Math 486. Febr. 4, 2010. Midterm 1.
5 problems, 15 pts each. Name_______Dr.V_______
m1 ..... /75. total ................./.....
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Find an equilibrium in pure strategies and the corresponding payoff. In Problems 1 and 2, the bet is $1.
1. Restricted Nim. Last move wins. Players alternate and can take 1,2,4, or 5 stones in a move from a pile. Initial position: 2 piles, 100 and 200 stones.
Solution. First we do one pile
|
stones |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
|
take |
L |
1 |
2 |
L |
1 or 4 |
2 or 5 |
L |
1 or 4 |
2 or 5 |
L |
1 or 4 |
2 or 5 |
Period is 3. A winning strategy: keep the number of stones divisible by 3. You loose (L) against a perfect player if you start with the number divisible by 3.
Now we consider two piles. Still we have period 3 for W and L in both directions.
A winning strategy: keep the numbers of stones congruent each other modulo 3. It suffices to take 1 or 2 stones in a move to win.
In particular, in the position (100,200) take either 1 or 4 from the pile of 200 or 2 or 5 from pile of 100.
2. Blackjack. Player (P) has hard 17. Dealer (D) shows 10. Cards remaining including D's face-down are 4, 4,4.7, 7,7.
see pdf by H.K.
D's face down card is 7 D stands at 17 payoff is $0.
/
0.5
/
P stands at 17 - --0.5-- -> D's face down card is 4 D drws at 14 D wins. P's payoff is -$1.
stand / expected payoff is -$0.5.
P -$0.15
draws \
-$0.15 P draws at 17 --- 0.5---> P gets 7. The payoff is -$1. D gets 7 and stands at 17. $1.. D gets 7 and ties. $0.
\ 0.5 / 0.6 | 0.75
$0.7 P gets 4. P has 21. P cannot do better. P stands at 21.D draws. ---0.4--> D gets 4. D draws at 14. $0.25.
| 0.25
D gets 4 again and stands at 18. $1
The value of game is -$0.15. Optimal solution is to draw.
3. 2 player game in normal form.
|
7, 1 |
4,0 |
-1, 3 |
0,0 |
3, 3 |
-3,4 |
|
5,-1 |
5,0 |
-2, 5 |
1, 4 |
3,1 |
0,0 |
|
0, 5 |
4,-1 |
-2, 4 |
6, 0 |
0, 3 |
1,1 |
|
1,1 |
5,0 |
0, 3 |
6,0 |
3,3 |
3,3 |
Solution.
|
7*, 1 |
4,0 |
-1, 3 |
0,0 |
3*, 3 |
-3,4* |
|
5,-1 |
5*,0 |
-2, 5* |
1, 4 |
3*,1 |
0,0 |
|
0, 5* |
4,-1 |
-2, 4 |
6*, 0 |
0, 3 |
1,1 |
|
1,1 |
5*,0 |
0*, 3* |
6*,0 |
3*,3* |
3*,3* |
There are 3 equilibria. All in the last row, in columns 3,5,6. They are positions with two *.
4. Extensive form, 3 players, A B, C.
initial position B
/ \
/ \
A B
/ \ / \
B C B C
/ \ / \ / \ / \
1,2,3 0,-1,0 -1,-2,1 -1,0,1 2,-1,0 0,0,2
Solution. Double line indicate a strategy profiles which are equilibria.
initial position B
(1,2,3) the payoff at both equilibria
/ / \
// \
A (1,2,3) B (0,0,2) or (-1,0,1)
/ / \ / / \ \
B (1,2,3) C (-1,-2,1) (-1,0,1) B C (0,0,2)
/ / \ / \ \ / / \ / \ \
1,2,3 0,-1,0 -1,-2,1 -1,0,1 2,-1,0 0,0,2
5. Game with 3 players, A, B, C. in normal form.
strategy payoff
A B C A B C
1 1 1 0 -1 1
1 1 2 1 1 -2
1 2 1 1 0 0
1 2 2 0 0 1
2 1 1 0 -1 1
2 1 2 1 1 -2
2 2 1 1 0 -1
2 2 2 -1 1 0
3 1 1 0 -1 1
3 1 2 -1 1 -2
3 2 1 1 -1 0
3 2 2 -1 0 0
Solution.
strategy payoff
A B C A B C
1 1 1 0 -1 1
1 1 2 1 1 -2
1 2 1 1 0 0
1 2 2 0 0 1
2 1 1 0 -1 1
2 1 2 1 1 -2
2 2 1 1 0 -1
2 2 2 -1 1 0
3 1 1 0* -1* 1* equilibrium
3 1 2 -1 1 -2
3 2 1 1* -1* 0* equilibrium
3 2 2 -1 0 0
*maximal payoff for this player if the other players do not change.