Expected number of flips

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Bread10

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Bread10

#1 h Apr 15, 2025, 8:58 PM • 1 Y

Y by WiseTigerJ1

An unfair coin has a $\frac{4}{7}$ probability of coming up heads and $\frac{3}{7}$ probability of coming up tails. The expected number of flips necessary to first see the sequence $HHTHTHHT$ in that consecutive order can be written as $\frac{m}{n}$ for relatively prime positive integers $m$ , $n$ . Find the number of factors of $n$ .

$\textbf{(A)}~40\qquad\textbf{(B)}~42\qquad\textbf{(C)}~44\qquad\textbf{(D)}~45\qquad\textbf{(E)}~48$

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MathRook7817

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MathRook7817

#2 h Apr 15, 2025, 9:09 PM

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$n= 2^(10) * 3^3$
$11x4 = 44 [b]C[/b]$
44 C
latex aint working

This post has been edited 2 times. Last edited by MathRook7817, Apr 15, 2025, 9:10 PM

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maromex

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maromex

#3 h Apr 15, 2025, 9:10 PM

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I don't think this will work like that one MATHCOUNTS problem because the sequences are not actually disjoint.

$HHTHTHHTHTHHT$ has 2 copies of the sequence inside it which are not disjoint.

This post has been edited 1 time. Last edited by maromex, Apr 15, 2025, 9:11 PM

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Bread10

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Bread10

#4 h Apr 15, 2025, 9:13 PM

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MathRook7817 wrote:

$n= 2^(10) * 3^3$
$11x4 = 44 [b]C[/b]$
44 C
latex aint working

How did you get that answer?

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MathRook7817

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MathRook7817

#5 h Apr 15, 2025, 9:28 PM

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Bread10 wrote:

MathRook7817 wrote:

$n= 2^(10) * 3^3$
$11x4 = 44 [b]C[/b]$
44 C
latex aint working

How did you get that answer?

i looked at the denominators and stuff

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Bread10

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Bread10

#6 h Apr 15, 2025, 9:28 PM

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I'm changing the problem because I don't like how easy it is to find the denominator

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Bread10

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Bread10

#7 h Apr 15, 2025, 9:32 PM

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An unfair coin has a $\frac{2}{3}$ probability of coming up heads and $\frac{1}{3}$ probability of coming up tails. The expected number of flips necessary to first see the sequence $HHTHTHHT$ in that consecutive order can be written as $n$ . Find the greatest positive integer less than $n$ .

$\textbf{(A)}~205\qquad\textbf{(B)}~211\qquad\textbf{(C)}~234\qquad\textbf{(D)}~256\qquad\textbf{(E)}~264$

This post has been edited 1 time. Last edited by Bread10, Apr 15, 2025, 9:35 PM

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Pengu14

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Pengu14

#10 h Apr 15, 2025, 10:35 PM

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I just spent 10 minutes bashing this with states and I just realized I misread the question :sob:

The answer would be 265 if the probabilities were both 1/2

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mathprodigy2011

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mathprodigy2011

#11 h Apr 16, 2025, 12:13 AM

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Pengu14 wrote:

I just spent 10 minutes bashing this with states and I just realized I misread the question :sob:

The answer would be 265 if the probabilities were both 1/2

ORZ

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Bread10

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Bread10

#12 h Apr 16, 2025, 4:09 AM

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Pengu14 wrote:

I just spent 10 minutes bashing this with states and I just realized I misread the question :sob:

The answer would be 265 if the probabilities were both 1/2

It would actually be 264 but close enough ig

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maxamc

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maxamc

#13 h Apr 16, 2025, 4:30 AM

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Bread10 wrote:

An unfair coin has a $\frac{2}{3}$ probability of coming up heads and $\frac{1}{3}$ probability of coming up tails. The expected number of flips necessary to first see the sequence $HHTHTHHT$ in that consecutive order can be written as $n$ . Find the greatest positive integer less than $n$ .

$\textbf{(A)}~205\qquad\textbf{(B)}~211\qquad\textbf{(C)}~234\qquad\textbf{(D)}~256\qquad\textbf{(E)}~264$

211.(78125)

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Bread10

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Bread10

#14 h Apr 16, 2025, 4:38 AM

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maxamc wrote:

Bread10 wrote:

An unfair coin has a $\frac{2}{3}$ probability of coming up heads and $\frac{1}{3}$ probability of coming up tails. The expected number of flips necessary to first see the sequence $HHTHTHHT$ in that consecutive order can be written as $n$ . Find the greatest positive integer less than $n$ .

$\textbf{(A)}~205\qquad\textbf{(B)}~211\qquad\textbf{(C)}~234\qquad\textbf{(D)}~256\qquad\textbf{(E)}~264$

211.(78125)

correct!

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