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Difference between revisions of "2007 Alabama ARML TST Problems/Problem 13"

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Now, from the initial state the first throw takes us either to situation <math>H</math> or to situation <math>T</math>, with equal probability.  
 
Now, from the initial state the first throw takes us either to situation <math>H</math> or to situation <math>T</math>, with equal probability.  
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Thus, the answer is <math>\frac 12 \cdot p_H + \frac 12 \cdot p_T = \boxed{\frac{63}{78}}</math>.
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Thus, the answer is <math>\frac 12 \cdot p_H + \frac 12 \cdot p_T = \boxed{\frac{21}{26}}</math>.
  
 
==See also==
 
==See also==
 
{{ARML box|year=2007|state=Alabama|num-b=12|num-a=14}}
 
{{ARML box|year=2007|state=Alabama|num-b=12|num-a=14}}

Latest revision as of 10:48, 9 December 2022

Problem

Before he gets out of bed every morning, Calvin the Compulsive plays a game with a fair coin. He flips it until either he flips four consecutive heads or he flips six consecutive tails, then he immediately gets out of bed and brushes his teeth. If his last flip is a head, he eats two melons for breakfast. Otherwise, he eats just one. Find the probability that Calvin ate two melons for breakfast this morning.

Solution

We will say that Calvin wins the game if he eats two melons.

Consider these two cases:

  • $H$: The last run of equal throws contains exactly one head.
  • $T$: The last run of equal throws contains exactly one tail.

That is, situation $H$ occurs either after the very first throw (if it was a head), or after a sequence of throws that ends with "... tail head".

Let $p_H$ be the probability that Calvin wins the game if he is now in situation $H$, and similarly let $p_T$ be the probability of winning from $T$.

We can now make the following observations:

When in the situation $H$, we have probability $\frac 18$ of winning the game right away, by throwing three more heads in a row. With probability $\frac 78$ this does not happen, and we throw a tail. The first tail we throw takes us into the situation $T$.

Similarly, from situation $T$ either we lose right away, which happens with probability $\frac 1{2^5}=\frac 1{32}$, or we get into situation $H$.

This gives us two equations for $p_H$ and $p_T$:

\[p_H = \frac 18 \cdot 1 + \frac 78\cdot p_T\] \[p_T = \frac 1{32} \cdot 0 + \frac {31}{32} \cdot p_H\]

This solves to $p_H=\frac{32}{39}$ and $p_T=\frac{31}{39}$.

Now, from the initial state the first throw takes us either to situation $H$ or to situation $T$, with equal probability. Thus, the answer is $\frac 12 \cdot p_H + \frac 12 \cdot p_T = \boxed{\frac{21}{26}}$.

See also

2007 Alabama ARML TST (Problems)
Preceded by:
Problem 12 		Followed by:
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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