TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW01.5

Coin toss game

Anna and Nick play the following game: Anna repeatedly tosses an unfair coin with the probability $p,0<p\neq {\frac {1}{2}}$ of obtaining a head (H). The game ends the first time that two consecutive heads (HH) or two consecutive tails (TT) are observed. Anna wins if (HH) is observed and Nik wins if (TT) is observed. For example, if the outcome is HTHTT, Nik wins while if the outcome is HTHTHH, Anna wins. Given that Anna won the game, what is the probability

that the first coin toss resulted in head?

Dieses Beispiel ist als solved markiert. Ist dies falsch oder ungenau? Aktualisiere den Lösungsstatus (Details: Vorlage:Beispiel)

Lösungsvorschlag von Lessi

--Lessi 2024-02-07T13:04:11Z

We know:

${\begin{aligned}P(H)&=p\\P(A)&=p^{2}\\P(H|A)&={\frac {P(A|H)*P(H)}{P(A)}}\end{aligned}}$

$P(A|H)$ can be found by finding the geometric series out of all toss games where the first toss results in H (e.g. {HH, HTT, HTHH, HTHTT, HTHTHH, ...} with the probabilites $\{p^{2},(1-p)^{2}*p,(1-p)*p^{3},(1-p)^{3}*p^{2},(1-p)^{2}*p^{4},\dots \}$).

The sum of all events where Alice wins is: $p^{2}*\sum _{i=0}^{\infty }p^{i}*(1-p)^{i}$

This contains a geometric series with $q=p*(1-p)$ which is $<1$ given that $p<1$ and $(1-p)<1$.

The geometric series converges to $s_{n}={\frac {1}{1-q}}={\frac {1}{1-(p*(1-p))}}={\frac {1}{1-p+p^{2}}}$ and therefore $P(A|H)=p^{2}*{\frac {1}{1-p+p^{2}}}$

Turning that conditional probability around we now know that $P(H|A)=p*{\frac {1}{1-p+p^{2}}}$