Estimating Probabilities in Perudo - r

Question

I am new to coding and using R. I am working on a project to simulate the game Liar's Dice, also known as Perudo, and have some questions about creating the simulation.
Basically, the game consists of two or more players rolling five dice in a cup, turning it over, and and making bids on how many of a certain side they think is on the table. You can look at your own dice, but not anyone else's. To make bids, on your turn you would say "two 5's," which would mean there are at least two dice that landed on 5. Each bid will either increase the side or the amount. So if you said "two 5's," I could then say "two 6's" or "three 3's" on my turn.
When you believe the last bid is incorrect, you would say "Liar" on your turn, then everyone reveals their dice. If you were wrong, you lose a dice, but if you were right, the last bidder loses a dice. This continues until there is only one player left who has dice.
First, I decided to create a function called cup() which rolls a cup of five six-sided dice.
cup <- function(sides = 6, dice = 5){
sample(1:sides, size = dice, replace = TRUE)
}
Next, with a little assistance, I created a new function called cups() which rolls three cups for three players.
cups <- function(players = 3, sides = 6, dice = 5){
out <- cup(sides, dice)
for(i in 2:players){
out <- rbind(out, cup(sides, dice))
}
rownames(out) <- 1:players
rownames(out) <- c("P1", "P2", "P3")
return(out)
}
What I want to accomplish next is to create a table of probabilities of possible dice outcomes. In other words, what's the probability of there being at least two of a side given fifteen dice (five for each player) in play? And then the probability of there being three, four, five, etc. all the way up to fifteen in this case.
My question is how would I go about doing this in R? And what direction should I go in after getting the probabilities in R?

Answer 1

Here is an empirical process for determining the percentage outcomes of all the same, 4 the same, 3 the same, 2 the same, none the same upon rolling 5 die:
library(gtools) # package with permutations function
allcombos <- permutations(6, 5, repeats.allowed = TRUE) # all 6 choose 5 with replacment combos
alluniques <- apply(allcombos, 1, unique) # uniques for each combo
alllengths <- sapply(alluniques, length) # lengths for each combo imputes num repeats
alllengths2 <- as.factor(alllengths) # convert to factor to count unique
allsum <- summary(alllengths2) # sum by num uniques
allsum
1 2 3 4 5 # 1=all same, 2=4 same, 3=3 same, 4=2 same, 5=all different
6 450 3000 3600 720
totsum <- sum(allsum)
allfrac <- allsum / totsum
allpercent <- allfrac * 100
allpercent
1 2 3 4 5
0.07716049 5.78703704 38.58024691 46.29629630 9.25925926 # percentage breakout
There is no doubt an analytical solution but I don't know what it is. You could use standard probability calculations to estimate specific outcomes among multiple players. E.g. P(at least 1 4-same | 3 players) or run some simulations.

Answer 2

Here's likely more than you asked for but focusing on number of sides on the dice, total number of dice and probability of rolling Nrolled or more
dicegame <- function(Nsides = 6,
Ndice = 5,
Nrolled = 1,
verbose = FALSE)
{
total_possible_outcomes <- choose(Nsides + Ndice - 1, Ndice)
outcomes_matrix <- t(combn(Nsides + Ndice - 1,
Ndice,
sort)) - matrix(rep(c(0:(Ndice - 1)),
each = total_possible_outcomes),
nrow = total_possible_outcomes)
chances <- sum(apply(outcomes_matrix, 1, function(x) sum(x==2)) >= Nrolled) / total_possible_outcomes
if(verbose) {
cat(paste("Number of dice",
Ndice,
"each with", Nsides, "sides",
"chances of rolling", Nrolled,
"\n or more of any one side are:\n"))
}
return(chances)
# return(total_possible_outcomes)
# return(outcomes_matrix)
}
dicegame(verbose = TRUE)
#> Number of dice 5 each with 6 sides chances of rolling 1
#> or more of any one side are:
#> [1] 0.5
dicegame(6, 15, 10)
#> [1] 0.01625387

Answer 3

Using probability we can demonstrate that the probability to get a value n times is equal to :
we can easily write this into an R function:
prob_get_n <- function(ntimes, players=3, dice=5, sides=6){
if(missing(ntimes)) ntimes <- 0:(players*dice)
choose(players*dice,ntimes)*(1-1/sides)^((players*dice)-ntimes)*sides^(-ntimes)
}
Notice that this function is by construction vectorised ie it accepts 1:2, c(9,5) as valid inputs.
prob_get_n() -> probs
data.frame(ntimes=1:length(probs)-1, probs=probs,or_more= rev(cumsum(rev(probs))))
ntimes probs or_more
1 0 6.490547e-02 1.000000e+00
2 1 1.947164e-01 9.350945e-01
3 2 2.726030e-01 7.403781e-01
4 3 2.362559e-01 4.677751e-01
5 4 1.417535e-01 2.315192e-01
6 5 6.237156e-02 8.976567e-02
7 6 2.079052e-02 2.739411e-02
8 7 5.346134e-03 6.603585e-03
9 8 1.069227e-03 1.257451e-03
10 9 1.663242e-04 1.882242e-04
11 10 1.995890e-05 2.190005e-05
12 11 1.814445e-06 1.941153e-06
13 12 1.209630e-07 1.267076e-07
14 13 5.582909e-09 5.744548e-09
15 14 1.595117e-10 1.616385e-10
16 15 2.126822e-12 2.126822e-12
Edit
Or we can use R built in dbinom function to get the distribution and pbinom to get the cumulative probability function:
probs <- function(ntimes, players=3, dice=5, sides=6){
if(missing(ntimes)) ntimes <- 0:(players*dice)
data.frame(ntimes=ntimes, probs=dbinom(ntimes, players*dice, 1/sides), or_more=1-pbinom(ntimes-1, players*dice, 1/sides))
}
ntimes probs or_more
1 0 6.490547e-02 1.000000e+00
2 1 1.947164e-01 9.350945e-01
3 2 2.726030e-01 7.403781e-01
4 3 2.362559e-01 4.677751e-01
5 4 1.417535e-01 2.315192e-01
6 5 6.237156e-02 8.976567e-02
7 6 2.079052e-02 2.739411e-02
8 7 5.346134e-03 6.603585e-03
9 8 1.069227e-03 1.257451e-03
10 9 1.663242e-04 1.882242e-04
11 10 1.995890e-05 2.190005e-05
12 11 1.814445e-06 1.941153e-06
13 12 1.209630e-07 1.267076e-07
14 13 5.582909e-09 5.744548e-09
15 14 1.595117e-10 1.616385e-10
16 15 2.126822e-12 2.126743e-12