Difference between revisions of "2007 Alabama ARML TST Problems/Problem 13"
(New page: ==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 si...) |
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==Solution== | ==Solution== | ||
| − | We will say that Calvin wins the game | + | We will say that Calvin wins the game if he eats two melons. |
| − | Consider these two | + | Consider these two cases: |
* <math>H</math>: The last run of equal throws contains exactly one head. | * <math>H</math>: The last run of equal throws contains exactly one head. | ||
* <math>T</math>: The last run of equal throws contains exactly one tail. | * <math>T</math>: The last run of equal throws contains exactly one tail. | ||
| Line 27: | Line 27: | ||
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. | ||
| − | Thus the answer is <math>\frac 12 \cdot p_H + \frac 12 \cdot p_T = \boxed{\frac{63}{78}}</math>. | + | Thus, the answer is <math>\frac 12 \cdot p_H + \frac 12 \cdot p_T = \boxed{\frac{63}{78}}</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}} | ||
Revision as of 23:05, 17 January 2016
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:
: The last run of equal throws contains exactly one head.
: The last run of equal throws contains exactly one tail.
That is, situation 
occurs either after the very first throw (if it was a head), or after a sequence of throws that ends with "... tail head".
Let 
be the probability that Calvin wins the game if he is now in situation , and similarly let 
be the probability of winning from 
.
We can now make the following observations:
When in the situation , we have probability 
of winning the game right away, by throwing three more heads in a row. With probability 
this does not happen, and we throw a tail. The first tail we throw takes us into the situation .
Similarly, from situation either we lose right away, which happens with probability 
, or we get into situation .
This gives us two equations for and :
![\[p_H = \frac 18 \cdot 1 + \frac 78\cdot p_T\]](http://latex.artofproblemsolving.com/b/e/0/be0a930edd547a06cd683cf6062b9bc3c91c4860.png)
![\[p_T = \frac 1{32} \cdot 0 + \frac {31}{32} \cdot p_H\]](http://latex.artofproblemsolving.com/9/5/3/953a252c8fd86844883813f43937ce0659352c06.png)
This solves to 
and 
.
Now, from the initial state the first throw takes us either to situation or to situation , with equal probability.
Thus, the answer is 
.
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 | ||
