Petya is playing a computer game called "Pile of Stones." Initially, there are 16 stones in the pile. Players take turns removing 1, 2, 3, or 4 stones from the pile. The player who takes the last stone wins. Petya is playing for the first time and therefore takes a random number of stones each time, without violating the rules of the game. The computer plays according to the following algorithm: on each turn, it takes as many stones as needed to be in the most advantageous position. Petya always goes first. What is the probability that Petya will win?
Note that the player making the first move always has an advantage and wins with the correct strategy. Indeed, on the first step, one should take one stone from the pile, and on each subsequent step, take such a number of stones so that the remaining number of stones is divisible by 5. Since according to the rules of the game, on each step, it is allowed to take 1, 2, 3, or 4 stones, such a strategy is always feasible.

At the same time, if at any step the player making the first move deviates from this strategy, then his opponent has the opportunity to win the game by using the same strategy. Note also that if the player making the first move wins, he must make exactly 4 moves during the game.

Thus, Petya has only one chance not to lose at each move: if he takes one stone first, leaves the computer exactly 10 stones on the second move, exactly 5 stones on the third, and takes all the remaining stones on the fourth.

After each of Petya's moves, the computer, to minimize the probability of losing, must take one stone. Then Petya wins only if he takes 4 stones from the four remaining to maintain divisibility by 5.

Thus, the probability of Petya winning is $\left(\frac{1}{4}\right)^{4}=1 / 256$, where $1 / 4$ is the probability that Petya will take the correct number of stones on his turn.

## Answer

$1 / 256$.