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让我们将 $A$ 表示为 Alice 翻转硬币的次数，将 $B$ 表示为 Bob 翻转硬币的次数。我们给出了条件 $A < B$。找到 $P(A < B)$ 后，我们就可以形成条件期望。通过对称性 $P(A < B) = P(B < A)$，我们可以看到： $$P(A < B) = \frac{1 - P(A =B)}2$$ 有机会 $\frac12\cdot\frac12$ 两个硬币都翻转一次，$\frac14\cdot\frac14$ 两次，$\frac18\cdot\frac18$ 三次，依此类推。因此， $$P(A = B) = \frac14+\frac1{16}+\frac1{64}...$$ $$\rightarrow 4P(A = B) = 1 + \frac14+\frac1{16}+\frac1{64}...$$ 从底部减去顶部我们可以看到： $$3P(A = B) = 1$$ $$\rightarrow P(A = B) = \frac13$$ 使用它我们计算： $$P(A < B) = \frac{1 - \frac13}2 = \frac13$$ 我们现在可以计算 $P(A = n | A < B)$。如果爱丽丝抛硬币一次，我们知道她在第一次抛硬币时一定会得到正面，而鲍勃在第一次抛硬币时一定会得到反面。如果她掷硬币两次，她会得到正面，然后是反面，而鲍勃在前两次抛硬币时会得到反面，依此类推。我们的有条件期望： $$E[A | A < B] = \frac{\frac12\cdot\frac12}{\frac13}(1) + \frac{\frac14\cdot\frac14}{\frac13}(2) + \frac{\frac18\cdot\frac18}{\frac13}(3)... = \frac34(1) + \frac3{16}(2) + \frac3{64}(3)...$$ $$= \frac34(1) + \frac3{16}(2) + \frac3{64}(3)...$$ 我们可以将 $4E[A | A < B] - E[A | A < B]$ 计算为： $$4E[A | A < B] = 3(1) + \frac3{4}(2) + \frac3{16}(3) + \frac3{64}(4)...$$ $$4E[A | A < B] - E[A | A < B] = 3E[A | A < B] = 3 + \frac34 + \frac3{16} + \frac3{64}...$$ 同样我们计算： $$12E[A | A < B] = 12 + 3 + \frac34 + \frac3{16} + \frac3{64}...$$ $$12E[A | A < B] - 3E[A | A < B] = 9E[A | A < B] = 12$$ $$\Longrightarrow E[A | A < B] = \boxed{\frac43}$$

import random

flip_coin = lambda: True if random.randint(1,2) == 1 else False

alice_flip_counts = []
num_iters = 100_000
for i in range(num_iters):

    alice_flips = 0
    while(True):

        bob_flips_heads = flip_coin()
        if bob_flips_heads:
            break

        alice_flips += 1

        alice_flips_heads = flip_coin()
        if alice_flips_heads:
            alice_flip_counts.append(alice_flips)
            break

#expecting close to 4/3
print(sum(alice_flip_counts) / len(alice_flip_counts))

Original Explanation

Let's denote $A$ as the number of times Alice flips her coin and $B$ the number of times Bob flips his coin. We are given the condition $A < B$. Upon finding $P(A < B)$ we can form our conditional expectations. By symmetry $P(A < B) = P(B < A)$ so we see that: $$P(A < B) = \frac{1 - P(A =B)}2$$ There is a $\frac12\cdot\frac12$ chance both coins are flipped once, $\frac14\cdot\frac14$ twice, $\frac18\cdot\frac18$ three times and so on. Thus, $$P(A = B) = \frac14+\frac1{16}+\frac1{64}...$$ $$\rightarrow 4P(A = B) = 1 + \frac14+\frac1{16}+\frac1{64}...$$ Subtracting the top from bottom we see: $$3P(A = B) = 1$$ $$\rightarrow P(A = B) = \frac13$$ Using this we compute: $$P(A < B) = \frac{1 - \frac13}2 = \frac13$$

We can now compute $P(A = n | A < B)$. If Alice flips her coin once, we know that she must get a heads on the first flip and Bob must get a tails on his first flip. If she flips her coin twice, she gets a heads followed by tails and Bob gets tails for his first two flips and so on. Our conditional expectations: $$E[A | A < B] = \frac{\frac12\cdot\frac12}{\frac13}(1) + \frac{\frac14\cdot\frac14}{\frac13}(2) + \frac{\frac18\cdot\frac18}{\frac13}(3)... = \frac34(1) + \frac3{16}(2) + \frac3{64}(3)...$$ $$= \frac34(1) + \frac3{16}(2) + \frac3{64}(3)...$$

We can compute $4E[A | A < B] - E[A | A < B]$ as: $$4E[A | A < B] = 3(1) + \frac3{4}(2) + \frac3{16}(3) + \frac3{64}(4)...$$ $$4E[A | A < B] - E[A | A < B] = 3E[A | A < B] = 3 + \frac34 + \frac3{16} + \frac3{64}...$$ Similarly we compute: $$12E[A | A < B] = 12 + 3 + \frac34 + \frac3{16} + \frac3{64}...$$ $$12E[A | A < B] - 3E[A | A < B] = 9E[A | A < B] = 12$$ $$\Longrightarrow E[A | A < B] = \boxed{\frac43}$$

import random

flip_coin = lambda: True if random.randint(1,2) == 1 else False

alice_flip_counts = []
num_iters = 100_000
for i in range(num_iters):

    alice_flips = 0
    while(True):

        bob_flips_heads = flip_coin()
        if bob_flips_heads:
            break

        alice_flips += 1

        alice_flips_heads = flip_coin()
        if alice_flips_heads:
            alice_flip_counts.append(alice_flips)
            break

#expecting close to 4/3
print(sum(alice_flip_counts) / len(alice_flip_counts))