Suggestion for the Probability Theory Column

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Post originally by Steve at 2005-08-17 14:04:51
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Try putting the calculations in the footnotes. Some people might acutally want to see how it is done and having it in the foot notes can help them.

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Post originally by Stupendous Press at 2005-08-17 18:47:49
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I agree, it would be nice to see or at least have explained. I also wouldn't mind seeing slightly more explanation in layman's terms for some of the math. I'm not very familiar with deviation, and the article gave no indication of exactly what a 1.7 or 5.8 deviation (for d6 and d20, respectively) meant. Otherwise, very well-written and concise. I look forward to more complex probability schemes.

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Post originally by jay verkuilen at 2005-09-25 12:39:35
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Stupendous Press wrote:
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>>I agree, it would be nice to see or at least have explained. I also wouldn't mind seeing slightly more explanation in layman's terms for some of the math. I'm not very familiar with deviation, and the article gave no indication of exactly what a 1.7 or 5.8 deviation (for d6 and d20, respectively) meant. Otherwise, very well-written and concise. I look forward to more complex probability schemes.<<

I'll also thank Paul for his work.

The standard deviation tells you how spread out the points are. Anyway, to get some intuition for the standard deviation, consider the following:

The mean of a D4 is 2.5, right? That's (1 + 2 + 3 + 4)/4. This tells you where the distribution balances. The standard deviation of this is, following the definition:

SD = sqrt( .25* [(1-2.5)^2 + (2-2.5)^2 + (3-2.5)^2 + (4-2.5)^2] )

= sqrt( .25* [(-1.5)^2 + (-.5)^2 + .5^2 + 1.5^2] )

= 1.12.

These squared quantities are called the squared deviations, and we want to find the "standard" one, i.e., the typical. We're going to average. The reason why we have to do the square-square root and can't simply sum up the deviations directly is because we *always* know the answer in advance: it's 0! (Try it yourself.) By squaring, summing, and square rooting, we're actually computing the Euclidean distance, i.e., the distance based on the Pythagorean theorem.

Let's make a physical analogy:

If you made an object with four equal masses that were equally spaced and tried to hang it in the air, it would balance at the mean. If you could move the masses around you would shift where the distribution balances. Of course, if you're really clever, you can always counteract any move to keep the mean (the balance point) the same, but if you did this, you'd have to spread the masses out more.

This is like trying to balance a big kid and a little kid on a seesaw. The big kid has to move closer to the fulcrum or the little kid further away from the fulcrum to keep things balanced.

Let's try this with the D4 again. Let's say I take a magic marker and change all 2s to 1s and all 3s to 4s. This means we now have a die that rolls 1 with probability .5 and 4 with probability .4.

The mean does not change because I moved mass from 2 and 3 (near the mean) to the extremes symmetrically. This is two kids on a longer lever. But let's confirm:

mean = (1 + 1 + 4 + 4)/4 = 2.5.

However, the masses are now farther apart, which means that the average distance from the mean is bigger, i.e., the standard deviation goes up:

SD = sqrt( .25* [(1-2.5)^2 + (1-2.5)^2 + (4-2.5)^2 + (4-2.5)^2] )

= sqrt( .25* [(-1.5)^2 + (-1.5)^2 + 1.5^2 + 1.5^2] )

= 1.25.

What does this mean? In a system where the variation is larger, things are more unpredictable. Consider D20 vs. 3D6. Both have the same mean (10.5) but a D20 is more variable than 3D6, because a D20 gives equal probability (mass) to all points between 1 and 20, whereas 3D6 gives a lot more mass to points close to 10.5, and none at all to 1, 2, 19, and 20. The standard deviations reflect this.

Jay