chance of rolling a higher number (ties lose)

[shawny boy]

Subject says it all, what are the chances when we both roll a 6-sided die that I will roll a higher number than you?

[Gullanian]

If ties are ignored then it's 1/2.

With ties as a loss, you will tie 1/6 times. So 1/2 - 1/6 = your answer I beleieve.

0.5 - (1 / 6) = 0.333333333

[CT11]

I get 15/36~=0.417.

Proof(if you care):

Let X and Y be i.i.d. random variables defined on the same probability space (your dice and his dice). let P(X=1)=P(X=2)=...=P(X=6)=1/6 (ie a fair dice).

Let A be the event that X>Y

So, P(A) = P(X>Y)
the law of total probability implies
P(A) = P(X>Y|Y=1)P(Y=1)+...+P(X>Y|Y=6)P(Y=6)
=> P(A) = P(X>1)P(Y=1)+...+P(X>6)P(Y=6) by substitution law and independence of X and Y (two steps in one to save typing)
=> P(A) = (5/6+4/6+...0/6)*1/6 = 15/6/6 = 15/36

QED

[BruceZ]

[ QUOTE ]
If ties are ignored then it's 1/2.

With ties as a loss, you will tie 1/6 times. So 1/2 - 1/6 = your answer I beleieve.

0.5 - (1 / 6) = 0.333333333

[/ QUOTE ]

You remove the 1/6 ties from 1, and then split the remaining 5/6 to get 5/12, in agreement with CT11.

[shawny boy]

[ QUOTE ]
[ QUOTE ]
If ties are ignored then it's 1/2.

With ties as a loss, you will tie 1/6 times. So 1/2 - 1/6 = your answer I beleieve.

0.5 - (1 / 6) = 0.333333333

[/ QUOTE ]

You remove the 1/6 ties from 1, and then split the remaining 5/6 to get 5/12, in agreement with CT11.

[/ QUOTE ]

5/12 != 5/36

5/36 is correct ... (0 + 1/6 + 2/6 + 3/6 + 4/6 + 5/6)/6

[WhiteWolf]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
If ties are ignored then it's 1/2.

With ties as a loss, you will tie 1/6 times. So 1/2 - 1/6 = your answer I beleieve.

0.5 - (1 / 6) = 0.333333333

[/ QUOTE ]

You remove the 1/6 ties from 1, and then split the remaining 5/6 to get 5/12, in agreement with CT11.

[/ QUOTE ]

5/12 != 5/36

5/36 is correct ... (0 + 1/6 + 2/6 + 3/6 + 4/6 + 5/6)/6

[/ QUOTE ]

5/12 = 15/36

0+1+2+3+4+5 = 15 != 5

[shawny boy]

zomg, *redface*

5/12 == 15/36 whoops