Crazy Coin Flipping

Probability Level 4

There are two teams, The Eagles and The Hawks, consisting of $6$ people each.

Every person in The Eagles has a fair coin. But, every person in The Hawks has a biased coin which comes up heads with a probability of $3/4$ .

Everybody tosses their coin exactly once. The probability that The Hawks get more heads than The Eagles is $P$ .

Find $\lfloor 1000P \rfloor$

The answer is 735.

2 solutions

Brian Charlesworth
Oct 28, 2014

Let $E$ be the number of heads thrown by The Eagles and $H$ be the number of heads thrown by The Hawks. Then what we are needing to calculate is

$\sum_{k=0}^{5} (P(E \le k) * P(H = (k + 1)))$ .

Now $P(E \le k) = \sum_{j=0}^{k} \binom{6}{j} * (\frac{1}{2})^{j}(\frac{1}{2})^{6-j} = (\frac{1}{64})*\sum_{j=0}^{k} \binom{6}{j}$ ,

and $P(H = (k + 1)) = \binom{6}{k + 1} * (\frac{3}{4})^{k + 1} * (\frac{1}{4})^{6 - (k + 1)} = (\frac{1}{4096})*\binom{6}{k + 1} * 3^{k + 1}$ .

Thus the desired probability will be

$(\frac{1}{262144})*\sum_{k=0}^{5} ((\sum_{j=0}^{k} \binom{6}{j}) * \binom{6}{k + 1} * 3^{k + 1}) =$

$(\frac{1}{262144})*(6*3 + 7*15*9 + 22*20*27 + 42*15*81 + 57*6*243 + 63*729) =$

$\frac{192906}{262144} = \frac{96453}{131072} = 0.7358779...$ ,

giving a final solution of $\lfloor 1000*0.7358779... \rfloor = \boxed{735}$ .

Only one comment I would like to add about the statement of the problem. In such problems where big divisions are considered, only approximate value of the probability should be asked to state and not the integer solution. Sometimes even the Brilliant scratchpad gives wrong answer when big numbers are considered and then that may cause error in the final answer.

Snehal Shekatkar - 6 years, 7 months ago

i used brute force calculation and that took a lot of time!!

Cody Martin - 6 years, 3 months ago

Shrimat Kapoor
Oct 23, 2018

I just plugged numbers into the calculator lol

Crazy Coin Flipping

The answer is 735.

2 solutions

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