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# coryhofmann/yahtzee.R

Created February 26, 2017 22:47
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 #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%
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