P(N)為0到N之間猜數的最大勝率。邊界條件為P(1) = 1（實際上可以取任何值，因為P(2)固定為0）。算出比較小的項：P[2] = 0, i = 1P[3] = 1/2, i = 1P[4] = 2/3, i = 2P[5] = 1/2, i = 2P[6] = 2/5, i = 1P[7] = 1/2, i = 1P[8] = 4/7, i = 2P[9] = 1/2, i = 2P[10] = 4/9, i = 1P[11] = 1/2, i = 1P[12] = 6/11, i = 2P[13] = 1/2, i = 2P[14] = 6/13, i = 1P[15] = 1/2, i = 1P[16] = 8/15, i = 2P[17] = 1/2, i = 2P[18] = 8/17, i = 1P[19] = 1/2, i = 1P[20] = 10/19, i = 2不難發現勝率可以寫成：或者所有的奇數勝率相等，偶數中四的倍數先手勝率高，否則後手勝率高。選擇上則是四的倍數或四的倍數餘1選隔一個猜，另外兩種情況猜最小。反過來猜最大自然也可以。可以用數學歸納法證明。
P(N)為0到N之間猜數的最大勝率。邊界條件為P(1) = 1（實際上可以取任何值，因為P(2)固定為0）。算出比較小的項：P[2] = 0, i = 1P[3] = 1/2, i = 1P[4] = 2/3, i = 2P[5] = 1/2, i = 2P[6] = 2/5, i = 1P[7] = 1/2, i = 1P[8] = 4/7, i = 2P[9] = 1/2, i = 2P[10] = 4/9, i = 1P[11] = 1/2, i = 1P[12] = 6/11, i = 2P[13] = 1/2, i = 2P[14] = 6/13, i = 1P[15] = 1/2, i = 1P[16] = 8/15, i = 2P[17] = 1/2, i = 2P[18] = 8/17, i = 1P[19] = 1/2, i = 1P[20] = 10/19, i = 2不難發現勝率可以寫成：或者所有的奇數勝率相等，偶數中四的倍數先手勝率高，否則後手勝率高。選擇上則是四的倍數或四的倍數餘1選隔一個猜，另外兩種情況猜最小。反過來猜最大自然也可以。可以用數學歸納法證明。