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Unformatted text preview: Chap. 4, prob. 24. Game Theory: The Minimeor The orer‘n. We are given two players, A and 5', and A picks one of two numbers, 1 or 2.
B then has to gue: what number A picked. If B guesses number s correctly
[3' can be 1 or 2], then 5' gets t dollars from A. If E' is wrong, A gets it dollars from B. One strategyr is for B to randomize his guess, saying 1 with
probability p and 2 with probability 1 — p. The problem will be about what strategy the two players should follow. We have already said 5' will try to
randomize. [f A were to try to avoid losing too much to 3, you might think he should always pick 1, but then that gives 3 something to predict reliably,
and he can count on winning 1 dollar from A every turn. So you see there is a problem of picking a strategy without your opponent being able to read
exactly what it is and therefore being able to anticipate your move. That is
also the reason why a random strategy might be useful. Set G equal to the random variable "B’s gain. So, with 5' using this random
strategy, we get that stem chooses 13 = p  1 — Eu — p}. and 3
EEGA chooses?) = [l — p]  2 — EFL First notice that
E{GAchooses 1] = EEGA chooses 2) at p = %. Then min[E[GA chooses l],E{GAchooses2]} ={
El—rJJﬂ—ia :32 E. The maximum of this is attained at p = % and is equal to %. {You can see this by inspection from the graphs.) The “maximin” is %. Similarly, you can see that if we take the point of view of A trying to minimise
her loses, we would look at G as her loss, and calculate ﬁrst that q'1_g'll_‘1lr qEH,and
max{E{GEchooses 1], E(GBchooses2)} = Again, by inspection, the minimum of this is attained when :1 = %, and the
minimum is 1%, which is the “minimsx”, and is equal, as predicted by Von Neumann, to the maximin. F113 63 3
1:1}; t=§ 2hr:  E
Pissck e’“ : e “ maul fl: NIL17H: re Pf mans; r{ titre?) ~ Mei=9  rm rs =11)
. ,_ are as iii—e"! .___._I 31' a: :' Dillﬁr— Pg 175 #43.
(illlii‘JEEEl2 + EilLE‘JAEEII + [215 Pg 175 #41 [a] an: [b]: s 7 m .2 (E‘in‘u wig“. [ﬁll E {Emu m“. [iii] .2 (Elsa —pr*
1:5 1:5 1:4 1Where p = .17 for [a] and p = .3 for [b]. {c} 2 challenges. Pg 177 #53. {El The probability that an arbitrary couple were both bron on April ED is,
assuming independence and an equal chance of having being born on any given date. [1,3365]? Hence, the number of such couples is approximately Poisson with mean Sﬂﬂﬂﬂﬂﬂﬁﬁﬁ R: .5. Therefore1 the probability that at least one pair were both born on this date is approximately 1 — 34m. {bl
The probability that an arbitrary eouple were born on the same day of the
year is 13"335. Hence? the number of such couple is approximately Poisson with mean El]. ﬂﬂ'ﬂfﬂﬁE = 219.13. Hence, the probability of at least one such pair is 1 — E_21g.1a s: 1. Pg 178 #El]. P{2 beneﬁcialﬂftl
FH  beneﬁcial}3f£1 + P{2 not beneﬁcial}1l.l’£1
3393
E ii
s. + .:' P{beneﬁcial  2} Pg 179 #74. [a] Pp: =n} = {$7 [blPiX:>2}=1_ 54+“ “+94 a 1D Pg 18H #77
P{rejected} = 1 — [.Qj‘l' Pngl] #4
LetY=ex1then _ ‘3 yin ”(m—{Film} yen Pg 181 #13.
Easiest to ﬁrst take leg and then determine the 151 that maximizes leg P{X =
k}. 1']! legP{X = k} = log (1:) +8: lug]: + [n — k} lug[1 — 15} By taking derimtives with respect ten 13, we knew that when p = kfn?
ilegP{X = k} = i — E = ﬂ1mlugP{X = k} is going to be maxi—
m1zed. Pg 181 #18. lugP{X = k} = —J‘L +8: lug}; — lugl[k!]l %lugP{X=k}=—l+’f=ﬂ=>h=k Pg 181 #19
m '3'}
Em = Erew=zr‘eW
i=2 i=1
m . m '
= Zlﬂ_1e_"‘ﬁiﬂfii— 11!: Eu + 1]“‘154‘33Hfﬁ
i=1 i=0
DC“ .
= 121.3" + 11:“—le"‘w.:i! = mm: + 11“]
j=ﬂ
Hence
E[X3] = mux+1ﬂ
= A ZIEE—Allfil'FﬂZiE—AA‘IIII+ZE—A1hifll
i=0 1:” 1:” = Maxi] + 213m + 1} = m? +3.1 + 1) ...
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 Winter '08
 Buckingham
 Addition, Probability

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