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<blockquote data-quote="Cadence" data-source="post: 8928819" data-attributes="member: 6701124">--------------------Edit: See post #23 below for a much shorter way to do it.[URL unfurl=&quot;true&quot;]https://www.enworld.org/threads/introducing-the-countdown-dice-mechanic.695102/page-2#post-8928919[/URL]---------------------I think the attached R code using a transition matrix is correct for giving you what turn you run out on.[SPOILER=&quot;R Code for Using a Transition Matrix&quot;]#Code using transition matrix#Set the number of starting dicedat1&lt;-6#Set how many turns to check fornturn&lt;-50#Set the number of sides on the dicenside&lt;-6#Make the initial state vectors1&lt;-c(1,rep(0,dat1))#Transition matrixtmat&lt;-matrix(0,ncol=dat1+1,nrow=dat1+1)for (i in 1: (dat1+1)){&nbsp; temp&lt;-dbinom(((dat1-i+1):0),&nbsp; &nbsp; dat1-i+1,(nside-1)/nside)&nbsp; tmat[i,(i: (dat1+1))]&lt;-temp}&nbsp;#Get probability it has run out by that turncurrent&lt;-s1%*%tmatpendby&lt;-current[dat1+1]for (i in 1: (nturn-1)){&nbsp; current&lt;-current%*%tmat&nbsp; pendby&lt;-c(pendby,current[dat1+1])}#Get probability it has run out on that turnpendon&lt;-pendby[1]for (i in 2:nturn){pendon&lt;-c(pendon,pendby[ i] -pendby[i-1])}pendon&lt;-c(pendon,1-sum(pendon))x&lt;-cbind(1: (nturn+1),round(pendon,4),round(c(pendby,1),4))colnames(x)=c(&quot;Turn&quot;,&quot;Prob On&quot;,&quot;Prob By&quot;)x[/SPOILER]The first value of pendon is the probability of running out of dice on turn 1, etc...&nbsp; The last value is the probability it takes more than whatever you set the max number of turns to check for.&nbsp;&nbsp; pendby will give you the probability of running out on that turn or before.I checked it by writing one that did a simulation.[SPOILER=&quot;R code using a simulation.&nbsp; Make sure settings match the one using the transition matrix.&quot;]#Code using a simulation#Set the number of starting dicedat1&lt;-6#Set how many turns to check fornturn&lt;-50#Set the number of sides on the dicenside&lt;-6#Set number of simulationsnsims&lt;-100000simcount&lt;-rep(0,(nturn+1))for (i in 1:nsims){&nbsp; curcount&lt;-dat1&nbsp; turn&lt;-0&nbsp; while ((curcount&gt;0)&amp;(turn&lt;(nturn+1))){&nbsp; &nbsp; turn&lt;-turn+1&nbsp; &nbsp; curcount&lt;-rbinom(1,curcount,(nside-1)/nside)&nbsp; &nbsp; if (curcount==0){simcount[ i]&lt;-turn}&nbsp; &nbsp; if ((curcount&gt;0)&amp;(turn==nturn)){simcount[ i]&lt;-&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; (nturn+1)}&nbsp; }}table(simcount)/nsims#Compare them to make sure it workedcbind(1: (nturn+1),round(pendon,4),&nbsp;&nbsp; round(table(simcount)/nsims,4))[/SPOILER]Here are the probabilities for starting with three dice and going up to 20 turns.&nbsp; (Remember the last entry is &quot;More than 20&quot;).&nbsp; &nbsp; &nbsp; Turn Prob On Prob By&nbsp;[1,]&nbsp; &nbsp; 1&nbsp; 0.0046&nbsp; 0.0046&nbsp;[2,]&nbsp; &nbsp; 2&nbsp; 0.0239&nbsp; 0.0285&nbsp;[3,]&nbsp; &nbsp; 3&nbsp; 0.0462&nbsp; 0.0748&nbsp;[4,]&nbsp; &nbsp; 4&nbsp; 0.0640&nbsp; 0.1388&nbsp;[5,]&nbsp; &nbsp; 5&nbsp; 0.0752&nbsp; 0.2140&nbsp;[6,]&nbsp; &nbsp; 6&nbsp; 0.0802&nbsp; 0.2942&nbsp;[7,]&nbsp; &nbsp; 7&nbsp; 0.0805&nbsp; 0.3747&nbsp;[8,]&nbsp; &nbsp; 8&nbsp; 0.0773&nbsp; 0.4520&nbsp;[9,]&nbsp; &nbsp; 9&nbsp; 0.0720&nbsp; 0.5240[10,]&nbsp;&nbsp; 10&nbsp; 0.0655&nbsp; 0.5895[11,]&nbsp;&nbsp; 11&nbsp; 0.0586&nbsp; 0.6481[12,]&nbsp;&nbsp; 12&nbsp; 0.0517&nbsp; 0.6999[13,]&nbsp;&nbsp; 13&nbsp; 0.0451&nbsp; 0.7450[14,]&nbsp;&nbsp; 14&nbsp; 0.0391&nbsp; 0.7841[15,]&nbsp;&nbsp; 15&nbsp; 0.0336&nbsp; 0.8176[16,]&nbsp;&nbsp; 16&nbsp; 0.0287&nbsp; 0.8464[17,]&nbsp;&nbsp; 17&nbsp; 0.0244&nbsp; 0.8708[18,]&nbsp;&nbsp; 18&nbsp; 0.0207&nbsp; 0.8915[19,]&nbsp;&nbsp; 19&nbsp; 0.0175&nbsp; 0.9090[20,]&nbsp;&nbsp; 20&nbsp; 0.0148&nbsp; 0.9238[21,]&nbsp;&nbsp; 21&nbsp; 0.0762&nbsp; 1.0000Here it is for six dice&nbsp; &nbsp; &nbsp; Turn Prob On Prob By&nbsp;[1,]&nbsp; &nbsp; 1&nbsp; 0.0000&nbsp; 0.0000&nbsp;[2,]&nbsp; &nbsp; 2&nbsp; 0.0008&nbsp; 0.0008&nbsp;[3,]&nbsp; &nbsp; 3&nbsp; 0.0048&nbsp; 0.0056&nbsp;[4,]&nbsp; &nbsp; 4&nbsp; 0.0137&nbsp; 0.0193&nbsp;[5,]&nbsp; &nbsp; 5&nbsp; 0.0265&nbsp; 0.0458&nbsp;[6,]&nbsp; &nbsp; 6&nbsp; 0.0408&nbsp; 0.0866&nbsp;[7,]&nbsp; &nbsp; 7&nbsp; 0.0538&nbsp; 0.1404&nbsp;[8,]&nbsp; &nbsp; 8&nbsp; 0.0639&nbsp; 0.2043&nbsp;[9,]&nbsp; &nbsp; 9&nbsp; 0.0703&nbsp; 0.2746[10,]&nbsp;&nbsp; 10&nbsp; 0.0730&nbsp; 0.3475[11,]&nbsp;&nbsp; 11&nbsp; 0.0725&nbsp; 0.4201[12,]&nbsp;&nbsp; 12&nbsp; 0.0697&nbsp; 0.4898[13,]&nbsp;&nbsp; 13&nbsp; 0.0652&nbsp; 0.5550[14,]&nbsp;&nbsp; 14&nbsp; 0.0597&nbsp; 0.6148[15,]&nbsp;&nbsp; 15&nbsp; 0.0538&nbsp; 0.6685[16,]&nbsp;&nbsp; 16&nbsp; 0.0478&nbsp; 0.7163[17,]&nbsp;&nbsp; 17&nbsp; 0.0419&nbsp; 0.7583[18,]&nbsp;&nbsp; 18&nbsp; 0.0365&nbsp; 0.7948[19,]&nbsp;&nbsp; 19&nbsp; 0.0315&nbsp; 0.8263[20,]&nbsp;&nbsp; 20&nbsp; 0.0271&nbsp; 0.8534[21,]&nbsp;&nbsp; 21&nbsp; 0.1466&nbsp; 1.0000Edit: I think I fixed all the auto-emoji and mark-up making in the code snippets.</blockquote>

[QUOTE="Cadence, post: 8928819, member: 6701124"] -------------------- Edit: See post #23 below for a much shorter way to do it. [URL unfurl="true"]https://www.enworld.org/threads/introducing-the-countdown-dice-mechanic.695102/page-2#post-8928919[/URL] --------------------- I think the attached R code using a transition matrix is correct for giving you what turn you run out on. [SPOILER="R Code for Using a Transition Matrix"]#Code using transition matrix #Set the number of starting dice dat1<-6 #Set how many turns to check for nturn<-50 #Set the number of sides on the dice nside<-6 #Make the initial state vector s1<-c(1,rep(0,dat1)) #Transition matrix tmat<-matrix(0,ncol=dat1+1,nrow=dat1+1) for (i in 1: (dat1+1)){ temp<-dbinom(((dat1-i+1):0), dat1-i+1,(nside-1)/nside) tmat[i,(i: (dat1+1))]<-temp } #Get probability it has run out by that turn current<-s1%*%tmat pendby<-current[dat1+1] for (i in 1: (nturn-1)){ current<-current%*%tmat pendby<-c(pendby,current[dat1+1])} #Get probability it has run out on that turn pendon<-pendby[1] for (i in 2:nturn){ pendon<-c(pendon,pendby[ i] -pendby[i-1])} pendon<-c(pendon,1-sum(pendon)) x<-cbind(1: (nturn+1),round(pendon,4),round(c(pendby,1),4)) colnames(x)=c("Turn","Prob On","Prob By") x [/SPOILER] The first value of pendon is the probability of running out of dice on turn 1, etc... The last value is the probability it takes more than whatever you set the max number of turns to check for. pendby will give you the probability of running out on that turn or before. I checked it by writing one that did a simulation. [SPOILER="R code using a simulation. Make sure settings match the one using the transition matrix."]#Code using a simulation #Set the number of starting dice dat1<-6 #Set how many turns to check for nturn<-50 #Set the number of sides on the dice nside<-6 #Set number of simulations nsims<-100000 simcount<-rep(0,(nturn+1)) for (i in 1:nsims){ curcount<-dat1 turn<-0 while ((curcount>0)&(turn<(nturn+1))){ turn<-turn+1 curcount<-rbinom(1,curcount,(nside-1)/nside) if (curcount==0){simcount[ i]<-turn} if ((curcount>0)&(turn==nturn)){simcount[ i]<- (nturn+1)} }} table(simcount)/nsims #Compare them to make sure it worked cbind(1: (nturn+1),round(pendon,4), round(table(simcount)/nsims,4))[/SPOILER] Here are the probabilities for starting with three dice and going up to 20 turns. (Remember the last entry is "More than 20"). Turn Prob On Prob By [1,] 1 0.0046 0.0046 [2,] 2 0.0239 0.0285 [3,] 3 0.0462 0.0748 [4,] 4 0.0640 0.1388 [5,] 5 0.0752 0.2140 [6,] 6 0.0802 0.2942 [7,] 7 0.0805 0.3747 [8,] 8 0.0773 0.4520 [9,] 9 0.0720 0.5240 [10,] 10 0.0655 0.5895 [11,] 11 0.0586 0.6481 [12,] 12 0.0517 0.6999 [13,] 13 0.0451 0.7450 [14,] 14 0.0391 0.7841 [15,] 15 0.0336 0.8176 [16,] 16 0.0287 0.8464 [17,] 17 0.0244 0.8708 [18,] 18 0.0207 0.8915 [19,] 19 0.0175 0.9090 [20,] 20 0.0148 0.9238 [21,] 21 0.0762 1.0000 Here it is for six dice Turn Prob On Prob By [1,] 1 0.0000 0.0000 [2,] 2 0.0008 0.0008 [3,] 3 0.0048 0.0056 [4,] 4 0.0137 0.0193 [5,] 5 0.0265 0.0458 [6,] 6 0.0408 0.0866 [7,] 7 0.0538 0.1404 [8,] 8 0.0639 0.2043 [9,] 9 0.0703 0.2746 [10,] 10 0.0730 0.3475 [11,] 11 0.0725 0.4201 [12,] 12 0.0697 0.4898 [13,] 13 0.0652 0.5550 [14,] 14 0.0597 0.6148 [15,] 15 0.0538 0.6685 [16,] 16 0.0478 0.7163 [17,] 17 0.0419 0.7583 [18,] 18 0.0365 0.7948 [19,] 19 0.0315 0.8263 [20,] 20 0.0271 0.8534 [21,] 21 0.1466 1.0000 Edit: I think I fixed all the auto-emoji and mark-up making in the code snippets. [/QUOTE]

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