ECE 594D Game Theory and Multiagent Systems
Homework #2 Solutions
1. Craps: To evaluate the probability of winning on the pass line in craps we have the following:
pwin = Pr (7, 11) + Pr (win|4) + ... + Pr (win|10)
where Pr (win|10) indicates the probability of a win given that my first roll is a 10. Let
O = {2, 3, 4, 5, 6, 8, 9, 11, 12}. To evaluate the Pr (win|10), consider all events and associated
probabilities that result in a win
Pr ({10}) = p10
Pr ({O, 10}) = (1 − p7 − p10 )p10
Pr ({O, O, 10}) = (1 − p7 − p10 )2 p10
:
:
n
Pr ({O , 10}) = (1 − p7 − p10 )n p10
Adding the probabilities of these disjoint events gives us
Pr (win|10) =
∞
X
(1 − p7 − p10 )k p10
k=0
=
p10
p7 + p10
The alternative winning probabilities can be solved for in the same fashion.
2. FALSE. If A and B are disjoint,
Pr (A ∩ B) = 0 6= Pr (A) Pr (B)
Think of a single coin, with A being “heads”, and B being “tails”.
3. (a) Pr (X = 3) = 1 − (.4 + .2 + .1) = 0.3
(b) Pr (X is odd) = 0.2 + 0.3 = 0.5
(c) E [X] = 0 · 0.4 + 1 · 0.2 + 2 · 0.1 + 3 · 0.3
h
i
1
1
1
1
(d) E (X+1)
= (0+1)
· 0.5 + (1+1)
· 0.2 + 2+1
· 0.1 +
1
3+1
· 0.3
4. (a) Pr (first toss is H) = Pr (HH ∪ HT ) = 3/9 = 1/3
(b) Pr (second toss is T ) = Pr (HT ∪ T T ) = 3/9 = 1/3
(c) Pr (first toss is H|second toss is T ) =
Pr ((HH ∪ HT ) ∩ (HT ∪ T T ))
1/9
=
= 1/3
Pr (HT ∪ T T )
1/9 + 2/9
(d) Yes. Verify that
Pr (HH) = Pr (H) · Pr (H) , Pr (HT ) = ...
5. (a) Pr (X = 2) = 1/8 + 2/8
(b) Pr (Y = 2) = 2/8 + 1/8
(c) Pr (Y = 3|X = 3) =
2/8
2/8+1/8
= 1/3
(d) E [max X, Y ] =
2/8 · 1 + 1/8 · 2 + 2/8 · 2 + 1/8 · 3 + 2/8 · 3
(e) No. For example
Pr (X = 1 ∩ Y = 1) = 2/8 6= Pr (X = 1) Pr (Y = 1) = (2/8) · (3/8)
6. (a) In each case, pick “top” if Left is greater, “bottom” if Right is greater.
i. BoS:
(2p + 0(1 − p) vs 0p + 1p
ii. Stag hunt:
2p + 0(1 − p) vs 1p + 1(1 − p)
iii. Typewriter:
3p + 0(1 − p) vs 0p + 1(1 − p)
(b) Indifference occurs at equality
i. BoS: p = 1/3
ii. Stag hunt: p = 1/2
iii. Typewriter: p = 1/4