MHB What Is the Probability John Gets More Heads Than Mary?

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John tosses 6 fair coins while Mary tosses 5, and the discussion focuses on calculating the probability that John gets more heads than Mary. The problem is framed as a Bernoulli process with 11 trials, where a "good" outcome is defined as either a head for John or a tail for Mary. The solution indicates that John will win if there are 6 good outcomes out of 11, leveraging the symmetry of the binomial distribution. Consequently, the probability that John gets more heads than Mary is determined to be 1/2. This analysis highlights the equal likelihood of outcomes in this coin-tossing scenario.
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Here is this week's POTW:

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John tosses $6$ fair coins, and Mary tosses $5$ fair coins. What is the probability that John gets more "heads" than Mary?

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to castor28 for his correct solution to this week's POTW, which was Problem 99 in the MAA Challenges. His solution follows:

[sp]Let us call a "good" outcome either a head by John or a tail by Mary. Good / bad outcomes constitute a Bernoulli process with $11$ trials and probability $\dfrac12$.

John will win (get more heads than Mary) if there are $6$ good outcomes out of $11$. Because of the symmetry of the binomial distribution, this will happen with probability $\dfrac12$.[/sp]
 

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