关于公平硬币抛掷中首次正面出现在第N次的条件概率计算问询

Hey there! Let's break down this conditional probability problem clearly—you've got a solid start, but let's untangle the part that was tripping you up with $P(B|A)$.

First, let's reaffirm your event definitions to stay aligned:

• $A$: The first head appears on the $N$th toss (so the first $N-1$ tosses are all tails, and the $N$th is heads)
• $B$: At least one head appears in the first $N+M$ tosses

The key simplification for $P(B|A)$ #

You overcomplicated $P(B|A)$ earlier, and that's totally understandable! Let's think about what $P(B|A)$ means: it's the probability that $B$ happens given that $A$ has already occurred.

If $A$ is true, we know two things:

1. The first $N-1$ tosses are all tails
2. The $N$th toss is heads

Wait a second—if the $N$th toss is heads, that already satisfies the requirement for $B$ ("at least one head in the first $N+M$ tosses")! The remaining $M$ tosses (from $N+1$ to $N+M$) can be anything—tails, heads, whatever—because we already have a head at the $N$th position.

So $P(B|A) = 1$. That's it! No need to sum over those $M$ tosses; the event $A$ itself guarantees that $B$ is true.

Let's fix and simplify your original formula #

Now let's plug this back into the Bayesian formula, and also simplify the other terms to make the calculation cleaner:

1. $P(A)$: As you correctly wrote, this is the probability of $N-1$ tails followed by 1 head: $P(A) = (1/2)^{N-1} * (1/2) = (1/2)^N$
2. $P(B)$: Instead of summing all the "at least one head" cases, use the complement rule (it's way easier!). The only way $B$ doesn't happen is if all $N+M$ tosses are tails. So $P(B) = 1 - P(\text{all tails}) = 1 - (1/2)^{N+M}$

Now substitute into Bayes' theorem:
$$
P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} = \frac{1 \cdot (1/2)^N}{1 - (1/2)^{N+M}}
$$

We can clean this up by multiplying numerator and denominator by $2^{N+M}$ to eliminate the fractions:
$$
P(A|B) = \frac{2^M}{2{N+M} - 1}
$$

Quick sanity check #

Let's test with a small example to make sure this makes sense. Suppose $N=2$, $M=1$ (so we're asking: what's the probability the first head is on the 2nd toss, given there's at least one head in the first 3 tosses?).

• $P(A) = (1/2)^2 = 1/4$
• $P(B) = 1 - (1/2)^3 = 7/8$
• $P(A|B) = (1/4)/(7/8) = 2/7$
  Using our simplified formula: $2^1/(2{3}-1) = 2/7$—matches perfectly!

Why your original $P(B|A)$ guess was overcomplicating it #

You tried to calculate $P(B|A)$ by accounting for heads in the $M$ remaining tosses, but that's unnecessary. Since $A$ already ensures there's a head at the $N$th toss, $B$ is automatically true regardless of what happens in the later $M$ tosses. That's why $P(B|A)$ is just 1.

Hope that clears things up—let me know if you want to walk through any more edge cases or simplifications!

Greetings

备注：内容来源于stack exchange，提问作者daniel