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Question Number 164995 by Mathematification last updated on 24/Jan/22

Answered by aleks041103 last updated on 25/Jan/22

$${Let}^{\prime }slabelthetopvertex{V}^{+},thebottom$$$$one-V^{-},andthevertecesonthebase-O.$$$$Notation:$$$$p(n\mid A\to B)-theprobabilityofarriving$$$$atBafternedgesstartingfromA.$$$$ex.p(10\mid V^{+}\to V^{+})istheprobabilitywewant.$$$$$$$$Itisobvious,thatifafternsteps$$$$theantiseitheron{V}^{+}oron{V}^{-},then$$$$theantmusthavebeenonOthepreviousstep.$$$$IftheantisonO,2of4timesshewill$$$$chosetoremainonOand1/4timesshe$$$$willgoto{V}^{+}or{V}^{-}.Then:$$$$p(n\mid V^{+}\to V^{+})=p(n\mid V^{+}\to V^{-})=\frac{1}{4}p(n-1\mid V^{+}\to O)$$$$\Rightarrow p(n\mid V^{+}\to O)=4p(n+1\mid V^{+}\to V^{+})$$$$Butthetotalprobabilityafternsteps$$$$mustbe1:$$$$p(n\mid V^{+}\to V^{+})+p(n\mid V^{+}\to V^{-})+p(n\mid V^{+}\to O)=1$$$$\Rightarrow 2p(n\mid V^{+}\to V^{+})+4p(n+1\mid V^{+}\to V^{+})=1$$$$letp(n\mid V^{+}\to V^{+})=p_n,then:$$$$2p_n+4p_{n+1}=1$$$$\Rightarrow p_{n+1}=\frac{1}{4}-\frac{1}{2}{p}_n$$$$let{p}_n=a+b_n\Rightarrow$$$$a+b_{n+1}=\frac{1}{4}-\frac{a}{2}-\frac{1}{2}{b}_n$$$$\Rightarrow leta=\frac{1}{4}-\frac{a}{2},i.e.a=\frac{2}{3}\frac{1}{4}=\frac{1}{6},then$$$$b_{n+1}=-\frac{1}{2}{b}_n$$$$geometricprogression:$$$$b_n={(-\frac{1}{2})}^{n-1}{b}_1$$$$\Rightarrow p_n-\frac{1}{6}={(-\frac{1}{2})}^{n-1}(p_1-\frac{1}{6})$$$$But{p}_1=p(1\mid V^{+}\to V^{+})=0$$$$\Rightarrow p_n=\frac{1}{6}(1-{(-\frac{1}{2})}^{n-1})$$$$Thereforetheanswermustbe:$$$$p_{10}=\frac{1}{6}(1+\frac{1}{2^9})=\frac{513}{3072}=\frac{171}{1024}=p_{10}$$

Commented by aleks041103 last updated on 25/Jan/22

$$Itisworthnoting,that{p}_n\to \frac{1}{6},when$$$$n\to \mathrm{\infty }.Thisisconnectedto{Markov}^{\prime }srule$$$$thatafterenoughrandomwalksonagraph,$$$$theprobabilityofbeingonanyvertex$$$$isproportionaltothenumberofedges$$$$connectedtothegivenvertex.Inour$$$$caseallverteceshave4edgesconnected$$$$tothem,andthereforetheprobability$$$$isthesameforall6verteces,i.e.{it}^{\prime }s$$$$\frac{1}{6}.$$$$$$

Commented by Tawa11 last updated on 25/Jan/22

$$\mathrm{Great}\mathrm{sir}$$

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