MHB Probability Challenge: Jason's 2010 Coin Flips

[anemone]

Jason has a coin which will come up the same as the last flip $\dfrac{2}{3}$ of the time and the other side $\dfrac{1}{3}$ of the time. He flips it and it comes up heads. He then flips it 2010 more times. What is the probability that the last flip is heads?

 

[chisigma]

Jason has a coin which will come up the same as the last flip $\dfrac{2}{3}$ of the time and the other side $\dfrac{1}{3}$ of the time. He flips it and it comes up heads. He then flips it 2010 more times. What is the probability that the last flip is heads?

[sp]It's a simple Markov chain problem and the transition matrix is...

$\displaystyle A = \left | \begin{matrix} \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} \end{matrix} \right |\ (1) $

The solution is the term of first line and first column of $ A^{n}$ for n=2010. This term is the solution of the difference equation...

$\displaystyle a_{n+1} = \frac{a_{n}}{3} + \frac{1}{3}\ a_{1}= \frac{2}{3}\ (2)$

... which is...

$\displaystyle a_{n} = \frac {1 + 3^{- n}}{2}\ (3)$

The requested value is therefore $P = \frac{1 + 3^{- 2010}}{2}$, very close to $\frac{1}{2}$ of course (Wink)...[/sp]

Kind regards

$\chi$ $\sigma$

 

[anemone]

[sp]It's a simple Markov chain problem and the transition matrix is...

$\displaystyle A = \left | \begin{matrix} \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} \end{matrix} \right |\ (1) $

The solution is the term of first line and first column of $ A^{n}$ for n=2010. This term is the solution of the difference equation...

$\displaystyle a_{n+1} = \frac{a_{n}}{3} + \frac{1}{3}\ a_{1}= \frac{2}{3}\ (2)$

... which is...

$\displaystyle a_{n} = \frac {1 + 3^{- n}}{2}\ (3)$

The requested value is therefore $P = \frac{1 + 3^{- 2010}}{2}$, very close to $\frac{1}{2}$ of course (Wink)...[/sp]

Kind regards

$\chi$ $\sigma$

Well done, chisigma, well done!