Question

Two players roll a 3-sided dice. The player who fails to get better rolls than previous ones loses. How likely is it for the first player to win?

Answer 1

Is it 0.666666 ?

Answer 2

Or 0.7037

Answer 3

@satellite73 if u r finished there PLZ see this

Answer 4

@lgbasallote can also do probability

Answer 5

i highly doubt he can

Answer 6

what does a 3-sided dice look like?

Answer 7

i don't understand the question
do they take turns?

Answer 8

I believe they are taking turns...the first one to roll lower than the previous one's turn loses.

Answer 9

I think if the first one roll is lower than the second one loses.

Answer 10

What is the probability that the second roll will be higher than the 1st

Answer 11

@BTaylor 3 sided dice looks like this.

Answer 12

Just saying.

Answer 13

if you list out the options:
1) Player 1 rolls a 1. Player 2 rolls either a 2 or a 3. If P2 rolls a 2, then P1 must roll a 3 to win. If P2 rolls a 3, then P1 loses. P=?
2) P1 rolls a 2. Player 2 must role a 3 to win. If P2 rolls a 3, then P1 loses. P=?
3) P1 rolls a 3. P2 automatically loses.
(thanks @micahwood50 very nice)

Answer 14

suppose the first player rolls a 2 and the second rolls a 2, what happens then ?

Answer 15

First one wins

Answer 16

"fails to get Higher"

Answer 17

so first player wins with these combinations:
(3,1)
(3,2)
(2,1)

that is probability of 3/9 or 1/3 chance player 1 wins after 1 turn

Answer 18

ok so we can work in cases then right? we are only interested in the probability the first player wins

Answer 19

(1,1) ,(2,1) ,(3,1) ,(2,2),(3,2) ,(3,3)

Answer 20

rolls a 3 with probability \(\frac{1}{3}\) wins
rolls a 2 with probability \(\frac{1}{3}\) player 2 rolls 1 or 2 with probability \(\frac{2}{3}\) wins
rolls a 1 with probability \(\frac{1}{3}\) player 2 rolls a 1 with probability \(\frac{1}{3}\) wins

Answer 21

Then it is 6/3^2 = 0.66666

Answer 22

That makes sense...to me anyway.

Answer 23

there are 18 total possibilities of Roll Number 2 in terms of Roll Number 1. Of those 18, 3 lead to number two winning. So the answer is 15/18

Answer 24

unless i do not understand the question (which is possible) then it should be
\[\frac{1}{33}+\frac{2}{9}+\frac{1}{9}\]

Answer 25

oops typo

Answer 26

U mean 1/3 + 2/9 + 1/9

Answer 27

15/18 = 5/6

Answer 28

\[\frac{1}{3}+\frac{2}{9}+\frac{1}{9}\]yeah but that is \(\frac{2}{3}\) which is not the answer given above

Answer 29

2/3=0.666666

Answer 30

so maybe i am wrong
\[(3,1)(3,2)(3,3),(2,1)(2,2)(1,1) \] i still count 6 wins out of of 9

Answer 31

For each of number one's rolls number 2 gets 3 rolls, that's 9 possibilities... of those 3 lead to winning so its 6/9 or 2/3. (i miscalculated earlier :/)

Answer 32

So, the final answer is 2/3

Answer 33

Ide bet my bippy its 2/3

Answer 34

if i understand the question correctly, yes

Answer 35

Thanx guys

Answer 36

you bet your what?
http://www.youtube.com/watch?v=HbvQa-7u3ss

Answer 37

Just for kicks I did ran a simulation of 1000 games...player 1 won the game about 57-60% of the time (I ran the whole thing a few times).

Answer 38

well theres your answer than 60%

Answer 39

Being that the answer looked too low...I checked the code and found an error.

Now I'm getting a solid 66% Player 1 win average across 1,000 test runs...with each test run averaging the results of 100 games.

So yes, 66% must be correct :)