# Markov inequality with coin flips

Let's say Dave has a biased coin with P(heads)=0.6. He tosses this coin $N$ times and, out of the $N$ times, the coin lands on heads 134 times. He says that the probability of seeing at least this many heads is at most 0.8. The best lower bound, using Markov inequality, on the number of times John tossed the coin = $0.8 * \frac{134}{0.6}$. However, I do not quite understand the intuition behind this. Given that he saw 134 heads, it makes sense to say that the total number of flips is 268. However, I don't quite understand how the information about the probability of seeing at least 134 heads can be factored in. Can someone explain this on a high level and explain how it leverages the markov inequality?

• "Doesn't that make it unbiased? " Yes. Are you sure you got it right? Commented Nov 26, 2016 at 1:56
• Yes, I typed it word for word. Perhaps the value P(heads) is set to random values using a random number generator and that's why the wording is "baised" Commented Nov 26, 2016 at 2:02
• I edited the question and reassigned the $P(heads)$ so that it is less confusing. Commented Nov 26, 2016 at 2:36

I think I understand now. What we have is the following: $$P(X \geq 134) = \frac{N*P(\text{heads})}{134}$$ $$0.8 = \frac{N*0.5}{134}$$ $$N = \frac{0.8*134}{0.5}$$