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Title: Hard Dice Question [ .01 btc prize] Post by: RoooooR on January 01, 2014, 09:01:17 PM 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 Title: Re: Hard Dice Question [ .01 btc prize] Post by: Sheldor333 on January 02, 2014, 12:15:43 AM 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 Title: Re: Hard Dice Question [ .01 btc prize] Post by: RoooooR on January 02, 2014, 08:15:30 AM Could you please clarify, do dice1 and dice2 have 6 or 12 faces? Yeah, dice1 and 2 normal, dice3 has 12 faces.Title: Re: Hard Dice Question [ .01 btc prize] Post by: SaltySpitoon on January 02, 2014, 08:31:03 AM 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. Title: Re: Hard Dice Question [ .01 btc prize] Post by: RoooooR on January 02, 2014, 10:25:22 AM Hey
I asked a mathematic teacher and he said 1/12 . (6 * 6) / (6 * 6 * 12) = 1/12 Title: Re: Hard Dice Question [ .01 btc prize] Post by: Frost000 on January 02, 2014, 02:50:58 PM Yup, I just did the calculation and it'd be 36 out 432, or 1 out of 12.
Title: Re: Hard Dice Question [ .01 btc prize] Post by: quone17 on January 02, 2014, 05:51:19 PM 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 Title: Re: Hard Dice Question [ .01 btc prize] Post by: jjc326 on January 02, 2014, 06:02:26 PM my guess is
3/40, 3 outta 40 tries. Title: Re: Hard Dice Question [ .01 btc prize] Post by: quone17 on January 02, 2014, 06:10:44 PM Wait a minute. Did OP say there was a prize and then not pay it? wtf?
Title: Re: Hard Dice Question [ .01 btc prize] Post by: RoooooR on January 02, 2014, 10:43:39 PM Wait a minute. Did OP say there was a prize and then not pay it? wtf? I sent you ! Thanks ! Title: Re: Hard Dice Question [ .01 btc prize] Post by: Hippie Tech on January 04, 2014, 02:46:52 AM Wrong wrong wrong... Its 50/50.
It either WILL or it WILL NOT happen. :P Title: Re: Hard Dice Question [ .01 btc prize] Post by: deepceleron on January 04, 2014, 03:51:50 AM 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.
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... |