BGYD45 Problems
Information Theory
1. Lion Roars: Lions live in large groups called “prides” each of which has its own
hunting territory. Prides rarely hunt as a unit, but instead break up into smaller hunting
groups. The size of a hunting unit is thus highly variable from day to day. Hunting groups
will often roar at dawn as they get ready to hunt. If there are hunting groups from other
prides nearby, they will roar in reply. Each group compares the amplitude and persistence
of roars in the other group to that of its own. These roaring contests thus give each group
an opportunity to decide whether its group has more members than the other one. In all
cases, one or the other group will decide it is smaller, quit roaring first, and slink away to
avoid having to fight.
Suppose there is a 50:50 chance on average that the other group in such a roaring contest
is in fact larger than yours. Because the propagation of the roars over a given distance
may differ depending upon whether the roaring lions are on a high or low location, in the
open or in forest, etc., the roars do not provide perfect information about the size of their
group. A larger group will persist in roaring longest 70% of the time, but will erroneously
quit first 30% of the time. A smaller group will sometimes roar longer than the bigger
group and thus win 30% of the time, but will correctly quit first and leave 70% of the
time.
Question: Calculate the average amount of information (in bits) that would be required to
always know which group was larger (e.g. what is the amount of perfect information
needed here?). Then compute the amount of information (in bits) actually transferred by
the average roaring contest. Dividing the actual amount by the amount of perfect
information (e.g. divide the last computation by the first one) and multiplying by 100%
gives the percentage of perfect information provided by an average roaring contest.
Which of the following values for that percentage is closest to the true value given the
numbers above?
a) 2% b) 6% c) 12% d) 21%
Justify your choice quantitatively (by now you should have already calculated the correct
answer and no longer need the multiple choice -)
The amount of perfect information needed to completely clarify which group is larger is
H(C) = – (0.5 log2 0.5 + 0.5 log2 0.5) = 1.0 bit.
After the signal, there will be some residual uncertainty about which group is larger. This
is H(C|S) and is computed from the following two tables:
Signals:
P(C i and S j)
Other Group
roars
shorter/softer
Other Group
roars
longer/louder
P(C i |S j)
Contexts:
Own group
Own group
is larger
is smaller
Contexts:
Own group
Own group
is larger
is smaller
0.5 x 0.7 =
0.35
0.5 x 0.3 =
0.15
0.50
0.35/0.5 =
0.70
0.15/0.5 =
0.30
0.5 x 0.3 =
0.15
0.5 x 0.7 =
0.35
0.50
0.15/0.5 =
0.30
0.35/0.5 =
0.70
0.50
0.50
Because H(C|S) = – S (Pi and Cj) log2 (Pi | Cj), we get :
H(C|S) = – ( 0.35 log2 0.70 + 0.15 log2 0.30 + 0.15 log2 0.30 + 0.35 log2 0.70 )
= 0.881bits.
Thus the amount of information transferred on average, HT = H(C) – H(C|S) = 0.118
bits.
The fraction of the maximum possible which was transferred is thus HT/H(C) = 0.118/1.0
= 0.118 which converted to a percentage and rounding to the second place gives 12%.
Thus of the choices offered, the closest one is c).
Game Theory
2. Legislative Scumbags: Suppose a US politician, Senator Ripoff, has damning
evidence in his files that proves he personally benefitted from the award of a large
defense contract to a specific company. Suppose that a second US politician, Senator
Slime, was also in on the deal, made a tidy profit himself, and has equivalent
documentation in his files. Each knows what is in the other's files. Suppose that if both of
them keep their mouths shut, the chances of either being found out and indicted by the
Senate Anti-Scumbag Committee is only 10%. On the other hand, if one of them shreds
his own files and mentions to the New York Times that it might be fun to look in the
other's files, he looks so honest and righteous that his chances of being indicted drop to
5%. Were he to do so, the chances of the other being indicted, once the files are
examined, rise to 30%. If both of them shred their own files and then rat on the other,
they both look like such crooks that the chances they will be indicted rise to 15% (for
each). Assuming they want to avoid indictment, what is the ESS? (Is there a moral here?)
g
g
y
1
Minimize the probability of indictment.
BGYD45 Problems
3. Birth Control Strategies: Two lovers meet periodically for a tryst. Neither wants the
woman to get pregnant. Either of them could be the party responsible for buying some
birth control device. If one of them does buy a device, it costs $4 whether the man or the
woman purchases it. If the woman gets pregnant, it costs each of them $100. Both of
them want to minimize the net costs. For example, if one of them bought a device and the
woman got pregnant anyway, the one who bought the device would lose $4 + $100 and
the one who did not would lose $100. Let p = the probability that the woman would still
get pregnant if one of them uses a birth control device, and p2 = the probability that the
woman will get pregnant if both bring and use birth control devices on the same occasion.
(You should be able to figure out why this is the appropriate term). Let q = the
probability that the woman will get pregnant if neither of them uses a device. Consider
the following four sets of values for p and q, determine the optimal strategy in each case
(including the equilibrium values of buying or not buying a device if a mixed ESS), and
explain what agreement might be set up by the pair to insure meeting this optimal
strategy:
Case I: All available birth control devices are only 95% effective (p = .05). The couple is
ignorant of methods for computing when the greatest risks of pregnancy occur (e.g.,
taking daily temperatures) and there is a 25% chance that if they make love without a
device, the woman will get pregnant (q = .25).
*Case II: All available birth control devices are only 95% effective. How-ever, this
couple is aware of how to reduce chances of getting pregnant by avoiding certain days
each month. In his case, the risk of getting preg-nant even without any device is only 6%.
Case III: All available birth control devices are 99% effective. The couple is unaware of
how to avoid the most risky occasions and hence there is a 25%chance the woman will
get pregnant if no device is used.
*Case IV: All available birth control devices are 99% effective. The couple is very good
at monitoring risky times and in this case, there is only a 2%chance the woman will get
pregnant even though they use no device.
*When they use a device, they don't bother to count.
q (100) +
(1-q)(0) =
100 q
As with Problem 1, the cell values are costs and the players wish to minimize them.
The outcomes depend on the values of p and q. The four cases are summarized
below:
Case I:
Clearly, USES is a pure ESS since it gives the lowest costs regardless of what the partner
does.
Case II:
This is a Mixed Unstable system. There will be an unstable equilibrium point when USE
is adopted 80% of the time. However, the chances of drifting to either side of this
equilibrium will cause either 100% USE or 100% DON'T USE to be the stable ESS. If
we let f be the fraction of players playing USE, then the equilibrium point is
f=
9.00-6.00
= 0.80
(9.00-6.00)+(5.00-4.25)
If we had to predict which strategy was more likely to be found, it would be DON’T USE
because this is the rare strategy at the equilibrium point.
Case III: The table appears as follows:
This is a Mixed Stable System. If f is the fraction of the population playing USE, then
the equilibrium value will be
f=
25.00-5.00
= 0.869
(25.00-5.00)+(4.01-1.00)
Once everyone starts using a device about 87% of the time, the average payoff of playing
USE equals that of playing DON'T USE. We can thus compute the global average payoff
by just evaluating the payoff to USErs at the equilibrium f:
average PO = (0.869) ($4.01) + (1 - 0.869) ($5) = $4.14
The optimum is thus for everyone to buy a device 87% of the time at an average cost of
$4.14. (Now don't use this advice: we left AIDS out of the game).
Case IV: The table is
In this case, the outcome is a pure ESS for DON'T USE.
Bonus: Value of Information
Billy Seppo is the prime minister of a rich and powerful country, and wants to win reelection. He has two strategies to increase his support: 1) improve the economy by
concentrating on domestic policy; or 2) raise patriotic sentiment by starting a war. The
voting population is made up of “hawks”, “economists” and “pacifists”, with
probabilities P(hawk)=0.5, P(economist)=0.4 and P(pacifist)=0.1. In the event of a war,
hawks will vote for the incumbent prime minister (Billy Seppo) with P=0.9, and
economists would do so with P=0.3. If Billy Seppo works on the domestic agenda, hawks
will vote for him with P = 0.5 and economists will vote for him with P=0.7. Pacifists will
only vote to express their disapproval of a war, which depends on how many people die.
If each hawk and each economist represents 1 million votes, and there is one pacifist vote
(against Billy Seppo) for each person killed in a war, then how many people have to die
for Billy Seppo to lose the election if he goes to war?
This is a value of information question (like the students going or not to lectures question).
VI = ΣPi∆QiPOi - Ki
But the trick is that the cost only comes from the pacifists, so the K term must be discounted by a
factor of 0.1 (proportion of pacifists in population). Then you solve for the break-even cost.
Therefore:
VI = 0.5(0.9 – 0.5)(106) + 0.4(0.3 – 0.7)(106) – 0.1K
P(hawk)(votesifwar – votesifpeace)(1 million per hawk)
+ P(economist) (votesifwar – votesifpeace)(1 million per economist)
- P(pacifist)(number of dead)
Billy Seppo loses when
ΣPi∆QiPOi = 0.1Ki
(unless it’s the US and he doesn’t really need more than 50% of the vote).
0.5(0.9 – 0.5)(106) + 0.4(0.3 – 0.7)(106) = 0.1K
K = 400,000