1) A and B are playing a game where they each write their name on a card. A goes first and chooses one of the two cards with equal probability. If A chooses her own card then she wins. If she chooses B’s card, then it is B’s turn to play. He chooses one of the two cards randomly with equal probability, and he wins if gets his own card. Otherwise, the turn passes back to A and she tries again. This pattern continues until one or the other has won. Calculate the probability that A will win this game. (it is not 1/2.)

I'm assuming the way to do this is geometric distribution, but I'm still a but confused on this. I was trying to use sigma notation to denote how to calculate it, but I don't know how to go about this. I'm getting 1/2 which obviously isn't the answer.

2) C walks up and wants to join the game so she adds a card with her name into the pile. Now, if A draws B’s card then he goes next and if she draws C’s card then C goes next. At each turn, the person drawing a card will win if they draw their own card, and otherwise the turn will pass to whoever is named on the card. Every card is drawn independently with all three cards having equal probability. They continue to take turns until somebody wins. Calculate the probability of A winning, the probability of B winning, and the probability of C winning.

Also assuming this is the same approach as the last problem, but instead of using 1/2, we use 1/3.

  • $\begingroup$ Both 1) & 2) can also be solved using self-similarity. $\endgroup$ Aug 23 '16 at 1:12

The two questions can be very easily solved using recursion.


Let P(A ultimately wins) $=p$

P(A wins on first turn) $= \frac12$

P(A and B both lose on their turn) $= \frac12\cdot\frac12 = \frac14$

Now this brings us back to square $1$, so A's probability from here is again $p$

Thus $p = \frac12 + \frac14\cdot p\;\;$ which yields $ p = \frac23$


Let P(A ultimately wins) $=p$

If A doesn't win on the first toss, the two others have a probability $\frac{1-p}2$ each of ultimately winning.

Either $A$ wins on first turn with probability $p$, or, so to say, becomes one of the "others".

Thus $p = \frac13 + \frac23\cdot\frac{1-p}2,$ which yields $p = \frac12$

P(A wins) $= \frac12\;\;$ P(B wins) $=$ P(C wins) $= \frac14$

  • $\begingroup$ Should B's probability really be the same as C's? Why does A become one of the others? $\endgroup$ Aug 23 '16 at 10:01
  • $\begingroup$ @QuinnCulver: 1. "Every card is drawn independently with all three cards having equal probability", so if A doesn't get, B and C are equally likely to get, and thus will have equal probabilities of winning. 2. At each stage, there is a "player" whose turn it is to draw, and "others", so if A doesn't get , she becomes one of the "others". $\endgroup$ Aug 23 '16 at 11:00
  • $\begingroup$ Yes, your are right! $\endgroup$ Aug 23 '16 at 15:18

Regarding the first question: Let $W$ be the event that $A$ wins. This event can be "A draws A's card on the first draw"; "A draws B's card on the first draw, then B draws A's card, then A draws A's card"....

If we let the string of $C$ and $I$s denote the outcomes, where the $ith$ $C$ means the $i$th play was the player drawing his/her own card, and the $i$th $I$ means the $i$th play was the player drawing the other player's card (e.g. $IIC$ means A drew B's card, then B drew A's card, then A drew A's card), $W=\{C\}\cup \{IIC\} \cup \{IIIIC\} \cup ...$. Note these events are mutualy disjoint, so by countable additivity of probability measures, $P(W)=P(C)+P(IIC)+P(IIIIC)+...=\frac{1}{2}+\frac{1}{8}+...=\sum_{n=1}^\infty \frac{1}{2^n}-\sum_{n=1}^\infty \frac{1}{2^{2n}}$.

For the second question, you could describe the situation as a Markov chain, and then find the probability of absorption. This method is explained carefully in Karlin and Pinsky's An Introduction to Stochastic Modeling. I'll try to think of a way that doesn't use Markov chains as well.

  • $\begingroup$ What would be in place of n? Since we dont have the number of trials (or we do, its infinity) until A wins. $\endgroup$
    – Deegeeek
    Aug 22 '16 at 23:32
  • $\begingroup$ Sorry, not following the question. I'm just using $n$ as an index in the sum. The number of trials until $A$ wins is unknown; it's either a positive odd number (1, 3, 5, 7, ...) or undefined ($A$ might not win). $\endgroup$
    – manofbear
    Aug 22 '16 at 23:45
  • $\begingroup$ Is it possible to take the limits of the equations from 1 to infinity? $\endgroup$
    – Deegeeek
    Aug 23 '16 at 0:02
  • $\begingroup$ Yes, the sums are geometric series (en.wikipedia.org/wiki/Geometric_series#Example) with $r=\frac{1}{2}$ and $r=\frac{1}{4}$, using the notation of the Wikipedia page. So we get $P(W)=1-\frac{1}{3}=\frac{2}{3}.$ $\endgroup$
    – manofbear
    Aug 23 '16 at 0:28

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