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#Price is Right, The Big Wheel
#By: Cory Hofmann, Updated: 2017-12-05
#
#Simulates the Big Wheel to determine percentage of winning based on all possible decisions
#Objective is to determine the probability of winning based on first spin value
#
#In all cases, Player 3 spins in the most logical way (stops if winning, spins a second time if not winning no matter what)
#If Player 3 ties one player, he will spin again if he has less than 0.50, otherwise go to spin-off
#If Player 3 ties both players, he will spin again if he has less than 0.70, otherwise go to spin-off
#These above two conditions are based on the odds of winning a two-man (50%) vs. three-man spin-off (33.3%)
#
#Possible decisions for Player 1 and 2:
#Condition A: Player 1 stays, Player 2 stays only if better
#Condition B: Player 1 stays, Player 2 stays if better or tied
#Condition C: Player 1 stays, Player 2 spins again even if better or tied
#
#Condition D: Player 1 Spins twice, Player 2 Stays only if better
#Condition E: Player 1 spins twice, Player 2 stays if better or tied
#Condition F: Player 1 Spins twice, Player 2 Spins again even if better or tied
n <- 1000000 #number of simulations
spin = matrix(nrow = n, ncol = 12)
for (i in c(1:n)){
for (j in c(1:6)){
spin[i,j] <- 0.05*sample(1:20, 1, replace = T) #Randomly draw 0.05 to 1.00 equal probability
}
score = matrix(nrow = 3, ncol = 6)
###Player 1 potential scores are simple, but make sure they don't spin again on 1.00
if (spin[i,1] == 1.00){
score[1,1:6] <- spin[i,1]
} else {
score[1,1:3] <- spin[i,1] #represents player 1 score for one spin
score[1,4:6] <- spin[i,1] + spin[i,2] #represents player 1 score for two spins
}
#Check to see if Player 1 busted after two spins
if (score[1,4] > 1.00){
score[1,4:6] <- 0
}
###Player 2 has lots of options for his spin decision tree
if (spin[i,3] > score[1,1] || spin[i,3] == 1.00){
score[2,1] <- spin[i,3] #represents conditions A and D
score[2,4] <- spin[i,3]
} else {
score[2,1] <- spin[i,3] + spin[i,4]
score[2,4] <- spin[i,3] + spin[i,4]
}
#check if busted
if (score[2,1] > 1.00){
score[2,1] <- 0
score[2,4] <- 0
}
if (spin[i,3] >= score[1,1] || spin[i,3] == 1.00){
score[2,2] <- spin[i,3] #represents conditions B and E
score[2,5] <- spin[i,3]
} else {
score[2,2] <- spin[i,3] + spin[i,4]
score[2,5] <- spin[i,3] + spin[i,4]
}
#check if busted
if (score[2,2] > 1.00){
score[2,2] <- 0
score[2,5] <- 0
}
#For condition C and F, Player two spins again in all scenarios except where they spin 1.00
if (spin[i,3] != 1.00){
score[2,3] <- spin[i,3] + spin[i,4]
score[2,6] <- spin[i,3] + spin[i,4]
} else {
score[2,3] <- spin[i,3]
score[2,6] <- spin[i,3]
}
#check if busted
if (score[2,3] > 1.00){
score[2,3] <- 0
score[2,6] <- 0
}
###Player 3 stops if winning and spins again if not winning; spins again on ties depending on his total (see above for details)
for (k in c(1:6)){ #for all 6 conditions, player three will act the same
if (spin[i,5] > score[1,k] && spin[i,5] > score[2,k]){
score[3,k] <- spin[i,5]
} else if (spin[i,5] == score[1,k] && spin[i,5] == score[2,k] && spin[i,5] >= 0.70){
score[3,k] <- spin[i,5]
} else if (spin[i,5] == max(score[1,k],score[2,k]) && score[1,k] != score[2,k] && spin[i,5] >= 0.50){
score[3,k] <- spin[i,5]
} else {
score[3,k] <- spin[i,5] + spin[i,6] #all other scenarios, spin the second time!
}
#check to see if Player 3 busted
if (score[3,k] > 1.00){
score[3,k] <- 0
}
#For all six conditions, see who is the winner!
temp <- c(score[1,k], score[2,k], score[3,k])
if ( length(which(temp == max(temp))) == 1){
spin[i,k+6] <- which.max(temp) #winner has highest score
} else {
spin[i,k+6] <- sample(which(temp == max(temp)),1) #randomly draw winner amongst ties for highest score
}
}
}
##Calculate the win percentage for each Player, for all six conditions, based on value of first spin
##Also, check to see scenarios of first player and second player
player1 = matrix(nrow = 20, ncol = 7)
player2 = matrix(nrow = 20, ncol = 7)
player3 = matrix(nrow = 20, ncol = 7)
for (i in c(1:20)){
player1[i,1] <- i*0.05
player2[i,1] <- i*0.05
player3[i,1] <- i*0.05
for (j in c(2:7)){ #cycles through all six conditions (A-F)
player1[i,j] <- 100*sum(spin[,j+5] == 1 & spin[,1] == i*0.05)/sum(spin[,1] == i*0.05) #percentage won by Player 1 for each spin
player2[i,j] <- 100*sum(spin[,j+5] == 2 & spin[,3] == i*0.05)/sum(spin[,3] == i*0.05)
player3[i,j] <- 100*sum(spin[,j+5] == 3 & spin[,5] == i*0.05)/sum(spin[,5] == i*0.05)
}
}
##Player 2 odds of winning based on first spin vs. Player 1, for each of the six conditions
player1v2_a = matrix(nrow = 21, ncol = 21)
player1v2_a[2:21,1] <- 0.05*c(1:20)
player1v2_a[1,2:21] <- 0.05*c(1:20)
player1v2_b = matrix(nrow = 21, ncol = 21)
player1v2_b[2:21,1] <- 0.05*c(1:20)
player1v2_b[1,2:21] <- 0.05*c(1:20)
player1v2_c = matrix(nrow = 21, ncol = 21)
player1v2_c[2:21,1] <- 0.05*c(1:20)
player1v2_c[1,2:21] <- 0.05*c(1:20)
player1v2_d = matrix(nrow = 21, ncol = 21)
player1v2_d[2:21,1] <- 0.05*c(1:20)
player1v2_d[1,2:21] <- 0.05*c(1:20)
player1v2_e = matrix(nrow = 21, ncol = 21)
player1v2_e[2:21,1] <- 0.05*c(1:20)
player1v2_e[1,2:21] <- 0.05*c(1:20)
player1v2_f = matrix(nrow = 21, ncol = 21)
player1v2_f[2:21,1] <- 0.05*c(1:20)
player1v2_f[1,2:21] <- 0.05*c(1:20)
for (i in c(2:21)){
for (j in c(2:21)){
player1v2_a[i,j] <- 100*sum(spin[,7] == 2 & spin[,1] == (i-1)*0.05 & spin[,3] == (j-1)*0.05) / sum(spin[,1] == (i-1)*0.05 & spin[,3] == (j-1)*0.05)
player1v2_b[i,j] <- 100*sum(spin[,8] == 2 & spin[,1] == (i-1)*0.05 & spin[,3] == (j-1)*0.05) / sum(spin[,1] == (i-1)*0.05 & spin[,3] == (j-1)*0.05)
player1v2_c[i,j] <- 100*sum(spin[,9] == 2 & spin[,1] == (i-1)*0.05 & spin[,3] == (j-1)*0.05) / sum(spin[,1] == (i-1)*0.05 & spin[,3] == (j-1)*0.05)
player1v2_d[i,j] <- 100*sum(spin[,10] == 2 & spin[,1] + spin[,2] == (i-1)*0.05 & spin[,3] == (j-1)*0.05) / sum(spin[,1] + spin[,2] == (i-1)*0.05 & spin[,3] == (j-1)*0.05)
player1v2_e[i,j] <- 100*sum(spin[,11] == 2 & spin[,1] + spin[,2] == (i-1)*0.05 & spin[,3] == (j-1)*0.05) / sum(spin[,1] + spin[,2] == (i-1)*0.05 & spin[,3] == (j-1)*0.05)
player1v2_f[i,j] <- 100*sum(spin[,12] == 2 & spin[,1] + spin[,2] == (i-1)*0.05 & spin[,3] == (j-1)*0.05) / sum(spin[,1] + spin[,2] == (i-1)*0.05 & spin[,3] == (j-1)*0.05)
}
}
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