Squid game Episode 7 with simulation - r

Question

Last night I saw the episode 7 of the Squid game tv series. The episode has a game with binomial distribution in the bridge.
Specifically there are 16 players and a bridge with 18 pair of glasses (one pure glass and one safe glass).If one player happened to choose the pure glass then the glass couldn't stand the weight of the player and the glass broke. The next player had the advantage that he/she was starting from the position that the last player had and continues the binomial search.At the end 3 players happened to cross the bridge.
So i was wondering: It is like, I have 16 euros in my pocket and I play head or tails with p = 1/2. Every time I bet on heads. If the coin flip is head then I earn 0 and if is tails I lose 1 euro. What is the probability of hitting 18 times (consecutive or not) heads and to be left 3 euros in my pocket.
I tried to simulate this problem in R:
squid_bridge = function(a,n,p) {
players = a
while (position > 0 & position < n) {
jump = sample(c(0,1),1,prob=c(1-p,p))
position = position + jump
}
if (position == 0)
return(1)
else
return(0)
}
n = 18
trials = 100000
a = 16
p = 1/2
set.seed(1)
simlist = replicate(trials, squid_bridge(a, n, p))
It does not seem to work. Any help?

Answer 1

Here is a Monte Carlo experiment in R returning the distribution of the number of failures.
apply(apply(matrix(rgeom(16*1e6,.5)+1,nc=16),1,cumsum)>18,1,mean)
#with details:
#rgeom(16*1e6,.5)+1 for 16 x 10⁶ geometric simulations when
#the outcome is the number of attempts till "success",
# "success" included
#,1,cumsum) for the number of steps till 16th "success"
#)>18 for counting the cases when a player manages to X the bridge
#1,mean) for finding the probability of each of the players to X
This is not a Binomial but a truncated Negative Binomial experiment in that the number of new steps made by each player is a Geometric Geom(1/2) variate unless the 18 steps have been made. The average number of survivors is thus
sum(1-pnbinom(size=1:16,q=17:2,prob=.5))
#Explanation:
#pnbinom is the Negative Binomial cdf
#with size the number of "successes"
#q the integer at which the cdf is computed
#prob is the Negative Binomial probability parameter
#Because nbinom() is calibrated as the number of attempts
#before "success", rather than until "success", the value of
#q decreases by one for each player in the game
whose value is 7.000076, rather than 16-18/2=7!

Answer 2

Here is how I think you can model the game in R. The first version is similar to what you have: there's a 50% chance of guessing correctly and if the guess is correct, the players advance a tile. Otherwise they do not, and the number of players decrements by 1. If the number of players reaches 0, or they advance to the end, the game ends. This is shown in squid_bridge1().
squid_bridge1 <- function(players, n, prob) {
if (players == 0 | n == 18) {
# All players have died or we have reached the end
return(players)
}
jump <- rbinom(1, 1, prob)
if (jump == 0) {
# Player died
return(squid_bridge1(players - 1, n, prob))
}
if (jump == 1 & n < 18) {
# Player lives and advances 1 space
return(squid_bridge1(players, n + 1, prob))
}
}
However, this does not accurately depict the game since a wrong guess gives the remaining players additional information. If a player chooses wrong, the probability of the next guess being correct is not 50%, it's 100%. However, after that point the probability of a correct guess decreases to 50%. This can be accounted for with another argument to keep track of the correctness of the previous guess.
squid_bridge2 <- function(players, n, prob, previous) {
if (players == 0 | n == 18) {
# The game ends if there are no players or they have reached the end
return(players)
}
if (previous == 0) {
# The previous guess was wrong, but now the players know where to go next
return(squid_bridge2(players, n + 1, prob, previous = 1))
}
jump <- rbinom(1, 1, prob)
if (jump == 0) {
# Player died
return(squid_bridge2(players - 1, n, prob, previous = 0))
}
if (jump == 1 & n < 18) {
# Move is correct. Advance 1 space
return(squid_bridge2(players, n + 1, prob, previous = 1))
}
}
However, there's a catch. It wasn't quite that simple in the show, and players fell for reasons other than an incorrect guess (being pushed, jumping on purpose, etc.). I don't know what a reasonable probability of doing something like this is, but it is likely low, let's say 10%.
not_stupid <- function() {
x <- runif(1, 0, 1)
if (x <= 0.1) {
return(FALSE)
} else {
return(TRUE)
}
}
Since emotions spike just before each move, we will test this prior to each move.
squid_bridge3 <- function(players, n, prob, previous) {
if (players == 0 | n == 18) {
# The game is over because there are no players left or they reached the end
return(players)
}
if (previous == 0) {
# The previous guess was wrong, but now the players know where to go next
return(squid_bridge3(players, n + 1, prob, previous = 1))
}
if (!not_stupid()) {
return(squid_bridge3(players - 1, n, prob, previous = 1))
}
jump <- rbinom(1, 1, prob)
if (jump == 0) {
# Player died because of either choosing wrong or a self-inflicted loss
return(squid_bridge3(players - 1, n, prob, previous = 0))
}
if (jump == 1 & n < 18) {
# Move is correct. Advance 1 space
return(squid_bridge3(players, n + 1, prob, previous = 1))
}
}
Then running some simulations:
set.seed(123)
trials <- 10000
players <- 16
squid1 <- replicate(trials, squid_bridge1(players, 0, 0.5))
squid2 <- replicate(trials, squid_bridge2(players, 0, 0.5, 1))
squid3 <- replicate(trials, squid_bridge3(16, 0, 0.5, 1))
df <- tibble(squid1 = squid1,
squid2 = squid2,
squid3 = squid3) %>%
pivot_longer(cols = c(squid1, squid2, squid3))
ggplot(data = df,
aes(x = value)) +
geom_histogram(bins = 10,
binwidth = 1,
fill = "cornflowerblue",
color = "black") +
facet_wrap(~name,
nrow = 3) +
xlab("# of players to make it to the end") +
scale_x_continuous(breaks = seq(0, 16, by = 1),
labels = seq(0, 16, by = 1))
As you can see below, the first situation is heavily skewed to the left. Since the players are essentially "blindly guessing" at each tile, it is unlikely that any will make it to the end. However, after accounting for the information gained from a wrong guess, it averages somewhere around 7 players making it. By adding in a random chance of falling for another reason, the distribution skews to the left some.
Average for first situation: 1.45
Average for second situation: 7.01
Average for third situation: 4.99
To answer the question of the probability of only 3 players making it, I get ~ 10.8% for the last case
Edit: As requested, here is the code to generate the plots. I also fixed the various functions that had some naming issues (went through a few different names when I made them). It looks like it resulted in a slight bug for the 3rd function, but I have fixed it throughout.

Answer 3

○ △ □
##########
# Game ○ △ □
##########
squidd7<-function(N_Fields,N_Players,p_new_field){
Players<-data.frame(id = 1:N_Players, alive=rep(1,N_Players),Field=0)
for(i in 1:N_Players){
while (Players[i,"alive"]==TRUE && max(Players$Field)< N_Fields) {
Players[i,"Field"]=Players[i,"Field"]+1 # Jump onto the next Field
Players[i,"alive"]=rbinom(1,1,p_new_field)# Fall or repeat
}
Players[i+1,"Field"]=Players[i,"Field"] # next player starts where prior player died
}
Players<-Players[1:N_Players,] # cosmetic because i do i+1 in the prior line
# Print me some messages
if(!any(Players$alive>0)){
cat("Players loose!")
} else {
cat(" \n After", max(Players$Field),"goal was reached! ")
cat("Players",Players[Players$alive==1,"id"], "survive")
}
return(Players)
}
squidd7(18,16,0.5)
###########
# simulation ○ △ □
###########
results<-data.frame(matrix(0, nrow = 100, ncol = 20))
for(x in 1:ncol(results)){
for (i in 1:nrow(results)) {
Players<-squidd7(x+7,16,0.5)
results[i,x]<-sum(Players$alive)
}
}
###########
## Results ○○□□○ △ □
sdt<-apply(results,2,sd) # standart devation
mn<-apply(results,2,mean) # ○ △ □
boxplot(results,xlab ="n Steps ",names = 8:27,ylab="N Survivors of 16 ")
points(mn,type="l")
points(sdt,type="l")
colors<-colorRampPalette(c(rgb(0,1,0,0.4),
rgb(1,1,0,0.4),
rgb(1,0,0,0.4)), alpha = TRUE)(21)
plot(density(results$X1),type="n",xlim=c(-1,17),ylim=c(0,0.30),
main="○ △ □ ",
sub="○ △ □ ○ △ □ ○ △ □",
xlab="number of survivors")
for( i in 1:21){
polygon(density(results[,i]),col= colors[i])
}
legend(15,0.31,title="Steps",legend=8:28,fill=colors,border = NA,
y.intersp = 0.5,
cex = 0.8, text.font = 0.3)

Answer 4

well to simulate this game. you will need 50%/50% chance 16 times. meaning all you have to code is 50% chance of not losing and run it 16 times than if you lose you it will do a -1 form a varible of 18. that will create a perfect digital recreation of squidgame bridge

Answer 5

If I'm understanding correctly, I believe the other answers are complicating the simulation.
We can simulate draws from a binomial distribution of size 18. Every 1 kills someone (assuming there is anyone left to kill). Thus we can calculate the number of survivors by subtracting the number of 1s drawn from the number of players, truncated at 0 (any negative results are counted as 0, via pmax()).
set.seed(47)
n_sim = 1e4
survivors = pmax(0, 16 - rbinom(n_sim, size = 18, prob = 0.5))
mean(survivors)
# [1] 7.009
The mean seems to approach Xi'an's answer of 7.000076 as n_sim increases. At 500M simulations (which still runs rather quickly with this method!) we get a mean of 7.000073.
Plotting these results, they appear basically identical to cazman's squid2 scenario.
ggplot(data.frame(survivors), aes(x = survivors)) + geom_bar() +
scale_y_continuous(limits = c(0, 6000))