Hard Dice Question [ .01 btc prize]

[RoooooR]

Roll a random dice1
Roll a random dice2
Roll a random dice3 that has 12 faces.

What is the possibility of this equation ?

Code:

dice1+dice2 = dice3

[Sheldor333]

Check these forum posts:
http://stackoverflow.com/questions/5229745/c-question-can-anyone-help
http://www.coderanch.com/t/530468/java/java/Dice-Rolling-Simulation

[RoooooR]

Could you please clarify, do dice1 and dice2 have 6 or 12 faces?

Yeah, dice1 and 2 normal, dice3 has 12 faces.

[SaltySpitoon]

Wouldn't it be (1/6) (1/6) (1/12)

You have a 1 in 6 chance of getting a number on Die 1

and a 1 in 6 chance of getting a number on Die 2

And you have a 1 in 12 chance of getting a number on Die 3


giving you a 0.23% chance of any given combination of getting any one instance. I'm not entirely sure if thats what you are going for, if you are just talking about what your chances of rolling two dice,and the chances of the sum equaling the same number rolled on the 12 sided die, your answer is roughly 1 in 12, as unless you roll a 1 on the 12 sided die, you have a 1 in 12 chance of that number matching the sum of the two six sided dice.

*edit* whoops I was misunderstanding you. If no one has answered this by the time I wake up tomorrow, I'll calculate it.

[RoooooR]

Hey

I asked a mathematic teacher and he said 1/12 .

(6 * 6) / (6 * 6 * 12) = 1/12

[Frost000]

Yup, I just did the calculation and it'd be 36 out 432, or 1 out of 12.

[quone17]

Hmm I think the thing we're missing is that dice 3 can roll a 1, but dice 1 plus dice 2 cannot equal 1.  So the answer must be something less than 1/12.

I'll say it's 1/13, or 1 in 13 tries.


1EZM51qQuhyk8mrUk9ffvRY7aPT9RxMUiT

[jjc326]

my guess is

3/40, 3 outta 40 tries.

[quone17]

Wait a minute.  Did OP say there was a prize and then not pay it? wtf?

[RoooooR]

Wait a minute.  Did OP say there was a prize and then not pay it? wtf?

I sent you ! Thanks !

[Hippie Tech]

Wrong wrong wrong... Its 50/50.

It either WILL or it WILL NOT happen.

[deepceleron]

Lets view it starting with the d12 - it has a 1 in 12 chance of being any particular number.
Then we need to look at the probability that the two dice will roll a number that matches.
Then we evaluate the probability of the rolls being this match outcome for each d12 case. Then sum.

d12	d12 prob	2xd6's match prob	prob of this match
1	1/12	0/36	0/432
2	1/12	1/36	1/432
3	1/12	2/36	2/432
4	1/12	3/36	3/432
5	1/12	4/36	4/432
6	1/12	5/36	5/432
7	1/12	6/36	6/432
8	1/12	5/36	5/432
9	1/12	4/36	4/432
10	1/12	3/36	3/432
11	1/12	2/36	2/432
12	1/12	1/36	1/432
	Probability of match		36/432

The probability of a match is 8.333% - also 1/12

Hmm I think the thing we're missing is that dice 3 can roll a 1, but dice 1 plus dice 2 cannot equal 1.  So the answer must be something less than 1/12.

I'll say it's 1/13, or 1 in 13 tries.

Wait a minute.  Did OP say there was a prize and then not pay it? wtf?

I sent you ! Thanks !

Oops...