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#7 | |
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Join Date: Sep 2011
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Quote:
We have three dice, each of which has six faces, numbered 1 through 6, and we are interested in the sum appearing on those three dice. The possible sums vary from a low of 3 (each die shows a 1) to a high of 18 (each die shows a 6). The range of possible sums is 3 to 18, or 16 possible sums. The problem, as it were, is that the sums do not have an equal probability of occurring. To make this clearer, suppose that we can identify which number appears on which die. For our purposes, we will designate one die as being red, one die as being yellow and one die as being blue (all with white pips). All the die rolls are independent, i.e., what number appears on the red die has no effect on what number appears on either the yellow die or the blue die. We can now enumerate all the possible sums that appear on the three dice. Imagine that wr have rolled 1 on the red die and 1 on the yellow die, we can still roll any of the numbers 1 through 6 on the blue die, meaning that there are 6 possible sums when the red die is 1 and the yellow die is 1. Now assume that the red die is still 1 but the yellow die is 2. The blue die can now show any value of 1 through 6 for another 6 possibilities. We can do this for the yellow die being 3, 4, 5 and 6 as well. Each case gives us six possible outcomes for a total of 36 possible sums when the red die is 1 and we have run the yellow die and the blue die through all of their possible outcomes. If we rotate the red die to 2 we hve another 36 possible outcomes for the sum that appears on the yellow die and the blue die. This again holds true for each of the other values the red die might show, giving 36 possible sums for each value on the red die. Summing them together, there are 216 distinct and different ways to get our sum which still is confined to a range of 3 to 18. It is the weight of the different possible outcomes that gives us our bell curve. Of the 216 different rolls we can get on our three dice, only one, red=1, yellow=1, blue=1 will give us a sum of 3 and only one (red=6, yellow=6, blue=6) will give us a sum of 18. However, to get a sum of four, we need two dice to show 1s and one die to show a 2. Whether it is the red, yellow or blue die that shows the 2 is immaterial, any of those possibilities will give us a sum of 4. Likewise for 17, where we need two dice to show a 6 and one die to show a 5. Turning these 216 different possibilities into percentages, gives us roughly: Chance of rolling a 3 is 1/216 or a little less than 0.5%. Chance of rolling a 4, 3/216, or slightly less than 1.5%. Chance of rolling a 5, 6/216, or slightly less than 3%, and so on, until we get to chance of rolling an 18, 1/216 or slightly less than 0.5%. However, we aren't actually interested in rolling the exact number but in rolling a number or less, making the percentage chance, the sum of the the chance of rolling that number plus the chance of rolling each number that is less than that, which is why the table doesn't quite look like a bell curve. The rolls for an exact number do form a bell curve when plotted. For a number or less, the progression is 3- = chance of rolling exactly 3 , less than 0.5%. 4- = Chance of rolling a 3, less than 0.5% and chance of rolling a , less than 1.5%, totaling less than 2%. 5- = chance of rolling a 5, 10/216, slightly less than 5% and chance of rolling a 3 or 4, which we just calculated was slightly less than 2%, for a total of slightly less than 7%. 6- = chance of rolling a 6, 15/216, slightly less than 7.5%, and the chance of rolling a 3, 4 or 5, which we just calculated was slightly less than 7%, for a total of slightly less than 14.5% So, if you can manage an effective skill of 16-, you will critically hit about 14.5% of the time, not 40% of the time. I hope this helps clarify the matter for you. |
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| Tags |
| combat, defending, tactics, vtm |
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