BigWheel.R

#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)
   }
}