Chess Match

To find this probability we only focus what will happen when one of the two eventually win. On the next game that one of the two wins, the probability that it will be Kasparov is 3/8. This is found using conditional probability.

On the next three games on which one of the two win, it must be Kasparov if he is going to win the round. Thus, the answer should be (3/8)^3 = 27/512 ~ .0527.

To be sure, we will simulate chess rounds 1,000,000 times.

game <- function(N){
  Wins <- c(0,0)  ## They start off with 0 Rounds won each. 
  names(Wins) <- c("Karpov", "Kasparov")
  for(i in 1:N){  ## Simulate this any number of times. 
    chess <- c() ## Represents one chess round that will fill with 0's, 1's or a -3
    keepPlaying <- c(0,1,2)  ## While the sum of the chess vector is in this
                             ## they must continue play. 
    while(sum(chess) %in% keepPlaying){
      chess <- c(chess, sample(c(-3,0,1),1,prob = c(5/8,40/8,3/8)))
      # Fill up the round vector. If -3 is selected before three 1's are selected,
      # this would mean Karpov wins. 
    }
    # The sum of the chess vector is negative for Karpov, otherwise Kasparove wins. 
    ifelse(sum(chess)<0, Wins[1] <- Wins[1]+1, Wins[2] <- Wins[2]+1)
  }
  return(rbind(Wins))
}
set.seed(42)
game(1000000)
##      Karpov Kasparov
## Wins 947035    52965

The .052965 empirical result seems to confirm the theoretical answer.

Prismatic Problem

We would like to find whole numbers a, b, and c that satisfy the equation \[ abc = 2(ab+ac+bc). \]
In other words, we want to find solutions to the Diophantine equation \[ 2ab + 2ac + 2bc - abc = 0. \] I’m not one to try and find solutions to Diophantine equations. Let’s have a computer program do an exhaustive search instead.

VeqSA <- c()

for(a in 1:1000){
  for(b in a:1000){
    for(c in b:1000){
     if(a*b*c == 2*(a*b+a*c+b*c)){
       VeqSA <- rbind(VeqSA, c(a,b,c))
     } }
  }
    
}

VeqSA
##       [,1] [,2] [,3]
##  [1,]    3    7   42
##  [2,]    3    8   24
##  [3,]    3    9   18
##  [4,]    3   10   15
##  [5,]    3   12   12
##  [6,]    4    5   20
##  [7,]    4    6   12
##  [8,]    4    8    8
##  [9,]    5    5   10
## [10,]    6    6    6

The end result, which only has ten solutions with all side lengths less than or equal to 42, suggests that these may be the only ones. There may be a proof for this, but I’m unfamiliar with it if it exists.

Until then, we have come upon another possible great question to Life, the Universe, and Everything:

Among all the rectangular prisms with whole number sides that have a volume (in cubic units) equal to the surface area (in square units), what is the longest side of the one that has the largest volume/surface area?