4th IIGTVNMPCT 2012:

Fourth International Introduction-to-Game-Theory-VNM-POKER-Computer-Tournament 2012

A tournament for computer programs playing VNM POKER(5,1,1,2) is announced. The tournament is played in the knock-out system, where computer programs, from now on called "robots" play 200 rounds against other robots.

Every student should submit just one robot, and should provide

A nice sounding name for the computer player,
The probabilities by which the computer player, having the first move, should raise when facing
- the lowest value card "10",
- a jack "J",
- a queen "Q",
- a king "K",
- the highest value card "A",
The probabilities by which the computer player, having the second move, should call (after the other player raises) when facing
- the lowest value card "10",
- a jack "J",
- a queen "Q",
- a king "K",
- the highest value card "A",

Most of the games will be executed at the teacher's home computer, but cheating will not occur.

There is a prize for the creator of the winning computer player, maybe a box of self-made brownies, maybe a little book on Mathematics.

Here you can play the game

List of players and their strategies:

Name

Strategy

Dake Quinton

When starting, raise with these probabilities	0.15	0.2	0.5	0.65	0.9
When playing second, call with these probabilities	0	0.1	0.5	0.85	1
when facing a card of value	10	J	Q	K	A

RX-24

When starting, raise with these probabilities	1	1	1	0	0
When playing second, call with these probabilities	1	1	1	1	0
when facing a card of value	10	J	Q	K	A

Since I thought the (a little unreasonable) pure strategy was due to confusing raising and checking, as well as calling and folding, I reverted the order. I think this robot was intended:

RX-24rev

When starting, raise with these probabilities	0	0	0	1	1
When playing second, call with these probabilities	0	0	0	0	1
when facing a card of value	10	J	Q	K	A

R2-D2

When starting, raise with these probabilities	0	0.25	0.5	0.75	1
When playing second, call with these probabilities	0	0.25	0.5	0.75	1
when facing a card of value	10	J	Q	K	A

Jennirafe

When starting, raise with these probabilities	0.3	0.5	0.5	1	1
When playing second, call with these probabilities	1	1	0.5	1	1
when facing a card of value	10	J	Q	K	A

Brains

When starting, raise with these probabilities	0	0	0.5	0.75	1
When playing second, call with these probabilities	1	0.75	0.5	0.15	0
when facing a card of value	10	J	Q	K	A

Since I thought the (a little unreasonable) pure strategy for calling/folding was due to confusing calling and folding, I reverted the order there. I think this robot was intended:

Brainsrev

When starting, raise with these probabilities	0	0	0.5	0.75	1
When playing second, call with these probabilities	0	0.25	0.5	0.85	1
when facing a card of value	10	J	Q	K	A

Cantor

When starting, raise with these probabilities	1/3	0	0	0	1
When playing second, call with these probabilities	0	0.44	0.54	0.68	1
when facing a card of value	10	J	Q	K	A

Randy

When starting, raise with these probabilities	0.5	0.5	0.5	0.5	0.5
When playing second, call with these probabilities	0.5	0.5	0.5	0.5	0.5
when facing a card of value	10	J	Q	K	A

ImBad

When starting, raise with these probabilities	1	1	0	0	0
When playing second, call with these probabilities	1	1	0	0	0
when facing a card of value	10	J	Q	K	A

Here you can select two of these robots and let them play 200 rounds of the game all on their own.

Analysis of the different robots

The following table displays the expected payoff of the different pairings when playing 200 rounds. The higher it is, the greener it is, the lower, the reder.

Dake Quinton RX-24 R2-D2 Jennirafe Brains Cantor Randy Imbad RX-24rev Brainsrev

Dake Quinton 0 51.3 2.3 18.3 43,6 -3.14 26 49.8 0.8 0.1

RX-24 -51.3 0 -53.8 -68.5 24.5 -24.5 0 90 0 -52

R2-D2 -2.3 53.8 0 19.9 43.4 -2.4 25 50 -3.8 -1.6

Jennirafe -18.3 68.5 -19.9 0 41.7 -8.3 28 85 -12.5 -22

Brains -43.6 -24.5 -43.4 -41.7 0 -11.6 -29.5 -4.3 -34.5 -37.4

Cantor 3.1 24.5 2.4 8.3 11.6 0 12.3 21.5 0.1 1.6

Randy -26 0 -25 -28 29.5 -12.3 0 40 0 -24.5

Imbad -49.8 -90 -50 -85 4.3 -21.5 -40 0 10 -44.3

RX-24rev -0.8 0 3.8 12.5 34.5 -0.1 0 -10 0 8

Brainsrev -0.1 52 1.6 22 37.4 -1.6 24.5 44.3 -8 0

In the next table, the winning probability of the row player in the corresponding pairing is shown (assuming that draws are impossible, meaning that in case of a draw another 200 rounds are played, and so on, until a decision is reached). Again, red means low winning probability, and green means high winning probability. The numbers are somehow related to the number in the previous table, but not directly.

Dake Quinton RX-24 R2-D2 Jennirafe Brains Cantor Randy Imbad RX-24rev Brainsrev

Dake Quinton --- 99.5 55.1 81.9 99.4 42.9 92.4 99.8 52 50.3

RX-24 0.5 --- 0.3 0.1 91.8 9.9 50.1 100.0 50.3 0.2

R2-D2 44.9 99.7 --- 83.7 99.3 44.3 91.5 99.8 40.9 46.5

Jennirafe 18.1 99.9 16.3 --- 98.5 33.1 91.6 100.0 24.3 13.6

Brains 0.6 8.2 0.7 1.5 --- 24.8 4.4 39.5 1.8 1.7

Cantor 57.1 90.1 55.7 66.9 75.2 --- 75.9 89.6 50.4 53.8

Randy 7.6 49.9 8.5 8.4 95.6 24.1 --- 99.3 50.2 8.8

Imbad 0.2 0.0 0.2 0.0 60.5 10.4 0.7 --- 73.6 0.6

RX-24rev 48 49.7 59.1 75.7 98.2 49.6 49.8 26.4 --- 69.1

Brainsrev 49.7 99.8 53.5 86.4 98.3 46.2 91.2 99.4 30.9 ---

So what are the chances of the different robots of winning a 8-member knock-out tournament of 200 rounds for each pairing? These numbers are obtained by simulating a huge number of possible pairings and multiplying the corresponding winning probabilities. We get

Dake Quinton 33%

RX-24 0%

R2-D2 30%

Jennirafe 7%

Brains 0%

Cantor 28%

Randy 1%

Imbad 0%

There are only three likely candidates for the title left.

Why does Cantor, who would win each one of the others if the number of rounds being played would be very high (let's say 1,000,000,000 rounds) not have the highest chance of winning the tournament?

If the robots RX-24 and Brains are replaced by the more reasonable versions RX-24rev and Brainsrev, the probabilities that the robots win the whole tournament are:

Dake Quinton 21%

RX-24rev 18%

R2-D2 17%

Jennirafe 3%

Brainsrev 19%

Cantor 20%

Randy 1%

Imbad 0%

Thus we have 5 likely candidates for the title in that case.

Quarter-Final:

The (random) pairing was rather favorable for Dake Quinton. With this pairing, the winning probabilities for Dake Quinton increased to 48%. R2-D2 and Cantor still had probabilities for winning the tournament of 28% respectively 23%. All other robots had winning probabilities of 1% or less.

The results in the quarter-final were as follows:

Name Name Result (for Left) Expected Result Odds Left:Right

Dake Quinton Randy 55 26 92:8

RX-24 ImBad 70 90 100:0

R2-D2 Brains 65 43 99:1

Jennirafe Cantor 18 -8 33:67

As usual, what we expect is not what we get, but the numbers didn't deviate too much (thanks to the Law of Large Numbers). The deviations in the plays were 29, -20, 22, 26.

Semi-Final:

Before playing this semi-final, the winning probabilities changed further. The probabilities for winning the tournament are now 59% for Dake Quinton, 38% for R2-D2, 3% for Jennirafe, and less than 1% for RX-24.

Name Name Result (for Left) Expected Result Odds Left:Right

Dake Quinton RX-24 58 51.3 99:1

R2-D2 Jennirafe 37 19.9 84:16

Final

Question: Dake Quinton's chance for winning the tournament was 59% before the semi-final. Dake Quinton won, but still its (his?) winning chance decreased to 55%. How is this possible?

Name Name Result (for Left) Expected Result Odds Left:Right

Dake Quinton R2-D2 ... 2.3 55:44

The winner is ...

Name	Name	Result (for Left)	Expected Result	Odds Left:Right
Dake Quinton	Randy	55	26	92:8
RX-24	ImBad	70	90	100:0
R2-D2	Brains	65	43	99:1
Jennirafe	Cantor	18	-8	33:67