Board Game Designers Forum - Comments for "Probability Math Question"
https://www.bgdf.com/forum/game-creation/mechanics/probability-math-question
Comments for "Probability Math Question"enX3M wrote:...but 2 hits
https://www.bgdf.com/forum/game-creation/mechanics/probability-math-question#comment-91362
<p><div class="quote-msg"><div class="quote-author"><em>X3M</em> wrote:</div>...but 2 hits counts twice. Now we get 61,1% + 2 x 11,1% = 83,3%.<br />
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.</div><br />
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.</p>
<p>Still, it's good to know that my intuition was on the right path.</p>
Thu, 11 May 2017 10:56:57 +0000Desprezcomment 91362 at https://www.bgdf.comA different view on things
https://www.bgdf.com/forum/game-creation/mechanics/probability-math-question#comment-91337
<p>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:</p>
<p>4/6 die gives<br />
33,3%, 0 hits<br />
66,7%, 1 hit</p>
<p>5/6 die gives<br />
16,7%, 0 hits<br />
83,3%, 1 hit</p>
<p>4/6 and 1/6 dice gives<br />
27,8%, 0 hits<br />
61,1%, 1 hit<br />
11,1%, 2 hits</p>
<p>As you can tell, the total chance is always 100%. Now for the more complicated questions.</p>
<p>For an at least 1 hit, you simply add up 1 hit with 2 hits. Which gives 61,1% + 11,1% = 72,2%.<br />
Which is of course less than the expected 83,3% on the other upgrade.</p>
<p>If you wonder why I have chosen this method for calculations. In war games where health tracking is needed. You count a bit different.<br />
1 hit counts once, but 2 hits counts twice. Now we get 61,1% + 2 x 11,1% = 83,3%.<br />
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.</p>
<p>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.</p>
Wed, 10 May 2017 16:19:22 +0000X3Mcomment 91337 at https://www.bgdf.comGood point
https://www.bgdf.com/forum/game-creation/mechanics/probability-math-question#comment-91335
<p>I neglected to mention <em>why</em> it was easier to look at the probability of missing. It's because in order to miss you need to miss with die 1 <strong>and</strong> 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.</p>
Wed, 10 May 2017 15:31:30 +0000FrankMcomment 91335 at https://www.bgdf.comWhen you combine
https://www.bgdf.com/forum/game-creation/mechanics/probability-math-question#comment-91331
<p>When you combine probabilities, if you use an AND, you multiply both of them like for example:</p>
<p>0.6 * 0.4</p>
<p>When you use and OR, you multiply by the left over of the first probability:</p>
<p>0.6 + ( (1 - 0.6) * 0.4 )</p>
<p>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. </p>
<p>Each path has a probability using multiplication. Then add the the branches you want to get the final probability.</p>
<p>Else you can use combinations and arrangement to calculate the number of valid results over the total number of results.</p>
Wed, 10 May 2017 14:02:40 +0000lariennacomment 91331 at https://www.bgdf.comOk, so I'm doing the math
https://www.bgdf.com/forum/game-creation/mechanics/probability-math-question#comment-91287
<p>Ok, so I'm doing the math right, but just did it the long way. Good to know.</p>
<p>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.</p>
Tue, 09 May 2017 07:33:10 +0000Desprezcomment 91287 at https://www.bgdf.comBonus does seem small, but correct.
https://www.bgdf.com/forum/game-creation/mechanics/probability-math-question#comment-91284
<p>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.</p>
<p>It's a bit easier to parse if you look at the probability of <em>missing</em> on each die.</p>
<p>Hit with either or both of 4/6 and 1/6 chances:<br />
Miss on die 1: 2/6<br />
Miss on die 2: 5/6<br />
2/6 * 5/6 = 10/36 chance to miss (26/36 chance to hit)</p>
<p>Die 1 by itself already had a 24/36 chance to hit, so the increase seems really small.<br />
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).</p>
<p>If this seems underpowered, there are a couple other options:<br />
1. Roll two dice and compare the higher one against the to-hit number.<br />
2. Use your original system but give a bonus if both hit.</p>
Tue, 09 May 2017 04:05:23 +0000FrankMcomment 91284 at https://www.bgdf.com