Probability

[mastvikas]

Hi There ,

Please Help

Three traders, A, B and C -play a game of dice. Whoever throws a "6" first wins the game. A
starts the game. What is the probability of A winning the game?

a. 36/91
b. 25/91
c. 1/3
d. ½

 

[ShaktiRathore]

P(A wins)=P(A) or P(A'&B'&C'&A)+P(A'&B'&C'&A'&B'&C'&A)+...... assuming they continue sequentially until one wins in the order A,B,C till indefinitely until one emerges out as the winner
P(A)=P(B)=P(C)=1/6,P(A')=1-1/6=5/6=P(B')=P(C')
P(A wins)=P(A) or P(A')*P(B')*P(C')*P(A) or P(A')*P(B')*P(C')*P(A')*P(B')*P(C')*P(A)+...... as these are independent events occurrence of one is not affected by occurrence of another event
P(A wins)=1/6 +5/6*5/6*5/6*1/6+5/6*5/6*5/6*5/6*5/6*5/6*1/6+..............
P(A wins)=1/6+5^3/6^4+5^6/6^7+..... this is a infinite GP with first term 1/6 and common ratio of 5^3/6^3
P(A wins)=1/6/(1-125/216)=216/6*91=36/91. please check the answer

thanks

 

[mastvikas]

Hi Shakti ,

Thanks for the explanation - Your answer is correct .

Can you please elaborate the last calculation Part .

Thanks
Vikas

 

[ShaktiRathore]

GP: a=1/6 and r=5^3/6^3=125/216
P(A wins)
= a/1-r
=(1/6 )/(1-(125/216) )=(1/6 )/((216-125)/216)=(1/6 )/((91)/216)=216/6*91=36/91

thanks

 

[mastvikas]

Thanks Shakti , i Appreciate your help

 

[Kr1]

so why isnt it 1/3?
Whats the philosophy behind this, in layman terms?

 

[ShaktiRathore]

The order is sequential so that we need to consider every event when A can win out of a set of possible events where any of A,b,c can win. We just need to consider those events where A wins so we figure out the possible events as(we are given that orders of throw is sequential) P(A) or P(A'&B'&C'&A) or P(A'&B'&C'&A'&B'&C'&A) or.... weeding out other events where B or C can also win so we are not considering those events and since occurrence of these events where A can win are mutually exclusive we add them up. Its like finding probability of 5 or 6 showing in a die throe P(5 or 6)=P(5)+P(6)=2/6. Since any of the possible event as P(A) or P(A'&B'&C'&A) or P(A'&B'&C'&A'&B'&C'&A) ... can occur where A wins exclusively we need to be more specific. We cannot simply consider 1/3 its like asking whats the probability that any of A,b,c wins in a game of chance where the likelihood of each winning is the same.But here its the special case where order is important which affects probability of occurrence of any of these events.

thanks