Sunday, April 18, 2010

Another Game of Dice

Your friend offers to play a game of dice with you. He explains the game to you.

"We each get one die, the highest die wins. If we tie, I win, but since you always lose when you roll a one, if you roll a one you can roll again. If you get a one the second time you have to keep it."

What is each person's probability of winning?

What are the probabilities of winning if you can keep rolling until you get something besides a one?

In the first game the probabilities of winning are 37/72 for your friend and 35/72 for you. In the second game each player has a 50% probability of winning.

There are 36 combinations (6 x 6) for your die and your friend's die on the first roll. In the first game if you don't roll a one on the first roll then that roll is equally likely to be a 2, 3, 4, 5 or 6, which can beat 1, 2, 3, 4, 5 different numbers, respectively. That gives a

(1+2+3+4+5)/36 = 15/36

probability of winning on the first roll. There is a 1/6 chance of rolling a one, in which case you have a 15/36 probability of winning on the second roll, which gives a

1/6 * 15/36 = 15/216 = 5/72

probability of winning on the second roll for a total probability of winning of

15/36 + 5/72 = 30/72 + 5/72 = 35/72 = .486111...

In the second game we never keep a one, so there are only 30 combinations (5 x 6) for the final roll. As shown above, there are 15 ways to win, which gives a probability of

15/30 = 1/2

of winning the second game.


  1. The second time you roll a 1, you should keep it right? Why do you calculate further after calculating the .486111...?

  2. How is it equal in the second game? Isn't there still the condition that if they tie the friend wins?

  3. In second game, probability that you win is:
    5/12*(1 + 1/6 + 1/6^2 + ...) = 5/12*6/5 = 1/2

    Does it answer your question?


  4. I think in the solution for winning in first roll doesnt consider the winning in tie situation. 15 cases are there for 2,3,4,5,6 i.e
    2 can beat 1
    3 -- 2,1
    4 -- 3,2,1
    5 -- 4,3,2,1
    6 -- 5,4,3,2,1

    so total adds to 15 cases as above but here tie cases are not included i..e 2 can beat 2 , 3 can beat 3 like that so there are 5 more such cases one each for 2,3,4,5,6
    so total probability will be 20/36..

    1. Antariksh SrivastavaAugust 2, 2012 at 7:32 PM

      i agree with the last comment...!!!
      probability would be 20/36 and not 15/36

      tie cases should be included...!!

    2. That will be the case for your friend not you!