Stark Woods Solution Chp1-7 3ed 3c1d4y

Solution Manual

13

The three door game (Exercise 1.50 in text) Here, we take up the task of analyzing the problem, e.g., using Bayes’ theorem, to calculate the posterior conditional expectation.

There are three doors . Events are determined by where the treasure is, by which door we choose, and by which opens up after our first selection. I.e.,

if we designate by a triple the collection of outcomes: treasure in A, select A, open C which is empty, the event that treasure is behind door is simply

while the event that we select



where



run over all possibilities



is similarly denoted by

!

The space of outcomes is: "

# $% $% &% ' # $ ( &% ) &% $%(*) & (' &%* +, $%(' ', $%(' ' # $

and -

the family of its subsets /.0 " -

1 $1 2 31465%!

Following the description of the problem, not all elementary events have the same probability. We have 7

:+; # &% $<



:+; (+ ) &% $<



:+; (' ', $%(' ' # <



:+; % #<



:+; %* =<



:+; / #<



:+; / '*<



:+; /' =<



:+; /' *<

89

14 while 7

7

: # < ?@:+; <



: ( +, < ?@:+; (+ $<



: (' *, $< AB:+; (' ' # < &!

8 9 >



The above are due to the fact that, e.g., contains only one elemen $ tary outcome, because only can be opened, whereas contains two outcomes with equal probability. These facts define everything! From here on we can enumerate, we can check independence, etc. etc. or, we can compute the conditional probability for potential benefit in a “switching strategy”. This is what we do next. First, by direct enumeration: Given that we select , and that the door behind it, we need to compare :+; DC 2 $<



has been opened with no treasure

:+; *& $DCE # $< $!

and

:+; DC # < GF

:+; *& $DC 2# $<

If we can show that then a “switching strategy” helps. Using the probabilities of the elementary events listed earlier, we compute that 7

:+; < H 8

7

:+; # < H

8 9JI

:+; MLN 2 $< O



and

2# $

7

8 9 >



7 K

:+; # $< P

Thus,

7

7 Q

:+; < R:+; 2 $< &

are obviously independent, and : $<

:+; DC 2 $< S



:+; 2# $<

7 8

!

Similarly (though done a bit differently for variety), :+; *& $DC 2# $< S

:+; *& $MLT 2 $<

: # $<

15 :+; * # <



:+; # <

V;U W

XU >



!

8

Second, using Bayes’ theorem: We again enumerate possibilities, except that we turn things around, which sometimes makes it easier. We compute: :+; (' DC 2 $< S

: # DC (' < R:+; *& $/

:+; # <

: $DCE (' < R:+; ('$ $<



:+; # <

VU VU V;U W I

V;U W 9 U

8

> !

Here we used the fact that # YLT (' J $YLZ (' $





because if we selected and treasure is in , is the only choice for open * # ing. This is reflected in the fact that there is no event other than in ('& $ . Of course, if we realize that “no switching” gives us a V U probability of winning, then we would expect that “switching” would have an advantage, and it does. But it is not entirely obvious. The tools and concepts of probability allow a systematic approach to all such questions. The bottom line is that we have to enumerate possibilities. In this case it may look tedious, but in many cases we can save quite a bit of effort using such an approach (e.g., using Bayes’ theorem). Incidentally, as you can see, it is often the case that the right “language” and the right “notation” allows us to think and compute more easily than without it!

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