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Difference between revisions of "2007 iTest Problems/Problem 19"

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<cmath>E = 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{8} \cdots</cmath>
 
<cmath>E = 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{8} \cdots</cmath>
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Using the infinite series formula, <math>E = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1</math>, so Jason is expected to earn <math>\boxed{\textbf{(L) } \$ 1}</math>.
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Using the infinite series formula, <math>E = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1</math>, so Jason is expected to earn <math>\boxed{\textbf{(L) } 1 \text{ gold coin}}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 00:18, 29 October 2023

Problem

One day Jason finishes his math homework early, and decides to take a jog through his neighborhood. While jogging, Jason trips over a leprechaun. After dusting himself off and apologizing to the odd little magical creature, Jason, thinking there is nothing unusual about the situation, starts jogging again. Immediately the leprechaun calls out, "hey, stupid, this is your only chance to win gold from a leprechaun!" Jason, while not particularly greedy, recognizes the value of gold. Thinking about his limited college savings, Jason approaches the leprechaun and asks about the opportunity. The leprechaun hands Jason a fair coin and tells him to clop it as many times as it takes to flip a head. For each tail Jason flips, the leprechaun promises one gold coin. If Jason flips a head right away, he wins nothing. If he first flips a tail, then a head, he wins one gold coin. If he's lucky and flips ten tails before the first head, he wins $\textit{ten gold coins.}$ What is the expected number of gold coins Jason wins at this game?

$\textbf{(A) }0\qquad \textbf{(B) }\dfrac1{10}\qquad \textbf{(C) }\dfrac18\qquad \textbf{(D) }\dfrac15\qquad \textbf{(E) }\dfrac14\qquad \textbf{(F) }\dfrac13\qquad \textbf{(G) }\dfrac25\qquad$

$\textbf{(H) }\dfrac12\qquad \textbf{(I) }\dfrac35\qquad \textbf{(J) }\dfrac23\qquad \textbf{(K) }\dfrac45\qquad \textbf{(L) }1\qquad \textbf{(M) }\dfrac54\qquad$

$\textbf{(N) }\dfrac43\qquad \textbf{(O) }\dfrac32\qquad \textbf{(P) }2\qquad \textbf{(Q) }3\qquad \textbf{(R) }4\qquad \textbf{(S) }2007$


Solution

The probability of getting $1$ head is $\frac{1}{2}$. The probability of getting $1$ tail then $1$ head is $\frac{1}{4}$. From there, the probability of getting $n$ tails then $1$ head is $\frac{1}{2^{n+1}}$.

Let $E$ be expected value of earnings. Using expected value, \[E = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{8} + 3 \cdot \frac{1}{16} \cdots\] Multiply both sides by $2$ to get \[2E = 0 \cdot 1 + 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{8} \cdots\] Subtract original to get \[E = 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{8} \cdots\] Using the infinite series formula, $E = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$, so Jason is expected to earn $\boxed{\textbf{(L) } 1 \text{ gold coin}}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 18 		Followed by:
Problem 20
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