yahtzee.R

#Yahtzee.R
#By Cory Hofmann
#Last Updated: 2/26/2017
#
#This code will simulate 3 rolls of 5 dice, trying to get a Yahtzee.
#Will determine odds of achieving Yahtzee in 1, 2, or 3 rolls.
#Also, to see best hand remaining if attempted and failed.
#Code is set-up to run 1,000,000 times to determine probabilities.
#
#Outputs: 
#yahtzee -> # of Yahtzees in (1 Roll, 2 Roll, 3 Rolls)
#score -> Instances # of matching die (1, 2, 3, 4, 5)

remove(list=ls())

total <- 1000000 #Desired number of games to simulate
score <- rep(0,total)
yahtzee <- c(0, 0, 0)

for (j in 1:total){
	
   ###Roll 1
   roll <- sample(1:6, 5, replace = T)
   highest <- 0

   for (i in c(1:6)){
      if (sum(roll == i) > highest){
	   count <- sum(roll == i)
	   die <- i
	   highest <- sum(roll == i)
      }
   }

   if (count == 5){
	yahtzee[1] = yahtzee[1] + 1
	score[j] <- 5
	next
   }

   ###Roll 2
   roll2 <- sample(1:6, 5-count, replace = T)
   count2 <- count + sum(roll2 == die)

   if (count2 == 5){
	yahtzee[2] = yahtzee[2] + 1
	score[j] <- 5
	next
   }

   #If Roll 2 was 'better' than Roll 1, Reassign
   for (k in c(1:6)){
	if (sum(roll2 == k) > count2){
         count2 <- sum(roll2 == k)
	   die <- k #Reassign
	}	  
   }

   ###Roll 3
   roll3 <- sample(1:6, 5-count2, replace = T)
   count3 <- count2 + sum(roll3 == die)

   #If Roll 3 was 'better' than Roll 2
   for (l in c(1:6)){
	if (sum(roll3 == l) > count3){
         count3 <- sum(roll3 == l)
	   die <- l #Reassign
	}	  
   }

   if (count3 == 5){
	yahtzee[3] = yahtzee[3] + 1
	score[j] <- 5
   }else{
	score[j] <- count3 #Highest multiple
   }

}

#Output
#
#Yahtzee in 1, 2, 3 = .0745%; 1.1668%; 3.3620%;
#Total = 4.6033%
#
#Best Score:
#1: 0.0783%; 2: 25.6249%; 3: 45.2652%; 4: 24.4283%; 5: 4.6033%
#
#For Reference, Exact Probabilities:
#Yahtzee in 1, 2, 3 = .077%; 1.186%; 3.34%;
#Total = 4.603%