Difference between revisions of "2024 SSMO Speed Round Problems/Problem 8"
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| + | Let <math>P(f,k)</math> denote the fair coin shows <math>k</math> heads and <math>P(r,k)</math> denote the probability the rigged coin shows <math>k</math> heads. Note that the probability <math>k</math> heads are shown when Bob flips the coin is <math>\frac{P(f,k)+P(r,k)}{2}.</math> Now, if <math>k</math> heads are showing, the probability that Alice guesses correct is <math>\frac{\max(P(f,k),P(r,k))}{P(f,k)+P(r,k))}.</math> So, the answer is <cmath>\sum_{k=0}^{2}\left(\left(\frac{P(f,k)+P(r,k)}{2}\right)\left(\frac{\max(P(f,k),P(r,k))}{P(f,k)+P(r,k)}\right)\right).</cmath> It is easy to compute <cmath>P(f,0) = \frac{1}{4},P(f,1) = \frac{1}{2},P(f,2) = \frac{1}{4},P(r,0) = \frac{4}{9},P(r,1) = \frac{4}{9},\text{and }P(r,2) = \frac{1}{9}.</cmath> Therefore, the answer is <cmath>\left(\frac{\frac{4}{9}+\frac{1}{4}}{2}\right)\left(\frac{\frac{4}{9}}{\frac{4}{9}+\frac{1}{4}}\right)+\left(\frac{\frac{4}{9}+\frac{1}{2}}{2}\right)\left(\frac{\frac{1}{2}}{\frac{4}{9}+\frac{1}{2}}\right)+\left(\frac{\frac{1}{9}+\frac{1}{4}}{2}\right)\left(\frac{\frac{1}{4}}{\frac{1}{9}+\frac{1}{4}}\right) = </cmath><cmath>\frac{2}{9}+\frac{1}{4}+\frac{1}{8} = \frac{43}{72}\implies \boxed{115}.</cmath> | ||
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| + | ~SMO_Team | ||
Latest revision as of 14:28, 10 September 2025
Problem
Bob has two coins; one is fair, and one lands on heads with a probability of 
Bob chooses a random coin and flips it twice. Alice watches the two coin flips and guesses whether Bob flipped the fair or rigged coin. Given that Alice is a good mathematician and guesses the more likely option (guessing randomly when they are equally likely), the probability she guesses right can be expressed as 
for relatively prime positive integers \(m\) and \(n.\) Find 
Solution
Let 
denote the fair coin shows 
heads and 
denote the probability the rigged coin shows heads. Note that the probability heads are shown when Bob flips the coin is 
Now, if heads are showing, the probability that Alice guesses correct is 
So, the answer is ![\[\sum_{k=0}^{2}\left(\left(\frac{P(f,k)+P(r,k)}{2}\right)\left(\frac{\max(P(f,k),P(r,k))}{P(f,k)+P(r,k)}\right)\right).\]](http://latex.artofproblemsolving.com/8/3/2/8322b1581c7116b7b7f969a0f02b7081abfb4ab9.png)
It is easy to compute ![\[P(f,0) = \frac{1}{4},P(f,1) = \frac{1}{2},P(f,2) = \frac{1}{4},P(r,0) = \frac{4}{9},P(r,1) = \frac{4}{9},\text{and }P(r,2) = \frac{1}{9}.\]](http://latex.artofproblemsolving.com/b/c/5/bc52953f808993aef881cd5d2be5b5f4843d6069.png)
Therefore, the answer is ![\[\left(\frac{\frac{4}{9}+\frac{1}{4}}{2}\right)\left(\frac{\frac{4}{9}}{\frac{4}{9}+\frac{1}{4}}\right)+\left(\frac{\frac{4}{9}+\frac{1}{2}}{2}\right)\left(\frac{\frac{1}{2}}{\frac{4}{9}+\frac{1}{2}}\right)+\left(\frac{\frac{1}{9}+\frac{1}{4}}{2}\right)\left(\frac{\frac{1}{4}}{\frac{1}{9}+\frac{1}{4}}\right) =\]](http://latex.artofproblemsolving.com/3/8/1/3812d8454cf85d9a6cd4245cfba1b27b855853a7.png)
![\[\frac{2}{9}+\frac{1}{4}+\frac{1}{8} = \frac{43}{72}\implies \boxed{115}.\]](http://latex.artofproblemsolving.com/7/5/7/757392af059ab2599dea622019870e96d13aa8fe.png)
~SMO_Team
