My Math Forum probability/chances/expected value/variance question :)

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 October 29th, 2013, 04:05 PM #1 Newbie   Joined: Oct 2013 Posts: 2 Thanks: 0 probability/chances/expected value/variance question :) Hello everyone I'm trying to run a variance simulation of my own. In order to do it i created an empty Expected Value graph which i will be filling with results of throwing dice. I decided to start with using three dice. Each containing the digits from 1 to 6 That means the results can vary from minimum #3 to maximum #15 = 13 numbers. I decided that first 7 numbers (3,4,5,6,7,8,9) will give me a +EV result, while the remaining 6 numbers (10,11,12,13,14,15) will give me a -EV result. I would like to know if my assumptions below are correct: if all 13 numbers = 100% is chance to hit first 7 numbers 65% ? and is chance to hit remaining 6 numbers = 35% ? so is EV EDGE of choosing 7 numbers = +15% ? Thanks
October 29th, 2013, 04:12 PM   #2
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Re: probability/chances/expected value/variance question :)

Quote:
 Originally Posted by adasko99 I decided to start with using three dice. Each containing the digits from 1 to 6 That means the results can vary from minimum #3 to maximum #15 = 13 numbers.
If you have 3 dice and they all come up 6, isn't that a result of 18?

 October 30th, 2013, 01:09 PM #3 Newbie   Joined: Oct 2013 Posts: 2 Thanks: 0 Re: probability/chances/expected value/variance question :) that's right, sorry, my mistake :P so how can i estimate the percentage chances of hitting certain total numbers?
 October 31st, 2013, 02:44 AM #4 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: probability/chances/expected value/variance question :) The best way to calculate this is to look at the odds of each result then group and add up the probabilities of results that have the same score. I'd put it in a spreadsheet as it will be easier than on paper. In general, the odds of any throw is 1/216. So, the odds of throwing 6-6-6 is 1/216. But, throwing 4-5-6 is the same score as 4-6-5, 5-4-6, 5-6-4, 6-4-5 and 6-5-4. So, the odds of throwing one 4, one 5 and one 6 = 6/216. For two dice the same (e.g. two sixes and a 4) the odds are: 3/216. As it could be 6-4-4, 4-6-4 or 4-4-6 So, now you can calculate the odds of every throw from 1-1-1 to 6-6-6. Finally, you need to group throws with the same score and add the probabilities together. E.g. 2-3-4 (in any order) gives the same score as 1-4-4 (any order) and 1-3-5 etc. (I think this better than looking at all the ways to score 3-18, although you could do it that way).

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