"="T5". To show the selection dilemma I also allow "T10" to have 2 different possible strategies, optimal "o" and strong "s". In the following tables the left hand values assume a "o" "T10" strategy and, separated by a '|', the same values assuming a "s" strategy for "T10".

First I compute p("T10") given
"T1" "T6" "T7" "T8" are eliminated:

31.3 38.4 40.4 42.7 | 31.2 38.4 40.4 42.7

No big deal, it seems that "o" is close to "s" for "T10".

p("T10") can be computed as a symmetric matrix given the two teams selected for elimination:

32.5 32.6 32.7 | 32.4 32.5 32.6
32.5 39.3 40.2 | 32.4 39.3 40.2
32.6 39.3 41.4 | 32.5 39.3 41.4
32.7 40.2 41.4 | 32.6 40.2 41.4

Notice that if "T1" is selected, there is a high probability that "T1" will be eliminated.
"T10" will definitely not like that to happen.

p("T5") given "T1","T6","T7","T8" are eliminated:

14.7 13.2 14.0 15.0 | 14.3 13.2 14.0 15.0

"T5" is nearly insensitive to which team is eliminated. "T5" benefits relative to "T10" from the elimination of "T1" since this makes "T5" the weakest team.

p("T5") given the two teams selected for elimination:

14.4 14.6 14.7 | 14.1 14.2 14.4
14.4 13.6 13.9 | 14.1 13.6 13.9
14.6 13.6 14.5 | 14.2 13.6 14.5
14.7 13.9 14.5 | 14.4 13.9 14.5

Here the outcome depends on the strategy of "T10" in the following way

If "o": "T5" will select "T1" for elimination unless "T10" selects "T1" in which case "T5" will select "T8". In any case "T10" is trapped and
ends up with a lousy p("T10")=32.7.

If "s": There is no choice for "T10" except for "E1"="T8" and "T5" will choose "E2"="T7" simply because 14.5>14.4. This leaves a happy "T10" with
p("T10")=41.4.

Is it a contradiction that using the "o" strategy in this case leaves "T10" in a worse situation than using the "s" strategy? The answer is no and
most of the problem lies with the choice by "T5". "T5" is not allowed in the game definition to make a binding deal with "T10". "T5" does not gain
by the changed strategy of "T10" unless they make the deal that "T10" in the next round tries not to select "T5" for elimination.

The above is for me one example of how Nash equilibria often can appear contrived.

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Results from the [NORVIG] simultation and analysis:

I first redid Peter's analysis on the actual game allowing a mixture of global strategies for the teams. I chose only to look at "w" and "s" strategies.
Both the group of "W" teams and "S" group can have 0-3 teams with "s" strategies, while there is only one "M" team. For each of the combinations I show the probability that the teams eventually win in the same way as Peter did.

Teams | W W M M S S
with |
"s" |
W M S |p("w")|p("s")|p("w")|p("s")|p("w")|p("s")|
-------------------------------------------------
0,0,0 | 0.1 | X | 3.0 | X | 32.2 | X |
1,0,0 | 0.1 | 0.1 | 3.5 | X | 32.0 | X |
2,0,0 | 0.2 | 0.2 | 4.1 | X | 31.8 | X |
3,0,0 | X | 0.3 | 5.0 | X | 31.4 | X |
0,1,0 | 0.2 | X | X | 4.5 | 31.6 | X |
1,1,0 | 0.3 | 0.3 | X | 5.6 | 31.1 | X |
2,1,0 | 0.5 | 0.5 | X | 6.9 | 30.5 | X |
3,1,0 | X | 0.9 | X | 8.4 | 29.7 | X |
0,0,1 | 0.6 | X | 7.6 | X | 24.4 | 41.7 |
1,0,1 | 0.9 | 0.8 | 8.5 | X | 24.1 | 40.7 |
2,0,1 | 1.3 | 1.2 | 9.6 | X | 23.7 | 39.3 |
3,0,1 | X | 1.8 | 10.8 | X | 23.1 | 37.4 |
0,1,1 | 1.6 | X | X | 10.4 | 23.3 | 38.3 |
1,1,1 | 2.2 | 1.9 | X | 11.7 | 22.7 | 36.5 |
2,1,1 | 3.2 | 2.8 | X | 13.0 | 21.9 | 34.4 |
3,1,1 | X | 4.1 | X | 14.2 | 20.9 | 31.8 |
0,0,2 | 1.9 | X | 12.2 | X | 19.0 | 31.5 |
1,0,2 | 2.7 | 2.4 | 13.0 | X | 18.6 | 30.3 |
2,0,2 | 4.0 | 3.4 | 13.8 | X | 18.1 | 28.6 |
3,0,2 | X | 5.1 | 14.4 | X | 17.4 | 26.5 |
0,1,2 | 4.1 | X | X | 14.3 | 17.7 | 27.9 |
1,1,2 | 5.6 | 4.4 | X | 15.0 | 17.1 | 26.2 |
2,1,2 | 7.9 | 6.1 | X | 15.2 | 16.3 | 24.2 |
3,1,2 | X | 8.8 | X | 15.0 | 15.3 | 21.7 |
0,0,3 | 3.9 | X | 13.6 | X | X | 24.9 |
1,0,3 | 5.5 | 4.5 | 13.9 | X | X | 23.6 |
2,0,3 | 8.1 | 6.4 | 13.7 | X | X | 21.8 |
-------------------------------------------------
3,0,3 | X | 9.6 | 12.9 | X | X | 19.4 |--> A N-E
-------------------------------------------------
0,1,3 | 6.8 | X | X | 14.6 | X | 21.7 |
1,1,3 | 9.3 | 6.8 | X | 14.1 | X | 20.2 |
2,1,3 | 13.5 | 9.4 | X | 12.8 | X | 18.3 |
3,1,3 | X | 13.7 | X | 10.5 | X | 16.1 |
-------------------------------------------------

As Peter saw, the "S" group of teams is better off by all selecting "w" strategies. However, if one of them defects, that team will be better off,
in which another will defect, and even the third will eventually defect. The Nash equilibrium occurs for all the "W" and "S" teams having "s" strategy,
while "M" chooses the "w" strategy.

One could expect that all players using the optimal "o" strategy would be a N-E. But it turns out that a "W" team is slightly better off (probability goes from 4.5 to 4.6%) defecting to a "w" strategy. Then all of the "W" teams will follow suit, which is not at all good for the "W" group like it happened for
the "S" group above. Finally "M" is also slightly better off switching to the "w" strategy. The tabler with all the probabilities with one team defecting
from the optimal strategy is:

W M S | W | W | W-d | M-d | S | S | S-d |
-----------------------------------------------------------
"ooooooo"| 4.5 | 4.5 | 4.5 | 11.0 | 25.1 | 25.1 | 25.1 |
"oowoooo"| 3.5 | 3.5 | 4.6 | 26.1 | 26.1 | 26.1 | 26.1 |
"oosoooo"| 5.5 | 5.5 | 2.9 | 11.4 | 24.9 | 24.9 | 24.9 |
"ooowooo"| 3.2 | 3.2 | 3.2 | 9.7 | 26.7 | 26.7 | 26.7 |
"ooosooo"| 5.8 | 5.8 | 5.8 | 9.3 | 24.5 | 24.5 | 24.5 |
"oooooow"| 3.6 | 3.6 | 3.6 | 13.7 | 28.3 | 28.3 | 18.9 |
"oooooos"| 7.9 | 7.9 | 7.9 | 13.3 | 22.0 | 22.0 | 19.1 |
-----------------------------------------------------------

The N-E "wwwwooo" gives no improvements for any single team that chooses to switch strategy:

W M S | W | W | W-d | M-d | S | S | S-d |
-----------------------------------------------------------
"wwwwooo"| 2.2 | 2.2 | 2.2 | 11.7 | 27.2 | 27.2 | 27.2 |
"wwowooo"| 3.1 | 3.1 | 1.6 | 12.0 | 26.8 | 26.8 | 26.8 |
"wwswooo"| 3.4 | 3.4 | 1.2 | 12.1 | 26.5 | 26.5 | 26.5 |
"wwwoooo"| 2.9 | 2.9 | 2.9 | 11.2 | 26.7 | 26.7 | 26.7 |
"wwwsooo"| 3.5 | 3.5 | 3.5 | 9.7 | 26.5 | 26.5 | 26.5 |
"wwwwoow"| 1.2 | 1.2 | 1.2 | 10.3 | 29.6 | 29.6 | 26.8 |
"wwwwoos"| 3.3 | 3.3 | 3.3 | 11.0 | 29.3 | 29.3 | 20.6 |
-----------------------------------------------------------

I didn't look into all the revenge strategies. But, basically they all are sub-optimal in a one-go game. In general they would mostly bias the strategies
to become weaker in the first rounds of the game.

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PS.
I can provide the Python (3.0) code for the above if anyone is interested, but with the caveat that it looks like John cleese writing "FORTRAN" compared
to Peter's code...
PPS. Any simple way to format the tables in my comment?

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Re: Beauty and the Geek Game Theory: Answering the Freakonomics Challenge