A high school university entrance examination question:

Players A and B take turns in throwing two dice. The winner is the first player to throw a double six. Player A starts the Game.

  1. Find the probability that Player A wins on the first throw. Attempt: Obviously 1/36.

  2. What is the probability that Player A wins on the first or second throw? Attempt: Probability of win on the second throw is 35/36 times 1/6. Probability of win on first or second throw is therefore the sum of (i) and (ii), or 1/6(1+35/36).

  3. Find the probability that Player A eventually wins the game. Attempt: I tried a geometrical progression using 35/36 as the common ratio, but am stuck. Can you give me a prompt?

  • 1
    $\begingroup$ In part (ii), it looks like you've omitted B's first throw. $\endgroup$ – user21467 Dec 1 '13 at 9:10
  • 2
    $\begingroup$ Where do you get the 1/6 from? This is all a big bunch of 1/36 and 35/36 probabilities. $\endgroup$ – aaaaaaaaaaaa Dec 1 '13 at 9:21
  • $\begingroup$ Thank you! I had completely forgotten about the other player. $\endgroup$ – Guy Corrigall Dec 1 '13 at 9:27

You can solve this recursively. Noting that after A and B both didn't throw double six the game basically restarts, you get

$p={1\over 36} + {35 \over 36}{35 \over 36}p$.

Thus $p = \frac{{1\over 36}}{1-{35 \over 36}{35 \over 36}}= {36 \over 71}.$

  • $\begingroup$ Very succint. Thank you. $\endgroup$ – Guy Corrigall Dec 1 '13 at 9:33

In the (iii) part we can do in the following way :-

Let the event A : A wins in his turn

and the event B : B loses in his turn

$ P = P(A) + P(A'BA) + P(A'BA'BA) + \ldots $

$P=\frac{1}{36}+\frac{35^2}{36^2}*\frac{1}{36}+\frac{35^4}{36^4}*\frac{1}{36}+\ldots $

$ P = \frac{1}{36}*(1+\frac{35^2}{36^2}+\frac{35^4}{36^4}+\ldots)$

$ P = \frac{1}{36}*\frac{1}{1-\frac{35^2}{36^2}}$

$ P = \frac{36}{71}$

  • $\begingroup$ Yes, that was the GP I was looking for. $\endgroup$ – Guy Corrigall Dec 1 '13 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.