For a Schnapsen deck (20 cards, with 4 suits, each suit containing Ace, Ten, King, Queen, Jack), 3 people A, B, C takes turns drawing a random card (starting from A -> B -> C), returning it to the deck and reshuffling the cards until a K heart or Q heart is drawn. The first one to draw one of these two cards wins.
a) What's the probability of each outcome in the sample space? Show that they all add up to 1.
b) Calculate the probability that A wins, B wins, C wins and no one wins.
For this question, I think the sample space will be infinite, since we can't be sure the 3 people can draw the right card each time.
The sample space will look like {W, LW, LLW, LLLW, …}
Since each person will take a card and return it to the deck, each person has 2/20 chance of winning.
However, I don't know how to proceed from here! Any help will be appreciated. Thanks!
Best Answer
Yes, you are right that the sample space is infinite. The players may keep getting non-winning cards forever. To answer the first part, since the events are all mutually exclusive of each other the probability of all events will sum up to be $1$.
Now lets calculate the probability of $A$ winning. So either he wins his first try or he wins on the second try or he wins on the third try and so on till infinity-
$$\begin{align}P(A)&=\frac {2}{20}+\biggr(\frac {18}{20}\biggr)^3\frac {2}{20}+\biggr(\frac {18}{20}\biggr)^6\frac {2}{20}+\biggr(\frac {18}{20}\biggr)^9\frac {2}{20}\cdots\\ &=\frac {2}{20}\biggr(1+\frac {18}{20}^3+\frac {18}{20}^6\cdots\biggr) \\ &=\frac {2}{20}\Biggr({1\over 1-\bigr(\frac {18}{20}\bigr)^3}\Biggr) \\ &=\frac {1}{10}\times \frac {1000}{271} \\ P(A)&= \frac {100}{271} \end{align}$$
Can you try to solve for P(B) and P(C) and P(none) similarly?