Question

Alice, Bob, and Chris, in that order and starting with Alice, take turns drawing a random card from a 52 card deck and then returning it to the deck, continuing until an ace is drawn. The first one to draw an ace wins.

(a) Explain why Ω = {0 i1 | i ≥ 0} ∪ {the infinite string 000 · · ·} is a reasonable definition of the sample space of this experiment.

(b) What is the probability of each outcome in the sample space? Show that these probabilities add up to 1, as they must in any proper sample space.

(c) Let A be the event that Alice wins, B be the event that Bob wins, and C be the event that Chris wins. Let N be the event that no one wins, ever. Define each in terms of elements from Ω and calculate the probability of each. Give each probability exactly as a simplified fraction

Provide a brief explanation of 1-2 sentences for each question.

Answer 1

$(a)~\Omega=Sample~space~of~drawing~a~card~by~Alice,Bob~and~Chris~from\\ 52~cards~with~replacement~technique~until~one~of~them~get~ace\\ =\{0,0,0,1,0,1,0,0,1,.....\}=\{0,0,0,O_{1},O_{2},O_{3},....\}\\ Where,~O_{3i+1}=Event~of~getting~ace~by~Alice~at~i^{th}~trail\\ O_{3i+2}=event~of~getting~ace~by~Bob~at~i^{th}~trial\\ O_{3i+3}=event~of~getting~ace~by~Bob~at~i^{th}~trial,~i\geq 0\\(b)~If~No~success~(getting~ace)~occur~at~ith~trial,~then~probability~of~the~outcome\\ =\left(\frac{48}{52} \right )=\frac{12}{13}\\ Probability~getting~success~at~ith~trial=p_i=(1/13)(12/13)^{i-1},~i=1,2,....\\ \sum_{i=1}^\infty p_i=\frac{1}{13}\times \frac{1}{1-(12/13)}=1\\$

$(c)~A=\{O_{1},0,0,0,O_{4},0,0,0,0,0,0,O_{7}.....\}\\ B=\{0,O_{2},0,0,0,0,O_{5},0,0,0,0,0,0,0,O_{8},.....\}\\ C=\{0,0,O_3,0,0,0,0,0,O_6,0,0,0,0,0,0,0,0,O_9,....\}\\ N=\{0,0,0,...\}\\ P(A)=(1/13)+(1/13)(12/13)^3+(1/13)(12/13)^6+...\\ =(1/13)*(1/(1-(12/13)^3))=0.3603\\ P(B)=(12/13)(1/13)+(1/13)(12/13)^4+(1/13)(12/13)^7+...\\ =(12/13)*(1/13)*(1/(1-(12/13)^3))=0.3326\\ P(C)=(12/13)^2(1/13)+(1/13)(12/13)^5+(1/13)(12/13)^8+...\\ =(12/13)^2*(1/13)*(1/(1-(12/13)^3))=0.3070\\ P(N)=1-P(A)-P(B)-P(C)=1-0.3603-0.3326-0.3070\\ =0.0001$