Frank alternates in between flipping a weight coin that has actually a $2/3$ opportunity of landing heads and a $1/3$ opportunity of landing tails and also another weighted coin that has actually a $1/4$ chance of landing heads and also a $3/4$ possibility of landing tails.

The first coin toss is the "$2/3$-$1/3$" load coin. The probability the he sees two heads in a row before he sees two tails in a row is?

let H signify the occasion of landing a head and T because that a tail. So there room some possibilities: HH...,THH..TT...,HTHH...TT yet how to execute further?

You are watching: Probability of flipping two heads in a row

edited Jul 2 "20 in ~ 11:42

2,4131010 silver- badges2727 bronze title
request Jul 2 "20 in ~ 11:33

79644 silver- badges1616 bronze badges
$\begingroup$ perform you average exactly 2 heads in a row? $\endgroup$
Jul 2 "20 at 12:04
| show 3 more comments

3 answers 3

active oldest Votes

See more: Solved: What Is 20 Percent Of 14000 = 2800, How Much Is 20 Percent Of 14000

Denote the two coins by $A,B$. After ~ the an initial toss, every energetic state the the game can be defined by two parameters: what worth did you simply toss and also which coin space you about to toss? We brand the four energetic states together $(H,A),(H,B), (T,A), (T,B)$. We represent by, say, $p_H,A$ the probability that you"ll check out $HH$ before you check out $TT$ assuming girlfriend are at this time in state $(H,A)$. The starting state can never be got to again, we"ll signify it by $\emptyset$. Of course, $p_\emptyset$ is the answer us seek.

Now, consider the feasible results that the an initial toss. V probability $\frac 23$ you acquire an $H$ and move come state $(H,B)$. V probability $\frac 13$ you acquire a $T$ and also move to state $(T,B)$. Thus $$p_\emptyset=\frac 23\times p_H,B+\frac 13\times p_T,B$$

Similarly we obtain $$p_H,B=\frac 14\times 1 +\frac 34\times p_T,A\quad \quad \quad p_T,B=\frac 34\times 0+\frac 14\times p_H,A$$ $$p_H,A=\frac 23 \times 1 +\frac 13\times p_T,B\quad \quad \quad p_T,A=\frac 13\times 0 + \frac 23\times p_H,B$$

Barring arithmetic error (always possible), this system means $$\boxed p_\emptyset=\frac 1333$$

Note: i am somewhat surprised the this is much less than $\frac 12$. ~ all, girlfriend are an ext likely to throw $H$ originally so I believed that would provide $H$ part advantage. I imply checking the arithmetic really carefully.