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Probability Math Question

6 replies [Last post]
Mon, 05/08/2017 - 22:37
Desprez
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Joined: 12/01/2008

I'm looking at some probabilities, and I think I'm doing the math correct, but the results seem a little counter-intuitive. Perhaps someone can comment.

These examples are for a chance to score a hit on d6.

If you increase the chance to hit by one die face, it's a flat increase of 16.7 percentage points, no matter if increasing from 1/6 to 2/6, or from 4/6 to 5/6.

Easy enough.

But say you have a 4/6 chance to hit, and instead of increasing to 5/6, you add a second die with a 1/6 chance to score a hit as well, and roll both dice.

In this case the additional chance to hit is less than 16.7 points, meaning +1 die face on the original die is worth much more than adding a face on an additional die. And the added chance seems to vary depending on your original chance. (It's worth more if your original chance was 1/6 than if 5/6)

This seems a bit counter-intuitive to me, since at first blush both examples seem to add one scoring die face to the mix.

So, either my math is wrong, or probabilities are just weird that way.

Here's my math:

Example 1 (Add a scoring face on same die):
4/6 (66.7%) -> 5/6 (83.3%)
Increase of 16.6 points

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Example 2 (Add a scoring face on a new die):
4/6 (66.7%)
4/6 + 1/6 (72.3% of at least one hit)
Increase of 5.6 points

Example 2 breakdown of ways to hit with 2 dice:
hit on main and miss on secondary = 4/6 * 5/6 = 55.6%
miss on main and hit on secondary = 2/6 * 1/6 = 5.6%
hit on main and hit on secondary = 4/6 * 1/6 = 11.1%
Total = 72.3%

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But if the original began with 2/6 chance instead of 4/6, that additional die is now worth more, but still not as much as an increase to 3/6.

Example 3 (Add a scoring face on a new die):
2/6 (33.3%)
2/6 + 1/6 (44.4% of at least one hit)
Increase of 11.1 points

Example 3 breakdown of ways to hit with 2 dice:
hit on main and miss on secondary = 2/6 * 5/6 = 27.7%
miss on main and hit on secondary = 4/6 * 1/6 = 11.1%
hit on main and hit on secondary = 2/6 * 1/6 = 5.6%
Total = 44.4%

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And of course there's more disparity between the original starting with 1/6 compared to 5/6

Am I doing this correctly?

EDIT: Copied some numbers wrong from my notes.

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Tue, 05/09/2017 - 00:05
#1
FrankM
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Joined: 01/27/2017
Bonus does seem small, but correct.

The two dice are independent of one another, as as you noted there is a very real chance that both dice will yield the same result.

It's a bit easier to parse if you look at the probability of missing on each die.

Hit with either or both of 4/6 and 1/6 chances:
Miss on die 1: 2/6
Miss on die 2: 5/6
2/6 * 5/6 = 10/36 chance to miss (26/36 chance to hit)

Die 1 by itself already had a 24/36 chance to hit, so the increase seems really small.
In fact, of the 36 possible combinations of these two dice, there really are only two new ones that hit (1 and 6; 2 and 6).

If this seems underpowered, there are a couple other options:
1. Roll two dice and compare the higher one against the to-hit number.
2. Use your original system but give a bonus if both hit.

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Tue, 05/09/2017 - 03:33
#2
Desprez
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Joined: 12/01/2008
Ok, so I'm doing the math

Ok, so I'm doing the math right, but just did it the long way. Good to know.

I did, however, copy some numbers wrong from my notes which I fixed in the original post. Example 3 now shows a larger percentage point increase.

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Wed, 05/10/2017 - 10:02
#3
larienna
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Joined: 07/28/2008
When you combine

When you combine probabilities, if you use an AND, you multiply both of them like for example:

0.6 * 0.4

When you use and OR, you multiply by the left over of the first probability:

0.6 + ( (1 - 0.6) * 0.4 )

You can see this as a binary tree branch, you have 2 path, success or failure on the first die. Then for each path, you have have ausscess or failure of the second die.

Each path has a probability using multiplication. Then add the the branches you want to get the final probability.

Else you can use combinations and arrangement to calculate the number of valid results over the total number of results.

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Wed, 05/10/2017 - 11:31
#4
FrankM
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Joined: 01/27/2017
Good point

I neglected to mention why it was easier to look at the probability of missing. It's because in order to miss you need to miss with die 1 and miss with die 2. Working out the probabilities for "independent" conditions combined with "and" is the easiest math. They are "independent" if one has no impact on the other, and the "and" means you simply multiply the individual probabilities.

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Wed, 05/10/2017 - 12:19
#5
X3M
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Joined: 10/28/2013
A different view on things

I often tend to look a bit differently to things. And I got my own method for simple probability questions. I create a table and simply fill in how much hits might occur. So this is what I see:

4/6 die gives
33,3%, 0 hits
66,7%, 1 hit

5/6 die gives
16,7%, 0 hits
83,3%, 1 hit

4/6 and 1/6 dice gives
27,8%, 0 hits
61,1%, 1 hit
11,1%, 2 hits

As you can tell, the total chance is always 100%. Now for the more complicated questions.

For an at least 1 hit, you simply add up 1 hit with 2 hits. Which gives 61,1% + 11,1% = 72,2%.
Which is of course less than the expected 83,3% on the other upgrade.

If you wonder why I have chosen this method for calculations. In war games where health tracking is needed. You count a bit different.
1 hit counts once, but 2 hits counts twice. Now we get 61,1% + 2 x 11,1% = 83,3%.
As you can see, if you are going to track health. 4/6 with 1/6 will be the same as 5/6th in the long run.

On 1 health units, 2 hits is of course overkill. But it isn't if you allow the second hit to be done on another unit.

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Thu, 05/11/2017 - 06:56
#6
Desprez
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Joined: 12/01/2008
X3M wrote:...but 2 hits

X3M wrote:
...but 2 hits counts twice. Now we get 61,1% + 2 x 11,1% = 83,3%.
As you can see, if you are going to track health. 4/6 with 1/6 will be the same as 5/6th in the long run.

This, I think, is why it appeared counter-intuitive to me. I wasn't factoring in that two hits would count twice in the long run.

Still, it's good to know that my intuition was on the right path.

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