# tossing the two coins together, until for the first time either two heads appear ....

• Jun 1st 2010, 03:12 AM
acevipa
tossing the two coins together, until for the first time either two heads appear ....
Tom and Bob play a game by each tossing a fair coin. The game consists of tossing the two coins together, until for the first time either two heads appear when Tom wins, or two tails appear when Bob wins.

1) Show that the probability that Tom wins are or before the nth toss is $\displaystyle \frac{1}{2} - \frac{1}{2^{n+1}}$

2) Show that the probability that the game is decided at or before the nth toss is $\displaystyle 1-\frac{1}{2^n}$
• Jun 1st 2010, 03:41 AM
Quote:

Originally Posted by acevipa
Tom and Bob play a game by each tossing a fair coin. The game consists of tossing the two coins together, until for the first time either two heads appear when Tom wins, or two tails appear when Bob wins.

1) Show that the probability that Tom wins are or before the nth toss is $\displaystyle \frac{1}{2} - \frac{1}{2^{n+1}}$

2) Show that the probability that the game is decided at or before the nth toss is $\displaystyle 1-\frac{1}{2^n}$

The coins can come up TT, TH, HT, HH

The probability that Tom wins is

$\displaystyle \frac{1}{4}+\frac{2}{4}\ \frac{1}{4}+\frac{2^2}{4^2}\ \frac{1}{4}+\frac{2^3}{4^3}\ \frac{1}{4}+.......$

This is an infinite geometric series, with first term=a=0.25 and common ratio=r=0.5

Hence $\displaystyle S_n=\frac{0.25(1-0.5^n)}{1-0.5}$

The game is decided if two tails appear or two heads appear.

The probability of this is

$\displaystyle \frac{2}{4}+\frac{2}{4}\ \frac{2}{4}+\frac{2^2}{4^2}\ \frac{2}{4}+......$

This time "a" and "r" are both 0.5.