Two people A and B have 8 stones and 6 stones respectively. The first person A, rolls a die, and whatever outcome he gets {1,2,3,4,5,6}, he will take those many stones from person B. Then person B rolls a die, and whatever outcome B gets {1,2,3,4,5,6}, he will take those many stones from A.

After each game, who ever has more stones wins the game. If both have the same stones, then the game is repeated. We have to find the probability that B wins the game.

Solving the problem using Simulation

Stone_Dice_Problem <- function()
  {
A <- 8
B <- 6


repeat{
  
  x <- sample(1:6,1,rep=T)
  A <- (A + x)
  B <- (B - x)
  
  x <- sample(1:6,1,rep=T)
  B <- (B + x)
  A <- (A - x)  
  
  if( A > B)
    {
    
    return(0)
    
    }
  
   if( A < B)
    {
    
    return(1)
    
    }
    
}

}

Simulating the experiment 10000 times…

mean(replicate(10000,Stone_Dice_Problem()))
## [1] 0.3448

Solving the problem theoritically

r1 <- c('A','-','B','B','B','B')
r2 <- c('A','A','-','B','B','B')
r3 <- c('A','A','A','-','B','B')
r4 <- c('A','A','A','A','-','B')
r5 <- c('A','A','A','A','A','-')
r6 <- c('A','A','A','A','A','A')

df <- rbind(r1,r2,r3,r4,r5,r6)
df <- data.frame(df)
names(df) <- c('B1','B2','B3','B4','B5','B6')
row.names(df) <- c('A1','A2','A3','A4','A5','A6')
print(df)
##    B1 B2 B3 B4 B5 B6
## A1  A  -  B  B  B  B
## A2  A  A  -  B  B  B
## A3  A  A  A  -  B  B
## A4  A  A  A  A  -  B
## A5  A  A  A  A  A  -
## A6  A  A  A  A  A  A

Fig: 1 - Data frame showing the sample space

In the above data frame display, A1, A2, A3…A6 represent the events that person A gets 1,2,3,…,6 respectively, and B1, B2, B3 … B6 represent the events that person B gets 1,2,3,…,6 respectively. If an element of the data frame has A, it represents that A wins (for example, [A1,B1] element has A). If an element of the data frame has B, then it represents the event that B wins (for example, [A1,B3] has B). Similarly, if an element of the data frame has “-”, then it represents that neither A nor B wins the game…and the game has to be played once more to know the winner.

From Fig-1 (data frame), we can get the following probabilities:

Probability that A wins = P(A) = 21/36 Probability that B wins = P(B) = 10/36 Probability that neither A wins nor B wins = P(Draw) = 5/36

For B to win before A, we need the following events to happen:

B Wins in the first game (or) Neither A nor B wins the first game, and B wins the second game (or) Neither A nor B wins in the first 2 games, and B wins the third game …

Mathematically…

P(B) + P(Draw).P(B) + P(Draw).P(Draw).P(B) + … (10/36) + (5/36) . (10/36) + (5/36) . (5/36). (10/36) + …

Applying the sum of geometric series formula …

P(B wins before A) = (10/36)/(1 - 5/36) = (10/36)/(31/36) = 10/31

Note:

You may also apply the formula listed in Dobrow book page 74 (as given below).

A before B

In repeated independent trials, if A dn B are mutually exculsive events, the probability that A occurs before B is \[ \frac{P}{P+Q} \]

In repeated independent trials, if A dn B are mutually exculsive events, the probability that B occurs before A is \[ \frac{Q}{P+Q} \]

Where P = probability of event A, and Q = probability of event B

Therefore using the above formula also, the probability that person B wins before person A is:

\[ (\frac{10}{36}) / (\frac{10}{36} + \frac{21}{36}) = 10/31 \]

Therefore the probability that B wins before A = 10/31