TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2022W/HW01.2

Coin game

Two players, A and B, alternately and independently flip a coin and the first player to obtain a head wins. Assume player A flips first.

Problem 5a)

If the coin is fair, what is the probability that A wins?

Problem 5b)

Suppose that ${\textstyle P(head)=p}$, not necessarily ${\textstyle {\frac {1}{2}}}$. What is the probability that A wins?

Lösungsvorschlag von Friday

--Friday Sa 30 Jan 2021 14:08:31 CET

Solution 5a)

If ${\textstyle a}$ describes the posibility that A winns and ${\textstyle b}$ the winn of B, than we can create the following equation.

$$a={\frac {1}{2}}+{\frac {1}{2}}*(1-a)$$

First we have the probability that A winns imideatly which is ${\textstyle {\frac {1}{2}}}$, or A loses at first but wins in one of the later rounds. This second part can be described with the probability that first A has to lose but in the end B has to lose the game. ${\textstyle (1-a)}$ describes the probability that B will win the game.

So all we have to do now is to solve this equation:

$${\begin{aligned}a&={\frac {1}{2}}+{\frac {1}{2}}*(1-a)\\a&={\frac {1}{2}}+{\frac {1}{2}}-{\frac {a}{2}}\\{\frac {3a}{2}}&=1\\a&={\frac {2}{3}}\quad \Box \end{aligned}}$$

Solution 5b)

Let ${\textstyle a}$ be the probability that A wins. A detailed description as how the equation gets created can be found at the problem at 5a).

$${\begin{aligned}a&=p+(1-p)*(1-a)\\a&=p+1-a-p+ap\\a&=1-a+ap\\2a-ap&=1\\a*(2-p)&=1\\a&={\frac {1}{2-p}}\quad \Box \end{aligned}}$$