edit

94.tex

@@ -17,19 +17,19 @@ $S=\{(SS),(SDS),(SDD),(DSS),(DSD),(DDSS),(DDSD),(DDDS),(DDDD)\}$
 \medskip
 Map each event to a set of outcomes.
 
-$A=\{(SS),(SDS),(SDD)\}$
+First animal survives: $A=\{(SS),(SDS),(SDD)\}$
 
-$B=\{(SS)\}$
+Two animals survive: $B=\{(SS),(SDS),(DSS),(DDSS)\}$
 
-$C=\{(DDDS),(DDDD)\}$
+Third animal dies: $C=\{(DDDS),(DDDD)\}$
 
-$D=\{(SS),(SDS),(DSS),(DDSS)\}$
+At least two animals survive: $D=\{(SS),(SDS),(DSS),(DDSS)\}$
 
-$$A\cap B=\{(SS)\}$$
+$$A\cap B=\{(SS),(SDS)\}$$
 
 $$A^\prime\cup C=\{(DSS),(DSD),(DDSS),(DDSD),(DDDS),(DDDD)\}$$
 
-$$B\cup D=\{(SS)\}$$
+$$B\cup D=\{(SS),(SDS),(DSS),(DDSS)\}$$
 
 $$C\cap D=\{\}$$
 
@@ -98,22 +98,17 @@ $$P\{4\}=P\{10\}={1\over12}$$
 $$P\{5\}=P\{9\}={1\over9}$$
 $$P\{6\}=P\{8\}={5\over36}$$
 $$P\{7\}={1\over6}$$
-Probability of winning ``the point'' is computed by only
-considering rolls in which the outcome is the point itself or 7.
-$$P\{6|(6\cup 7)\}=P\{8|(8\cup7)\}={P\{8\}\over P\{7\}+P\{8\}}={5/36\over1/6+5/36}=5/11$$
-$$P\{5|(5\cup 7)\}=P\{9|(9\cup7)\}={P\{9\}\over P\{7\}+P\{9\}}={1/9\over1/6+1/9}=2/5$$
-$$P\{4|(4\cup 7)\}=P\{10|(10\cup7)\}={P\{10\}\over P\{7\}+P\{10\}}={1/12\over1/6+1/12}=1/3$$
+The probability of rolling point $n$ and then winning is
+$$P\{win|n\}=P\{n\}P\{n|(n\cup7)\}=P\{n\}{P\{n\}\over P\{n\}+1/6}$$
+Note that $(n\cup7)$ means that we only consider rolls of the
+dice that result in $n$ or 7.
+$$P\{win|6\}=P\{win|8\}=(5/36)(5/11)=25/396$$
+$$P\{win|5\}=P\{win|9\}=(1/9)(2/5)=2/45$$
+$$P\{win|4\}=P\{win|10\}=(1/12)(1/3)=1/36$$
 Overall probability of winning:
-$$\eqalign{
-P\{win\}&=P\{7\}+P\{11\}
-+2P\{8|(8\cup7)\}P\{8\}
-+2P\{9|(9\cup7)\}P\{9\}
-+2P\{10|(10\cup7)\}P\{10\}\cr
-&={1\over6}+{1\over18}
-+2\left({5\over11}\cdot{5\over36}\right)
-+2\left({2\over5}\cdot{1\over9}\right)
-+2\left({1\over3}\cdot{1\over12}\right)\cr
-&={244\over495}=0.492929
-}$$
+$$
+P\{win\}=P\{7\}+P\{11\}+\sum_n P\{win|n\}
+={244\over495}=0.492929
+$$
 
 \end