Board Game Designers Forum - Comments for "Odds calculations"
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations
Comments for "Odds calculations"enThanks gabrielcohn. You
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27962
<p>Thanks gabrielcohn. You helped me a lot.<br />
This is not the most intriguing combat system but it fast and simple.</p>
Thu, 10 Jan 2013 21:57:36 +0000svennecomment 27962 at https://www.bgdf.comthirds
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27950
<p>on one die you would get a hit 1/3 of the time.<br />
on 2 dice you would get ONE OR MORE hits 1/3 + 2/3*1/3 of the time (that is 5/9 total).<br />
on 3 dice you would get ONE OR MORE hits 1/3 + 2/3*1/3 + 4/9*1/3 of the time (19/27 total)<br />
on 4 dice you would get ONE OR MORE hits 1/3 + 2/3*1/3 + 4/9*1/3 + 8/27*1/3 of the time (65/81 total)<br />
...and so forth</p>
<p>An easier way to put this, is your CHANCE OF FAILING TO GET ONE OR MORE HITS = (2/3)^x (where x is the number of dice rolled)<br />
1 die = 2/3<br />
2 dice = 4/9<br />
3 dice = 8/27<br />
4 dice = 16/81<br />
...and so forth.</p>
Thu, 10 Jan 2013 12:29:08 +0000gabrielcohncomment 27950 at https://www.bgdf.comColor dice odds
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27949
<p>I am cutting in with a new question here because it seems to be nice odds thread.</p>
<p>I have 1d6 with two sides colored with red, the others are gray. Red is for hit and gray for misses.</p>
<p>If I just roll one of those dices the chance of a hit would be 33,33%, but if I am rolling two of them? Three of them and so on?</p>
<p>The thing is that I want a really easy fight system that smaller kids can handle.</p>
Thu, 10 Jan 2013 10:14:50 +0000svennecomment 27949 at https://www.bgdf.comclojure
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27369
<p>Could not resist trying to solve it in Clojure as well (probably far from the best solution... currently learning):</p>
<p>user=> (let [c (cartesian-product (range 12) (range 8))] (/ (count (filter (partial apply >) c)) (count c)))<br />
5/8</p>
<p>Somewhat cryptic, yes. It is nice that it gives the exact answer. :)</p>
Tue, 04 Dec 2012 21:35:20 +0000pellecomment 27369 at https://www.bgdf.comWell, R is for statistics
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27368
<p>Well, R is for statistics people, and was even created by statistics people. It is a weird language if you are used to programming normal programming languages. This is how the problem in this thread can be solved using R plus the optional prob library:<br />
> prob(probspace(expand.grid(1:8, 1:12)), Var2>Var1)<br />
[1] 0.625</p>
<p>There is probably (eh...) a better way to express that, but I am no R expert. For many other problems there is a very easy to use R dice library, but I don't think it can handle comparisons between different dice.</p>
<p>For an intellectual challenge I do enjoy solving probability problems manually, or (almost as manually) in a spreadsheet, but to get something done I can really recommend to try some scripting. It is useful for LOTS of other things when making boardgames anyway, even without doing much actual programming. </p>
<p>Here is a solution using python and a library for solving dice problems I wrote a few years ago:<br />
>>> dice(1, 12).p_gt_other(dice(1, 8))<br />
0.625</p>
Tue, 04 Dec 2012 21:22:36 +0000pellecomment 27368 at https://www.bgdf.comanydice.com
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27367
<p><div class="quote-msg"><div class="quote-author"><em>pelle</em> wrote:</div>anydice.com. I prefer to solve this kind of problem in R or Python, but that online solution seems popular.</div></p>
<p>Expecially for who knows nothing about programming :)</p>
Tue, 04 Dec 2012 19:50:12 +0000Izraphaelcomment 27367 at https://www.bgdf.comanydice.com. I prefer to
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27362
<p>anydice.com. I prefer to solve this kind of problem in R or Python, but that online solution seems popular.</p>
Tue, 04 Dec 2012 17:38:57 +0000pellecomment 27362 at https://www.bgdf.comonline
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27360
<p>Anyone knows if there are some online dice odds calculators?</p>
Tue, 04 Dec 2012 13:26:56 +0000Izraphaelcomment 27360 at https://www.bgdf.comstats
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27352
<p>First, I will confirm the results that<br />
<div class="quote-msg"><div class="quote-author">Quote:</div>the D12 will win 62.5% of the time<br />
the D8 will win 29.17% of the time<br />
and they will tie 8.33% of the time</div></p>
<p>Second, a brief statistics primer. </p>
<p>a) List, graph, etc the possible outcomes. This is called the SAMPLE SPACE. The number of possible outcomes is used as the DENOMINATOR when calculating possibilities. </p>
<p>(In this case, we need an 8x12 grid. Since 8*12 = 96, the sample space has size 96)</p>
<p>b) Determine what event you'd like to measure, and calculate how many ways there are for this to happen. This can get complicated and involve combinatorics, but it's simple in our case.<br />
e.g. We want to know in how many ways the 1d8 can equal the 1d12.<br />
11<br />
22<br />
33<br />
44<br />
55<br />
66<br />
77<br />
88<br />
So there are eight ways. The size of the event (X = Y) is 8. </p>
<p>c) Divide. The probability of an event is (size of event) / (size of sample space)<br />
In our example, 8/96</p>
<p>-----------------------<br />
But say you've got an 8x12 grid with all the "equals" spaces marked off. You want to know the remaining probabilities (like when X > Y or Y > X).</p>
<p>Well first you can trim off a 4x8 area to leave an 8x8 grid with a diagonal removed. You have something like a 7x7 triangle below and one above.<br />
The number of outcomes in this triangle is actually the seventh TRIANGULAR NUMBER = 28.<br />
Let's not forget the 4x8 = 32 outcomes that we trimmed off. </p>
<p>So<br />
(# of ways 1d8 > 1d12) = 28, and<br />
(# of ways 1d8 = 1d12) = 8, and<br />
(# of ways ad8 < 1d12) = 28 + 32 (from the 4x8) = 60<br />
----------------------<br />
28 + 8 + 60 = 96, the size of our sample space.</p>
Mon, 03 Dec 2012 19:47:57 +0000The Chazcomment 27352 at https://www.bgdf.comHere's something I do when
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27348
<p>Here's something I do when I'm trying to figure out the probability of die rolls. It takes a bit of work, but once its set-up its easy to tweak to figure things out:</p>
<p><a href="https://docs.google.com/spreadsheet/ccc?key=0ApUge-xbUlSOdHM5dl9OM1dON0tQckZ4Z1NQTlBMSnc&authkey=CODRhqYE" title="https://docs.google.com/spreadsheet/ccc?key=0ApUge-xbUlSOdHM5dl9OM1dON0tQckZ4Z1NQTlBMSnc&authkey=CODRhqYE">https://docs.google.com/spreadsheet/ccc?key=0ApUge-xbUlSOdHM5dl9OM1dON0t...</a></p>
<p>The first tab shows how competing 2d6 rolls happen and what their results are.</p>
<p>I added a tab for these results:</p>
<p>the D12 will win 62.5% of the time<br />
the D8 will win 29.17% of the time<br />
and they will tie 8.33% of the time</p>
Mon, 03 Dec 2012 15:53:29 +0000Yamahakocomment 27348 at https://www.bgdf.comYup. Again.
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27341
<p>Yeah, the 5:12 vs 7:12 result is correct.</p>
<p>Have to admit, though, I'm doing the calculations differently than you guys.</p>
<p>Assuming 'A' is the die that has to beat die 'B'...</p>
<p>When A is 1, B wins 6 times<br />
When A is 2, B wins 5 times<br />
...<br />
When A is 6, B wins 1 time</p>
<p>So B's total wins are 6 + 5 + 4 + 3 + 2 + 1 = 6 * 7 / 2 = 21 out of 36.</p>
<p>That leaves 15 / 36 for A.</p>
<p>As for the asymmetry, yeah, it's because A must *beat* B, so the dice aren't really equal.</p>
Mon, 03 Dec 2012 07:10:06 +0000MarkKreitlercomment 27341 at https://www.bgdf.com...?
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27340
<p><div class="quote-msg"><div class="quote-author"><em>The Chaz</em> wrote:</div>Why would you expect 50-50??</div></p>
<p>Just because the dice were the same, I thought the odds would be 50:50... But I understand that when the dice rolls are EQUAL, they are not favorable rolls. And therefore they need to be subtracted (6 out of 36). That means there are 30 out of 36, halved 15 out of 36 (5:12).</p>
Mon, 03 Dec 2012 04:04:36 +0000questccgcomment 27340 at https://www.bgdf.com??
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27339
<p>Why would you expect 50-50??</p>
<p>The probability that die A is higher than die B is the same as<br />
the probability that die B is higher than die A.</p>
<p>Hopefully that symmetry is obvious. Then there's the probability that they're equal. 11,22,33,44,55,66 --- six ways<br />
6/36 = 1/6.<br />
A > B half of the remaining 30/36 times.</p>
Mon, 03 Dec 2012 03:59:28 +0000The Chazcomment 27339 at https://www.bgdf.comI was not sure...
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27337
<p><div class="quote-msg"><div class="quote-author"><em>MarkKreitler</em> wrote:</div>That's what I get.</div></p>
<p>I was not sure because when I use the same calculations with 2d6 (1d6 vs. another 1d6), I expected to get the odds of 50:50... But it turns out 5:12 favorable and 7:12 unfavorable...</p>
<p>Am I correct with these odds???</p>
Mon, 03 Dec 2012 02:49:58 +0000questccgcomment 27337 at https://www.bgdf.comYup
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27336
<p>That's what I get.</p>
<p>M.</p>
Mon, 03 Dec 2012 02:33:38 +0000MarkKreitlercomment 27336 at https://www.bgdf.comAnd the odds would be:
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27335
<p>5:8 (Favorable) and 3:8 (Unfavorable).</p>
Mon, 03 Dec 2012 01:09:29 +0000questccgcomment 27335 at https://www.bgdf.comHmm...
https://www.bgdf.com/forum/game-creation/mechanics/odds-calculations#comment-27334
<p><div class="quote-msg"><div class="quote-author"><em>questccg</em> wrote:</div>I wanted to know what the odds would be that this occurs...</div></p>
<p>So I have calculated there are 60 favorable rolls, 36 unfavorable rolls for a total rolls of 96.</p>
<p>60 out of 96 = 62.5%</p>
<p>36 out of 96 = 37.5%</p>
<p>Is this correct? I'm not certain I did it right...</p>
Mon, 03 Dec 2012 00:07:12 +0000questccgcomment 27334 at https://www.bgdf.com