X, Y and Z in that order throws a tetrahedral die the first throws a 4 wins. The game continues indefinitely until someone wins. Find the probability that X wins, Y wins and Z wins.
outcomes: the last letter is a winner, Prob of each is found:
X = 1/4 = (1/4)
XY = (3/4)(1/4) = (1/4)3/4)
XYZ =(3/4)(3/4)(1/4) = (1/4)(3/4)^2
XYZX = (3/4)(3/4)(3/4)(1/4) = (1/4)(3/4)^3
XYZXY = ...
XYZXYZ = ..
XYZXYZX = (1/4)(3/4)^6
Prob that X wins
= (1/4) + (1/4)(3/4)^3 + (1/4)(3/4)^6
this is a GP with a = 1/4 , r = (3/4)^3
sum to infinity = a/(1-r)
= (1/4) / (1- 27/64)
= (1/4) / (37/64)
= (1/4)(64/37)
= 16/37
prob that Y wins
.....
follow my steps
To find the probability that X, Y, or Z wins the game, we can use the concept of geometric probability.
Let's start by finding the probability that X wins. X wins on the first throw by rolling a 4. The probability of rolling a 4 on a tetrahedral die is 1/4. Therefore, the probability that X wins on the first throw is 1/4.
If X doesn't win on the first throw, the game continues to the next player, Y. Now, for Y to win, X must not win, and Y must roll a 4 on their first throw. The probability that X doesn't win on the first throw is 3/4 (since there are three other numbers on the die). The probability of rolling a 4 on a tetrahedral die is still 1/4. Therefore, the probability that Y wins on their first throw is (3/4) * (1/4) = 3/16.
If neither X nor Y win on their first throws, the game moves to Z. Similar to before, for Z to win, X and Y must not win, and Z must roll a 4 on their first throw. The probability that neither X nor Y win on their first throws is (3/4) * (3/4) = 9/16. The probability of rolling a 4 on a tetrahedral die is still 1/4. Therefore, the probability that Z wins on their first throw is (9/16) * (1/4) = 9/64.
We can continue this pattern indefinitely to find the probabilities of X, Y, and Z winning on subsequent throws. The probabilities form a geometric sequence, where each term is obtained by multiplying the previous term by (3/4) * (3/4).
The probability that X wins can be calculated as follows:
P(X wins) = (1/4) + (3/4) * (3/4)^2 + (3/4) * (3/4)^4 + (3/4) * (3/4)^6 + ...
The probability that Y wins can be calculated similarly:
P(Y wins) = (3/4) * (1/4) + (3/4) * (3/4)^3 + (3/4) * (3/4)^5 + (3/4) * (3/4)^7 + ...
The probability that Z wins can be calculated as follows:
P(Z wins) = (9/16) * (1/4) + (9/16) * (3/4)^2 + (9/16) * (3/4)^4 + (9/16) * (3/4)^6 + ...
Note that the sum of all these probabilities should be equal to 1 since the game will eventually be won by one of the players (either X, Y, or Z).
To calculate the exact probabilities, you can use a calculator or write a program to compute the infinite geometric series.