AIME 1995 · 第 15 题

Solution

Solution 1

Think of the problem as a sequence of H's and T's. No two T's can occur in a row, so the successful sequences are composed of blocks of $1$ to $4$ H's separated by T's and end with $5$ H's. Since the probability that the sequence starts with TH is $1/4$, the total probability is that $3/2$ of the probability given that the sequence starts with an H.

The answer to the problem is then the sum of all numbers of the form $\frac 32 \left( \frac 1{2^a} \cdot \frac 12 \cdot \frac 1{2^b} \cdot \frac 12 \cdot \frac 1{2^c} \cdots \right) \cdot \left(\frac 12\right)^5$, where $a,b,c \ldots$ are all numbers $1-4$, since the blocks of H's can range from $1-4$ in length. The sum of all numbers of the form $(1/2)^a$ is $1/2+1/4+1/8+1/16=15/16$, so if there are n blocks of H's before the final five H's, the answer can be rewritten as the sum of all numbers of the form $\frac 32\left( \left(\frac {15}{16}\right)^n \cdot \left(\frac 12\right)^n \right) \cdot \left(\frac 1{32}\right)=\frac 3{64}\left(\frac{15}{32}\right)^n$, where $n$ ranges from $0$ to $\infty$, since that's how many blocks of H's there can be before the final five. This is an infinite geometric series whose sum is $\frac{3/64}{1-(15/32)}=\frac{3}{34}$, so the answer is $\boxed{037}$.

Solution 2

Let $p_H, p_T$ respectively denote the probabilities that a string beginning with H's(H-string) and T's(T-string) are successful. Notice that a T-string is just a T followed by an H-string. Thus,

$p_T = \frac 12p_H.$

A successful string can either be the string HHHHH, start with 1 to 4 H's and then revert to a T, start with a T and then continue with a string starting with H (as there cannot be $2$ T's in a row).

There is a $\frac{1}{2^5} = \frac{1}{32}$ probability we roll HHHHH. On the other hand, there is a $\frac{15}{16}$ probability we roll a H, HH, HHH, or HHHH, and then a probability of $p_T$ to follow that up with a string starting with T. Thus,

$p_H = \left(\frac{15}{16}\right) \cdot \left(\frac 12\right) p_H + \frac{1}{32} \Longrightarrow p_H = \frac{1}{17}.$

The answer is $p_H + p_T = \frac{3}{2}p_H = \frac{3}{34}$, and $m+n = \boxed{037}$.

Solution 3

For simplicity, let's compute the complement, namely the probability of getting to $2$ tails before $5$ heads.

Let $h_{i}$ denote the probability that we get $2$ tails before $5$ heads, given that we have $i$ consecutive heads. Similarly, let $t_{i}$ denote the probability that we get $2$ tails before $5$ heads, given that we have $i$ consecutive tails. Specifically, $h_{5} = 0$ and $t_{2} = 1$. If we can solve for $h_{1}$ and $t_{1}$, we are done; the answer is simply $1/2 * (h_{1} + t_{1})$, since on our first roll, we have equal chances of getting a string with "1 consecutive head" or "1 consecutive tail."

Consider solving for $t_{1}$. If we have 1 consecutive tail, then (a) rolling a head gets us to 1 consecutive head and (b) rolling a tail gets us to 2 consecutive tails. So, we must have:

$t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} h_{1}$

Applying similar logic, we get the equations:

$$\begin{aligned} h_{1} &= \frac{1}{2} h_{2} + \frac{1}{2} t_{1}\\ h_{2} &= \frac{1}{2} h_{3} + \frac{1}{2} t_{1}\\ h_{3} &= \frac{1}{2} h_{4} + \frac{1}{2} t_{1}\\ h_{4} &= \frac{1}{2} h_{5} + \frac{1}{2} t_{1} \end{aligned}$$ Since $h_{5} = 0$, we get $h_{4} = \frac{1}{2} t_{1}$ $\Rightarrow h_{3} = \frac{3}{4} t_{1}$ $\Rightarrow h_{2} = \frac{7}{8} t_{1}$ $\Rightarrow h_{1} = \frac{15}{16} t_{1}$ $\Rightarrow t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} \cdot \frac{15}{16} t_{1} = \frac{1}{2} + \frac{15}{32} t_{1} \Rightarrow t_{1} = \frac{16}{17}$ $\Rightarrow h_{1} = \frac{15}{16} \cdot \frac{16}{17} = \frac{15}{17}$.

So, the probability of reaching $2$ tails before $5$ heads is $\frac{1}{2} (h_{1} + t_{1}) = \frac{31}{34}$; we want the complement, $\frac{3}{34}$, yielding an answer of $3 + 34 = \boxed{037}$.

Note: The same approach still works if we tried solving the original problem rather than the complement; we would have simply used $h_{5} = 1$ and $t_{2} = 0$ instead. The repeated back-substitution is cleaner because we used the complement.

Solution 4(fast)

Consider what happens in the "endgame" or what ultimately leads to the end. Let A denote that a head has been flipped, and let B denote that a tail has been flipped. The endgame outcomes are AAAAA, BAAAAA, BB, ABB, AABB, AAABB, AAAABB. The probabilities of each of these are $\frac{1}{32},\frac{1}{64},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64}$ respectively. The ones where there is five heads are AAAAA and BAAAAA. The sum of these probabilities are $\frac{3}{64}$. The sum of all these endgame outcomes are $\frac{34}{64}$, hence the desired probability is $\frac{3}{34}$, and in this case $m=3,n=34$ so we have $m+n=\boxed{037}$ -vsamc

Solution 5

There can be 1-4 heads and 1 tails in any grouping, so we write $h_5,th_5,h_nth_5,th_nth_5,…$ etc., where $1\leq n\leq 4$ and subscript $n$ means there are $n$ of that state in a row. We see that this gives us 2 geometric sequences: one with first term $\frac{1}{2^5}$ and common ratio $\frac{1}{2}*\left(1/2+1/4+1/8+1/16\right)$ and one with first term $\frac{1}{2^6}$ and the same common ratio. Therefore we get $\frac{3}{34}$ or $\fbox{037}$ ~joeythetoey

Video solution

https://www.youtube.com/watch?v=Vo-4w5Cor9w&t