Another Game of Dice
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
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?
Answer
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.
Hide
Comments
fame   
Jan 14, 2009
| FIRST POST! |
tpg76
Jan 14, 2009
| Another fun one - a bit easier than the previous, but fun. Good explanation as always. Thanks. |
MagaJamba 
Mar 15, 2009
| I didnt understand a word you said sry |
opqpop
Sep 27, 2010
| Alternative way:
Condition on when you roll a 1 or not.
P(roll 1) = 1/6
P(win | you rolled 1) = (1 - 1/6) / 2 = 5/12
P(you don't roll 1) = 5/6
P(win | you don't roll 1) = P(win | you don't roll 1 and opponent rolls 1)P(opponent rolls 1) + P(win | you don't roll 1 and opponent doesn't roll 1)P(opponent doesn't roll 1) = 1 * 1/6 + ((1-1/5) / 2) * 5/6 = 1/2
Answer is 1/6 * 5/12 + 5/6 * 1/2 = 35 / 72.
The second part was already answered above: 1/2. |
opqpop
Sep 27, 2010
| The (1-1/6)/2 and (1-1/5)/2 comes from symmetry.
P(my # higher) = P(your # higher).
P(my # higher) + P(your # higher) + P(tie) = 1.
P(tie) is 1/6 and 1/5 for six and five sided die respectively. |
Stack1607
Jan 07, 2012
| Speaking of the first game, I reached a different result: 121/216
the probability of my getting a higher die is 15/36 and the probability of a tie but above 1 is 5/36 - (2,2),(3,3),(4,4),(5,5),(6,6). Since I win in case of such tie, it should be counted, and the total of the previous probabilities is alone 20/36 or 40/72 - higher than the answer.
The probability of a tie with both of us having 1 is 1/36. Multiply that by the probability of you getting another 1 on your second roll(1/6), since that's the only way i would win [i'm assuming that only you get to roll again while i keep my 1]. The result would be 1/216.
Then the total probability of my winning is: 20/36 + 1/216 = 121/216 |
javaguru
Jan 07, 2012
| @Stack: You're mixing the rules up. You don't ever win a tie. |
Stack1607
Jan 16, 2012
| It said "If we tie, I win, but since you always lose when you roll a one ...", which means, as I understood, in case of a tie other than (1,1), one person (i or you depending on the narrative) wins?! |
javaguru
Jan 16, 2012
| Yes, that's right. "I" win ties, "you" never win ties. |
Back to Top
| |
|
