Free Online Reading/Writing (Kumon Math Workbooks)Â . Kumon Math Workbooks (Under. Write the missing number below the number line.Q: Is the probability of winning a game at each move inversely proportional to the probability of winning an interval? Consider the following proposition: At move $n$ in a game between Player 1 and Player 2, Player 1 wins iff Player 2 played strictly better than Player 1 at move $n$ and at least one of the previous $n−۱$ moves. With this definition, we can partition the $2^{n-1}$ possible sequences (in other words, partition the set of sequences $\lbrace X_1,X_2, \ldots, X_{2^n}\rbrace$, where $X_j$ is the sequence that ends at move $j$) of length $n$ in two sets: The set of sequences that win the game The set of sequences that lose the game I am interested in the number of sequences that win the game (see also this question). Now if I denote by $p_n$ the probability of winning at move $n$ in a game, is there a closed formula for $p_n$? More generally, suppose that $\lbrace X_1,X_2, \ldots, X_{2^n}\rbrace$ is a set of sequences of length $n$: is there a closed formula for $\sum_{k=1}^{2^n} p_k \cdot |X_k|$? That is, I’m interested in the probability of a sequence wining the game. A: A sequence $X$ wins a game if $n_1,n_2,\dots,n_k$ occur, where the $n_i$ are the times the first player (the one whose turn it is to make a move) goes from $n_i$ to $n_{i+1}$. Then $$p_k=k\cdot\left(1-\sum_{i=1}^{k-1} p_{n_i}\right)\;.$$ This is obtained from the recurrence p_k=\sum_{i=1}^k\left