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1995 AIME Problems/Problem 15

Revision as of 19:30, 12 August 2020 by Angelalz (talk | contribs) (→‎solution 5 (has lots of variables and steps, but it is very easy to think up))

Problem

Let $p_{}$ be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of $5$ heads before one encounters a run of $2$ tails. Given that $p_{}$ can be written in the form $m/n$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n$.

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 sequence is blocks of $1$ to $4$ H's separated by T's and ending in $5$ H's. Since the first letter could be T or the sequence could start with a block of H's, the total probability is that $3/2$ of it has to start 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 and T's are successful. Thus,

$p_T = \frac 12p_H.$

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

There is a $\frac{1}{16}$ probability we roll $4$ consecutive H's, and there is a $\frac{15}{16}$ probability we roll a 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{align*} 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{align*}

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 (has lots of variables and steps, but it is very easy to think up)

Front note: The states are (h=heads t=tails)

$h_1$ is the probability of having exactly 1 H in a row. $h_2$ is the probability of having exactly 2 Hs in a row. $h_3$ is the probability of having exactly 3 Hs in a row. $h_4$ is the probability of having exactly 4 Hs in a row. $h_5$ is the probability of having exactly 5 Hs in a row. $t_1$ is the probability of having exactly 1 T in a row. $t_2$ is the probability of having exactly 2 Ts in a row.


(1)Since the probability to get one H is $\frac{1}{2}$, thus the probability of getting two H's in a row is $\frac{1}{2} \cdot \frac{1}{2}$. Hence we have $h_2 = \frac{h_1}{2}$. Similarly, $h_3 = \frac{h_1}{4}$, $h_4 = \frac{h_1}{8}$ and $h_5 = \frac{h_1}{16}$

(2)We can also see $h_1 = \frac{1}{2} + \frac{t_1}{2}$ because the first move has probability $\frac{1}{2}$ of landing heads, and any time there is tails, there is a $\frac{1}{2}$ chance of getting a head.

(3)Similar to part (1), $t_2 = \frac{t_1}{2}$

(4)We also have $t_1 = \frac{1}{2} + \frac{h_1}{2} +\frac{h_2}{2} +\frac{h_3}{2} + \frac{h_4}{2}$ and by then by using (1), $t_1 = \frac{1}{2} + \frac{15h_1}{16}$

(5)Using (2) and (4) we can solve for $t_1$ and $h_1$. We get $h_1 = \frac{24}{17}$, and $t_1 = \frac{31}{17}$.

Since $h_5 = \frac{h_1}{16}$, we now know $h_5 = \frac{3}{34}$, and since $t_2 = \frac{t_1}{2}$, $t_2 = \frac{31}{34}$.

We are looking for $\frac{h_5}{h_5+t_2}$, which gives the answer $\frac{3}{34}$. Hence $m+n=\boxed{37}$.


intelligence_20

Video solution

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

See also

1995 AIME (Problems • Answer Key • Resources)
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