Board Game Designers Forum - Comments for "Dice probabilities"
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0
Comments for "Dice probabilities"enIm happy it work a bit (I
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71946
<p>Im happy it work a bit (I know that for some systems or ipad and similar things it doesnt work because dont use the java environment, it uses instead html5).</p>
<p>I want to write just a last thing that maybe useful to anyone interested on define any type of probability of dice or others games.</p>
<p>The problem for 1's cancellation was really more hard to finalyze that I expected cause my poor understanding on the combinatorics laws.</p>
<p>The key in any problem on combinatorics is to create abstract concepts of independents things, something like a "high level language" to work with "low level" processes (in statistical maths these "low levels" process are named atomic or elemental events).</p>
<p>So if you can define some abstract sets with different composition you can work on this "high level" and the things become easier. In maths this is a norm since Hilbert when he "discover" that you can resolve really hard tasks if you think it from a very high and abstract level where you groups things, and these are the modern maths, i.e., abstract algebra.</p>
<p>In the formulas the binomials (or multinomial) coefficients means <em>permutations with repetitions</em>, i.e., how many shapes or forms I can order a list of elements (in this cases the elements maybe all diferents=normal permutations, or in groups=permutations with repetitions).</p>
<p>And the other big combinatoric concept is variations <em>into</em> a set... the multiplicity of the sets. These are the powers on equations.</p>
<p>So you have two things:</p>
<p>-permutations of sets of things ones with others -> binomial coefficients</p>
<p>-number of variations for each set -> powers</p>
<p>To understand this I have some good examples: variations are combinations from one class of things, a set, something like genetic differences between individuals of the same specie.</p>
<p>And permutations are the amount that different species can divide a territory.</p>
<p>I hope this maybe interesting or useful for someone. Sry if Im too "boring" with these things but I really like a lot maths... I should have been mathematician after all :P</p>
<p>EDIT: last words: the KEY to resolve it was to forget the numbers!!! I imagined the dice instead with numbers just sides of different colours, each colour with a different meaning related to the mechanic (this is why I named the things on the formulas by colours).</p>
<p>From here it becomes more easy...the numbers were confusing me because are not directly related to the meaning on the game or mechanic.</p>
<p>EDIT 2: fixed the representation of probabilities for a fixed value of <em>k</em> and reestructured a bit some info to be more clear.</p>
Mon, 01 Sep 2014 21:56:58 +0000Masacrosocomment 71946 at http://www.bgdf.comI can see it!! It looks
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71944
<p>I can see it!! It looks amazing! I am going to need some time to read it carefully to understand all the mechanics, but it is awesome! Thanks a lot!</p>
Fri, 29 Aug 2014 23:01:11 +0000apertotescomment 71944 at http://www.bgdf.comapertotes wrote:Sorry
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71924
<p><div class="quote-msg"><div class="quote-author"><em>apertotes</em> wrote:</div>Sorry Masacroso, but I can only see the initial explanation. I tried both in Firefox and Chrome, and also as java and html5, but it is not working.</div></p>
<p>Thank you for the feedback... I will try to see what is happening. This website have A LOT of applets and I see all ok but maybe Im doing something wrong. When possible I will see how they see in others computers.</p>
<p>EDIT: I think its fixed by now, maybe a problem with too high size on fonts. You cant see it correctly if dont have jre or some java installed. The applet load in two versions: HTML5 that dont charge any LATEX text and is buggy, and java.</p>
<iframe scrolling="no" src="https://www.geogebratube.org/material/iframe/id/147891/width/600/height/450/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/false/at/preferjava" width="600px" height="450px" style="border:0px;"> </iframe>
<p>P.S.: I updated the others as well.</p>
Fri, 29 Aug 2014 22:02:23 +0000Masacrosocomment 71924 at http://www.bgdf.comSorry Masacroso, but I can
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71923
<p>Sorry Masacroso, but I can only see the initial explanation. I tried both in Firefox and Chrome, and also as java and html5, but it is not working.</p>
Thu, 28 Aug 2014 15:40:29 +0000apertotescomment 71923 at http://www.bgdf.comI fixed the mechanic of
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71905
<p>EDIT: finally I fixed all mistakes and I have a deeper understanding on combinatorics.</p>
<p>Of course I did <a href="http://www.geogebratube.org/student/m147891">an applet</a> with all the info (I dont know why the applet dont work very fine on the forum... just use the link).</p>
<p>With some more time I will add the <em>sum&reroll</em> mechanic to the mix on the applet (this is very easy to add).</p>
<p>You can see, in general, that the cancel mechanic doesnt change too much the standard probabilities without cancelling, just narrow a bit. The mechanic of cancel compensate a bit the exponential nature of the mechanic without any cancelling.</p>
Thu, 28 Aug 2014 14:21:07 +0000Masacrosocomment 71905 at http://www.bgdf.comI must says other interesting
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71874
<p>I must says other interesting thing with the mechanic of <em>sum-and-reroll</em> design-oriented. You can calculates the probability for <em>j</em> rerolls happen and create something like a "average probability of rerolls" at P(j) probability:</p>
<p><a href="http://www.codecogs.com/eqnedit.php?latex=p^j(1-p)=P(j)&space;\rightarrow&space;j=\frac{lnP(j)-ln(1-p)}{lnp}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?p^j(1-p)=P(j)&space;\rightarrow&space;j=\frac{lnP(j)-ln(1-p)}{lnp}" title="p^j(1-p)=P(j) \rightarrow j=\frac{lnP(j)-ln(1-p)}{lnp}" /></a></p>
<p>Where P(j) is the probability for an exactly number <em>j</em> of rerolls happen. By example: with a probability of reroll of 70% we want to know what is the probability to 10 rerolls, that is P(j)=(0.7^10)*0.3=0.84% !!!</p>
<p>Inversely the number of <em>j</em> for a P(j)=0.5 (average of rerolls per throw) is (ln0.5-ln0.3)/ln0.7=<strong>-.43</strong>!!!, is negative because NEVER CAN HAPPEN cause is bigger that the probability of one reroll that is 0.7*0.3.</p>
<p>This happen because we are calculating <em>exact</em> number of <em>j</em> in a mechanic with a high probability for reroll (70%). If we calculate the same for the probability of <em>at least</em> a <em>j</em> amount of rerolls the numbers grows.</p>
<p><a href="http://www.codecogs.com/eqnedit.php?latex=P(\geq&space;j)=&space;p^j&space;\rightarrow&space;j=\frac{lnP(\geq&space;j)}{lnp}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?P(\geq&space;j)=&space;p^j&space;\rightarrow&space;j=\frac{lnP(\geq&space;j)}{lnp}" title="P(\geq j)= p^j \rightarrow j=\frac{lnP(\geq j)}{lnp}" /></a></p>
<p>Following this if you have a P(≥j)=50% when you p=70% then you have a probability of 50% for <em>at least</em> 2 rerolls per throw (j=ln0.5/ln0.7).</p>
<p>This maybe useful to predict the "annoyance level" of a game.</p>
<p>P.S.: still I dont know how to solve the problem of 1's, seems I fail in something related to dependency of events.</p>
Sun, 24 Aug 2014 12:21:42 +0000Masacrosocomment 71874 at http://www.bgdf.comapertotes wrote:Masacroso
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71865
<p><div class="quote-msg"><div class="quote-author"><em>apertotes</em> wrote:</div><div class="quote-msg"><div class="quote-author"><em>Masacroso</em> wrote:</div><a href="http://www.geogebratube.org/student/m146489">Here</a> you can see a java applet with this and more details as some graphs and tables.</div></p>
<p>damm, I can never see the full table. Although this one can be seen more than the others you linked, I still can't see the right part of it.</p>
<p>But I have a question, it looks that P(x>v) + P(x < v) is much higher than 1. For example, for D=12, n=1, V=5, P(x>5)=66.67% and P(x < V)=91.67%. That looks strange, right?</p>
<p>In any case, is there anyway to get an expected value from there?</div></p>
<p>Damn... I dont know how these applets of java really work (its an auto-export from the geogebra client, after you can "tweak" online but... anyway).</p>
<p>Mmmm... maybe something is wrong, I must check this... something maybe wrong.</p>
<p><div class="quote-msg"><div class="quote-author">Quote:</div><div class="quote-msg"><div class="quote-author"><em>Masacroso</em> wrote:</div>I leave here a general formula for the mechanic of "reroll and sum" for any type of dice, any number or whatever, its very simple:</p>
<p>...</p>
<p>So for the case that start this topic about average value from rolling a D12 dice and reroll and sum when 12:</p>
<ul>
<li>mR=12</li>
<li>mNR=6</li>
<li>pR=1/12 → <strong>X=12/11+6=7,1</strong></li>
</ul>
<p>LOL</div></p>
<p>That is awesome! This is always for rolling 1 die, right?</div></p>
<p>Yes. I leave the general formula to know any roll on any number of design of dice.</p>
<p>The best of this is that the formula for the average of ONE homogeneous dice is this</p>
<p><a href="http://www.codecogs.com/eqnedit.php?latex=\bar&space;X=\frac{D}{D-1}+\frac{D}{2}" target="_blank"><img src="http://latex.codecogs.com/png.latex?\bar&space;X=\frac{D}{D-1}+\frac{D}{2}" title="\bar X=\frac{D}{D-1}+\frac{D}{2}" /></a></p>
<p>and because this we can know, exactly, the ratio of the "boost" from a normal throw of a dice, i.e., the average of <em>roll-and-sum</em> between the standard average for a dice that is (D+1)/2</p>
<p><a href="http://www.codecogs.com/eqnedit.php?latex=r=\frac{\bar&space;X^*}{\bar&space;X}=\frac{D}{D-1}=&space;1+&space;\frac{1}{D-1}" target="_blank"><img src="http://latex.codecogs.com/png.latex?r=\frac{\bar&space;X^*}{\bar&space;X}=\frac{D}{D-1}=&space;1+&space;\frac{1}{D-1}" title="r=\frac{\bar X^*}{\bar X}=\frac{D}{D-1}= 1+ \frac{1}{D-1}" /></a></p>
<p>So you are "boosting" the average value in a fraction of D-1. So the boost is smaller when you have a dice with more sides.</p>
<p>P.S.: I updated the first "applet" about this.</p>
Sat, 23 Aug 2014 18:13:48 +0000Masacrosocomment 71865 at http://www.bgdf.comMasacroso
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71864
<p><div class="quote-msg"><div class="quote-author"><em>Masacroso</em> wrote:</div>[Here](<a href="http://www.geogebratube.org/student/m146489" title="http://www.geogebratube.org/student/m146489">http://www.geogebratube.org/student/m146489</a>) you can see a java applet with this and more details as some graphs and tables.</div></p>
<p>damm, I can never see the full table. Although this one can be seen more than the others you linked, I still can't see the right part of it.</p>
<p>But I have a question, it looks that P(x>v) + P(x < v) is much higher than 1. For example, for D=12, n=1, V=5, P(x>5)=66.67% and P(x < V)=91.67%. That looks strange, right?</p>
<p>In any case, is there anyway to get an expected value from there?</p>
<p><div class="quote-msg"><div class="quote-author"><em>Masacroso</em> wrote:</div>I leave here a general formula for the mechanic of "reroll and sum" for any type of dice, any number or whatever, its very simple:</p>
<p>...</p>
<p>So for the case that start this topic about average value from rolling a D12 dice and reroll and sum when 12:</p>
<p>* mR=12<br />
* mNR=6<br />
* pR=1/12 → **X=12/11+6=7,1**</p>
<p>LOL</div></p>
<p>That is awesome! This is always for rolling 1 die, right?</p>
Sat, 23 Aug 2014 17:35:26 +0000apertotescomment 71864 at http://www.bgdf.comkos wrote:Once you get more
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71863
<p><div class="quote-msg"><div class="quote-author"><em>kos</em> wrote:</div>Once you get more than 3 dice, it really heavily stacks the results around 10 and 11. That is, for any number of dice greater than 3, around 40% of your results will be 10 or 11 (no matter how many extra dice you add). The effect of adding more dice is that the number of results in the <=9 category decreases while the >=13 category increases, but the 10-11 category stays relatively stable.</div></p>
<p>Well, that sits perfectly with my intentions, which is that increasing the attack value of the unit would not create a huge swing in the results (as would do a simple sum of all dice rolled), but more a kind of security that you would hit your objective. For example, if you need an 8 to hit, rolling an 8 would be as good as rolling a 23. And rolling more dice would slowly increase your chance to succeed. But if you need a 25 to hit, well, you better have a very high attack value, or your chances will be very poor. But never 0, which is what I was looking for. Higher attack values have it easier, but even lower values could get lucky once in a while.</p>
<p><div class="quote-msg"><div class="quote-author"><em>kos</em> wrote:</div>Although I didn't model the 1's cancellation rule, I expect that the result would be an even greater concentration in the range 10-11. It wouldn't surprise me if the range 10-11 ended up with 75% of more of the results.</div></p>
<p>Yes, that "absurd level" rule will surely decrease the number of valid 12s, which will reduce immensely the probability of rolling higher than 12. Well, that is one of the intended consequences. Rolling higher than 12 should be a rare achievement. Indeed, my intuition tells me that with this absurd rule the probabilities of rolling a valid 12 is the same (or very similar) no matter how many dice you roll, which seems appropriate with the idea behind re-rolling 12s, which is a very lucky shot. And luck is not really affected by skill, is it?</p>
<p>The other intended consequence for the 1s cancelling high numbers is to offer the possibility of a complete disaster, regardless of the number of dice rolled.</p>
Sat, 23 Aug 2014 17:21:04 +0000apertotescomment 71863 at http://www.bgdf.comI reached the "absurd level
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71860
<p>I reached the "absurd level of task" xD.</p>
<p>I did a worksheet with the mechanic with the 1's and eliminations. It was hard and easy at the same time: hard because my knowledge of maths is so low in many areas, and easy because when you understand finally how to calculates these probabilities all is done.</p>
<p>The thing is two binomial distributions, one conditional to the other: the first one is the probability of take 1's, from zero to the maximum number of dice</p>
<p><a href="http://www.codecogs.com/eqnedit.php?latex=P(a)&space;=&space;\binom{n}{a}\frac{(D-1)^{n-a}}{D^n}" target="_blank"><img src="http://latex.codecogs.com/png.latex?P(a)&space;=&space;\binom{n}{a}\frac{(D-1)^{n-a}}{D^n}" title="P(a) = \binom{n}{a}\frac{(D-1)^{n-a}}{D^n}" /></a></p>
<p>where <em>a</em> is the number of 1's, <em>D</em> the number of sides of the dice and <em>n</em> the amount of dice you throw.</p>
<p>After is the other binomial conditional to the first .This is for know the probability to take a value V on the throw after eliminations (cause mechanic of 1's) and is</p>
<p><a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{80}&space;F_a(X\geq&space;V)=\frac{1}{(D-1)^{n-a}}\sum_{j=a+1}^{n-a}(D-V+1)^j(V-2)^{n-a-j}" target="_blank"><img src="http://latex.codecogs.com/png.latex?\dpi{80}&space;F_a(X\geq&space;V)=\frac{1}{(D-1)^{n-a}}\sum_{j=a+1}^{n-a}(D-V+1)^j(V-2)^{n-a-j}" title="F_a(X\geq V)=\frac{1}{(D-1)^{n-a}}\sum_{j=a+1}^{n-a}(D-V+1)^j(V-2)^{n-a-j}" /></a></p>
<p>This formula have the probability to have a throw with a <em>low bound</em> on value on <em>V</em> on a specific number <em>a</em> of 1's.</p>
<p>If you multiply the two binomials and sum over <em>a</em> you take the probability for have <em>at least</em> a value of <em>V</em> on the throw. Deleting just the "+1" on the second formula and repeating the sum above explained you take the probability to take <em>at maximum</em> a value of <em>V</em> on your throw.</p>
<p>And if you subtract these two bounded probabilities you take the probability to have <em>exactly</em> a value of <em>V</em> on the throw.</p>
<p><a href="http://www.geogebratube.org/student/m146489">Here</a> you can see a java applet with this and more details as some graphs and tables.</p>
<p>I leave here a general formula for the mechanic of "reroll and sum" for any type of dice, any number or whatever, its very simple:</p>
<p><a href="http://www.codecogs.com/eqnedit.php?latex=m_j=&space;m_{NR}&space;+&space;jm_R\\&space;\indent&space;p_j=p_{NR}*p_R^j\\&space;\indent&space;e_j=m_j*p_j\\&space;\indent&space;\bar&space;x=\sum_{j=0}^{\infty}e_j" target="_blank"><img src="http://latex.codecogs.com/gif.latex?m_j=&space;m_{NR}&space;+&space;jm_R\\&space;\indent&space;p_j=p_{NR}*p_R^j\\&space;\indent&space;e_j=m_j*p_j\\&space;\indent&space;\bar&space;x=\sum_{j=0}^{\infty}e_j" title="m_j= m_{NR} + jm_R\\ \indent p_j=p_{NR}*p_R^j\\ \indent e_j=m_j*p_j\\ \indent \bar x=\sum_{j=0}^{\infty}e_j" /></a></p>
<p>Where mNR is the average value of the group of events when no reroll happens (if it reroll with 6 in a D6 then mNR is the average from 1 to 5, i.e., 3), and mR is the average value of the events that reroll (if you reroll with a 6 in a D6 the average value is 6).</p>
<p>pNR is the probability of all group (layer) where no reroll happen, and pR when reroll (see that pNR+pR=1).</p>
<p><em>j</em> just indicates the layer that is evaluating, the only layer with no reroll is when j=0, j=1 is to evaluate the first reroll and so on.</p>
<p><em>ej</em> is the contributed value of each layer. I dont know (by now) if the infinite series have a explicit form... and it doesnt converges in all cases (when pR>=0.5)</p>
<p>EDIT: I did! :p The series converges completely no matter what because any probability is a number between 0 and 1.</p>
<p>And doing some maths over infinite geometric series the result is</p>
<p><a href="http://www.codecogs.com/eqnedit.php?latex=\bar&space;X=\frac{m_Rp_R}{1-p_R}+m_{NR}" target="_blank"><img src="http://latex.codecogs.com/png.latex?\bar&space;X=\frac{m_Rp_R}{1-p_R}+m_{NR}" title="\bar X=\frac{m_Rp_R}{1-p_R}+m_{NR}" /></a></p>
<p>So for the case that start this topic about average value from rolling a D12 dice and reroll and sum when 12:</p>
<ul>
<li>mR=12</li>
<li>mNR=6</li>
<li>pR=1/12 → <strong>X=12/11+6=7,1</strong></li>
</ul>
<p>LOL</p>
Sat, 23 Aug 2014 16:47:49 +0000Masacrosocomment 71860 at http://www.bgdf.comSimulation
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71855
<p>Using a spreadsheet, I simulated the results of the Advanced challenge (roll Nd12, reroll 12's, keep the highest).</p>
<p>Iterations: 1000<br />
Average result:<br />
- 1 die: 7.2<br />
- 2 dice: 9.8<br />
- 3 dice: 11.3<br />
- 4 dice: 12.5<br />
- 5 dice: 13.5<br />
- 6 dice: 14.2<br />
- 7 dice: 15.0<br />
- 8 dice: 15.5<br />
- 9 dice: 15.9<br />
-10 dice: 16.4</p>
<p>Of course those numbers do not exactly match the theoretical results, but it gives a rough idea of what the probabilities are.</p>
<p>Interesting to note is that the standard deviation slowly increases with the number of dice, starting around 5 and increasing to the low 6's. What this means is that the amount of variability in the rolls is gradually increasing, whereas in typical dice rolling (Nd12, add the results) the standard deviation decreases as the number of dice increases.</p>
<p>The other side effect to note is that the probability curve is a step function, meaning that there are "jumps" in the curve at 12, 24, 36. This is what you would expect, given the exploding dice rule.</p>
<p>Once you get more than 3 dice, it really heavily stacks the results around 10 and 11. That is, for any number of dice greater than 3, around 40% of your results will be 10 or 11 (no matter how many extra dice you add). The effect of adding more dice is that the number of results in the <=9 category decreases while the >=13 category increases, but the 10-11 category stays relatively stable.</p>
<p>---------------------------<br />
Although I didn't model the 1's cancellation rule, I expect that the result would be an even greater concentration in the range 10-11. It wouldn't surprise me if the range 10-11 ended up with 75% of more of the results.</p>
<p>---------------------------<br />
If you were using this dice rolling system as part of a game, I'd recommend that you keep the number of dice lower, say 3 dice for "average" and 5 dice for "excellent". Aside from the impact on the probability curve, the physical aspect of spotting the maximum dice from a group of 10 dice (and then rerolling them) can become time consuming.</p>
<p>Regards,<br />
kos</p>
Sat, 23 Aug 2014 03:26:44 +0000koscomment 71855 at http://www.bgdf.comWow, this is really
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71829
<p>Wow, this is really interesting, thanks for the job man.</p>
<p>(I really learned A LOT on mathematics doing these type of tasks.)</p>
<p>I can try about the "first cancel condition", this is all about combinations and it probabilities... the different combinations of 1's and 12's (or whatever other group of numbers).</p>
Fri, 22 Aug 2014 06:34:05 +0000Masacrosocomment 71829 at http://www.bgdf.comWell, I have made some
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71810
<p>Well, I have made some progress. I have created a Google Sheets document with the expected value of several dice.</p>
<p>The first page is a table with 12 rows (numbered 1-12, one for each side of the die) and multiple columns (34 at the moment, indicating the number of dice rolled).</p>
<p>That table has the following formula on each cell: P = (k^n - (k-1)^n)/s^n</p>
<p>where k=highest number, n=number of dice, s=sides of dice</p>
<p>What this table does is generate the posibilities for each side of the die to be the highest numbered side in relation to the number of dice rolled. For example, the firsts columns of the table are like this:<br />
1 2 3<br />
1 0.083333333 0.006944444 0.000578704<br />
2 0.083333333 0.020833333 0.004050926<br />
3 0.083333333 0.034722222 0.010995370<br />
4 0.083333333 0.048611111 0.021412037<br />
5 0.083333333 0.062500000 0.035300926<br />
6 0.083333333 0.076388889 0.052662037<br />
7 0.083333333 0.090277778 0.073495370<br />
8 0.083333333 0.104166667 0.097800926<br />
9 0.083333333 0.118055556 0.125578704<br />
10 0.083333333 0.131944444 0.156828704<br />
11 0.083333333 0.145833333 0.191550926<br />
12 0.083333333 0.159722222 0.229745370<br />
1.00000000 1.00000000 1.00000000</p>
<p>That was really the hardest part, mathematically speaking. After that, there was a lot of work to create a table on a different sheet referencing cells on the probabilities sheet, but the math was quite easy.</p>
<p>So, the tables on the other sheets are tailored for specific 12-sided dice. For example, a regular 1-12 dice, a 0-11 dice, a 0-11 dice where 0 are failures, a 0-11+ dice, a 2x dice (2, 4, 6,..., 24), a 0,0,0,0,0,0,0,0,0,0,0,12 dice, etc.</p>
<p>To do that, the rows are numbered as values of the side of the dice, in ascending order. The columns remain the same as before (number of dice rolled). And in each cell there is the following formula: i*P</p>
<p>where i is the value of the side of the dice in each row, and P is the value from that same row and column on the probabilities table.</p>
<p>Then we add all the values of each column to obtain the expected highest value when rolling n dice.</p>
<p>For example, the first columns of a dice with 11 sides as 0 and only one side as 12 looks like this:</p>
<p>0 0.00000000 0.00000000 0.00000000<br />
0 0.00000000 0.00000000 0.00000000<br />
0 0.00000000 0.00000000 0.00000000<br />
0 0.00000000 0.00000000 0.00000000<br />
0 0.00000000 0.00000000 0.00000000<br />
0 0.00000000 0.00000000 0.00000000<br />
0 0.00000000 0.00000000 0.00000000<br />
0 0.00000000 0.00000000 0.00000000<br />
12 1.00000000 1.91666667 2.75694444<br />
Exp. value 1.00000000 1.91666667 2.75694444</p>
<p>With this system is incredibly easy to calculate expected values for many different 12-sided dice. For example, if I wanted a system where any 0 would turn the roll into a failure, I would only have to number the rows on the second table as 1,2,3,4,5,6,7,8,9,10,11,0.</p>
<p>And I thought, well, if I want to calculate how the expected value would be affected by rolling an extra dice for every 12, I could use the expected value of a single die when re-rolling with every 12, and put in on the table.</p>
<p>So I created a table with the rows numbered like this: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, (11+6.091)</p>
<p>The first columns are:</p>
<p> 1 2 3<br />
0 0.00000000 0.00000000 0.00000000<br />
1 0.08333333 0.02083333 0.00405093<br />
2 0.16666667 0.06944444 0.02199074<br />
3 0.25000000 0.14583333 0.06423611<br />
4 0.33333333 0.25000000 0.14120370<br />
5 0.41666667 0.38194444 0.26331019<br />
6 0.50000000 0.54166667 0.44097222<br />
7 0.58333333 0.72916667 0.68460648<br />
8 0.66666667 0.94444444 1.00462963<br />
9 0.75000000 1.18750000 1.41145833<br />
10 0.83333333 1.45833333 1.91550926<br />
17.091 1.42425000 2.72981250 3.92657813<br />
Exp. value 6.00758333 8.45897917 9.87854572</p>
<p>I noticed that the expected value when increasing the number of dice was capped at aprox. 17, which did not feel right. I am sure that rolling 2000 dice and then rolling again as many dice as 12s, and then again, and again, etc, would almost always grant a value much higher than 17, so I made a change in the formula and decided to use the expected value of each column as the number to add to "11" on the next column. This created a table without any asymptote, which felt much more realistic.</p>
<p>These are the first columns of this table:</p>
<p> 1 2 3<br />
0 0.00000000 0.00000000 0.00000000<br />
1 0.08333333 0.02083333 0.00405093<br />
2 0.16666667 0.06944444 0.02199074<br />
3 0.25000000 0.14583333 0.06423611<br />
4 0.33333333 0.25000000 0.14120370<br />
5 0.41666667 0.38194444 0.26331019<br />
6 0.50000000 0.54166667 0.44097222<br />
7 0.58333333 0.72916667 0.68460648<br />
8 0.66666667 0.94444444 1.00462963<br />
9 0.75000000 1.18750000 1.41145833<br />
10 0.83333333 1.45833333 1.91550926<br />
17.091 1.42425000 2.71648900 4.46754936<br />
Exp. value 6.00758333 8.44565567 10.41951696</p>
<p>As a comparison, when using a constant 17.091 as highest row, the expected highest value when rolling 10 dice is 13.87967758, and if we use the expected value of the previous roll + 11, the expected value for 10 dice is 22.16154432, a very large difference.</p>
<p>Now, I am sure that this is probably not the most accurate way to calculate the values. But, how inaccurate would you say it is? Enough as to disregard the values completely?</p>
<p>Also, I still have not found any way to introduce the "1s cancel highest values" condition.</p>
<p>Anyway, I keep on researching. Thanks a lot for your help so far, guys!</p>
Thu, 21 Aug 2014 14:27:57 +0000apertotescomment 71810 at http://www.bgdf.comX3M wrote:Quote:No, no...
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71799
<p><div class="quote-msg"><div class="quote-author"><em>X3M</em> wrote:</div><div class="quote-msg"><div class="quote-author">Quote:</div><br />
No, no... the 5.5 is correct because when n=0 this is very different of the simple average. It is a different thing... the only real value of X is when n is infinite... in lower numbers the estimation is very poor and isnt the case for the simple average without reroll (that is 6.5).<br />
</div></p>
<p>I suspected as much that that 5.5 had notching to do with 6.5. :D Although, I couldn't see it yet. But now I do.</p>
<p>Perhaps it is better to post more tables with answers than just the formula's. After all, almost no one understands those formula's. Even some one like me (even though I know what you have posted) has trouble calculating with them if I can't use my calculator ^^.</div></p>
<p>Mmm... I dont think so because we are designing games... so we need formulas to fine-tune probabilities.</p>
<p>With formula I can setup variables faster to a concrete probability distribution. Variables are number of dice, number of sides of them or just values non-homogeneous of them, etc... And I can see if is better to "reroll", "sum", "subtract" or "choose" on a dynamic, or see distribution for concrete values or a range of them (greater or equal than...)</p>
<p>And many of these things are very useful for card games too, so I think formula are better to know some details that are not visible when you design the mechanic of a game.</p>
<p>The formula are not hard but I put here completely terminated so you cant "understand" what means these variables or where comes all of this.</p>
<p>I created a general formula for the "sum and reroll" of a die or any number of die. But I will post on a blog or something cause is a bit more long but is simpler and more beautiful that any formula of this topic (the mechanic itself of "reroll and sum" makes the distribution different of a standard binomial but is very similar and is very useful for any conditional event).</p>
Thu, 21 Aug 2014 14:17:46 +0000Masacrosocomment 71799 at http://www.bgdf.comI doubt that has anything to
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71807
<p>I doubt that has anything to do with his question.<br />
Further, that one is one of the oldest analysed dice rolling, kids get that one to do in school.<br />
However, the formula does look interesting.</p>
Thu, 21 Aug 2014 11:28:00 +0000X3Mcomment 71807 at http://www.bgdf.comLOL, see this fucking shit!!!
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71802
<p>LOL, see this fucking shit!!! This is what I was searching to calculate the probability of a sum on a throw of dice:</p>
<p><a href="http://mathworld.wolfram.com/Dice.html" title="http://mathworld.wolfram.com/Dice.html">http://mathworld.wolfram.com/Dice.html</a></p>
Thu, 21 Aug 2014 08:06:30 +0000Masacrosocomment 71802 at http://www.bgdf.comMasacroso wrote:[
2) Rolling
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71784
<p><div class="quote-msg"><div class="quote-author"><em>Masacroso</em> wrote:</div>[</p>
<p>2) Rolling _n_ dice and taking the maximum value for all of them (e.g.: rolling 3 D6 I get 4, 5 and 1... so the value of the play is 5 because is the maximum of all). On this topic _n_ means number of dice you throw.<br />
</div></p>
<p>But isn't that the formula I wrote several messages above?</p>
<p>12-(1^n+...+11^n)/12^n -------> this is for 12-sided die, write K instead of 12 and (K-1) instead of 11 to know the general case for fair dice of any number of sides.</p>
Wed, 20 Aug 2014 13:50:20 +0000apertotescomment 71784 at http://www.bgdf.comQuote:
No, no... the 5.5 is
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71779
<p><div class="quote-msg"><div class="quote-author">Quote:</div><br />
No, no... the 5.5 is correct because when n=0 this is very different of the simple average. It is a different thing... the only real value of X is when n is infinite... in lower numbers the estimation is very poor and isnt the case for the simple average without reroll (that is 6.5).<br />
</div></p>
<p>I suspected as much that that 5.5 had notching to do with 6.5. :D Although, I couldn't see it yet. But now I do.</p>
<p>Perhaps it is better to post more tables with answers than just the formula's. After all, almost no one understands those formula's. Even some one like me (even though I know what you have posted) has trouble calculating with them if I can't use my calculator ^^.</p>
Wed, 20 Aug 2014 11:36:27 +0000X3Mcomment 71779 at http://www.bgdf.comapertotes wrote:Masacroso
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71774
<p><div class="quote-msg"><div class="quote-author"><em>apertotes</em> wrote:</div><div class="quote-msg"><div class="quote-author"><em>Masacroso</em> wrote:</div></p>
<p>I dont follow you here guy. The applet that I post on commentary #11 is just for when you throw only ONE dice with the rule of reroll-and-sum-when-max. For a D12 de average value is about 7.1, for a D6 is about 4.2 and for a D20 is about 11.1.</p>
<p>For the case of more than one dice you need another formula. Maybe I will develop something related to this because seems interesting for fine tunning probabilities.</div></p>
<p>Ah! Now I understand! I thought n was the number of dice rolled, but it is the layers of repetitions for repeated 12s, right?</div></p>
<p>Yes but you are mixing topics... I put basically 2 formulas:</p>
<p>1) For throw ONE dice and reroll and sum if you get the maximum value (this is the gif). On this topic <em>n</em> dont means something real... is just a abstract number to order layers of probabilities. It doesn't means rolls or dice. The layers are partials calculus over different number of rerolls.</p>
<p>When n=0 there is the partial value to the average without taking in account any reroll (layer zero). For n=1 there are the values that are added to layer 0 for one reroll (without taking in account possible more rerolls). For n=2 for 2 rerolls, etc...</p>
<p>But any number of layers but infinity can show us real values of the average, but it falls so fast to numbers completely negligible so you can get easily a very good and useful estimation.</p>
<p>2) Rolling <em>n</em> dice and taking the maximum value for all of them (e.g.: rolling 3 D6 I get 4, 5 and 1... so the value of the play is 5 because is the maximum of all). On this topic <em>n</em> means number of dice you throw.</p>
<p>I created some formulas here to know the probability of get a specific value or the average value for the mechanic. Maybe more interesting to know the cumulative probability for some value (taking a value <em>greater or equal to</em>) but this is easy to track (you only must sum some individual probabilities).</p>
<p>3) I have for you (or me, or anyone :p) two more formula that maybe interesting. One to know the average of a throw of any amount of dice, and the other is a generalization of the throw discussed on 1): the reroll and sum of a dice if you get a value <em>V</em> or higher, by example taking a 4, a 5 or a 6 on a D6 (instead of just a 6).</p>
<p>But I will show this in another moment.</p>
Wed, 20 Aug 2014 07:58:38 +0000Masacrosocomment 71774 at http://www.bgdf.comMasacroso wrote:
I dont
http://www.bgdf.com/forum/game-creation/mechanics/dice-probabilities-0#comment-71765
<p><div class="quote-msg"><div class="quote-author"><em>Masacroso</em> wrote:</div></p>
<p>I dont follow you here guy. The applet that I post on commentary #11 is just for when you throw only ONE dice with the rule of reroll-and-sum-when-max. For a D12 de average value is about 7.1, for a D6 is about 4.2 and for a D20 is about 11.1.</p>
<p>For the case of more than one dice you need another formula. Maybe I will develop something related to this because seems interesting for fine tunning probabilities.</div></p>
<p>Ah! Now I understand! I thought n was the number of dice rolled, but it is the layers of repetitions for repeated 12s, right?</p>
Wed, 20 Aug 2014 03:24:46 +0000apertotescomment 71765 at http://www.bgdf.com