Whats the probability of $HTHT$ occuring before $HHTT$? What's the probability of $HTHT$ occuring before $HHTT$ in a stream of $H$ 's and $T$ 's (both equally likely) that will stop if either of those occur? What's the mean number of throws such that $HHTT$ occurs?
Occurrences of permutations in sequences of 4 coin flips I was reading this interesting article on hot hands and streaks in sports The article revolves around the 16 possible sequences of 4 coin flips (H = heads, T = tails): HHHH HHHT HHTH HHTT HTHH HTHT
Why is the expected number coin tosses to get $HTH$ is $10$? @Shaun well I was trying to use markov chains in order to solve this I know how to to draw the matrix but I don't remember how to calculate the expected number of times it takes to go from one state to another, here being going from state zero which is before we tossed any coins to state HTH (3 if we start from zero) I thought using stationary distributions but I'm not sure Also I was
Probability of exactly two heads in four coin flips? The derivation of binomial probability: Getting two heads out of 4 can be portrayed is, disregarding order: HHTT (H=heads and T=tails) Multiplying their probabilities will yield $ (0 5)^4$, but as for ordering, we get $4! (2!\cdot2!)$ due to repetition, which is the same as $4C2$ So our answer is $\binom42\cdot (0 5)^4$ which is $0
A fair coin is tossed four times. What is the probability. . . HW problem here I know the answer is 6 16 (per the back of the book) but I can't figure out how they got that A fair coin is tossed four times What is the probability that the number of heads