Board Game Designers Forum - Comments for "dice probability question (Can&#039;t Stop)"
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop
Comments for "dice probability question (Can't Stop)"enYes, moving twice on the same
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89834
<p>"For 4, 2, 2, you cannot move track 6 twice. But you could move track 4 twice, and I count that in my method (OP did not say otherwise I believe)."</p>
<p>Correct assumption- moving twice on the same track is allowed for what I have in mind so my movie example is flawed since I should have mentioned you can see the same movie twice if one wanted to.</p>
Wed, 08 Mar 2017 23:25:32 +0000NewbieDesignercomment 89834 at https://www.bgdf.comAlso hangs head in shame
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89832
<p>Well, I guess I am almost there now. By simply counting. And while most looks the same. A couple of numbers are different. I also looked if I had asymmetry, which should not be the case with this problem. A, for example, 6+6 has been counted twice though. While 9+3 and 3+9 was corrected.</p>
<p>I have spotted the 36, 30 and 25 now. It happens 6 times indeed for the single die.</p>
<p>This is the list that I have currently:</p>
<p>1-91<br />
2-107<br />
3-121<br />
4-137<br />
5-151<br />
6-167<br />
7-90<br />
8-76<br />
9-60<br />
10-46<br />
11-30<br />
12-16</p>
<p>A total of 1092</p>
<p>Obviously, 2 to 6 are different from your value's. I compared the two. And there is a neat little linear difference of 3, 6, 9, 12 and 15. So I was literately pulling my hair out right now. But then I also removed any double, like the example. Thus not moving twice on the same track.</p>
<p><strong>I got the same list and bow down.</strong></p>
Wed, 08 Mar 2017 22:50:16 +0000X3Mcomment 89832 at https://www.bgdf.comThe track choice doesn't
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89825
<p>The track choice doesn't matter (I hope). It is the stipulation that you have to take one sum and leave one dice alone.</p>
<p>For 4, 2, 2, you cannot move track 6 twice. But you could move track 4 twice, and I count that in my method (OP did not say otherwise I believe).</p>
<p>If OP doesnt want to allow the same track moved twice, even by two methods, then you would need to remove on rare occasion a few counts from the statistics I gave, but it wouldnt have a very significant effect as this is an overwhelmingly rare situation. </p>
<p>EDIT:<br />
NOPE. I did throw that out in the case of the third die having value between 2 thru 6, AND matching the value wanted. Went a bit too fast there, sorry.</p>
<p>So for row 2 through 6, if there is a value of 36, it needs to have added to it the following values:</p>
<p>2: 1<br />
3: 2<br />
4: 3<br />
5: 4<br />
6: 5</p>
<p>Its not a huge change to the statistics, the final tally of permutations is now 1062.</p>
Wed, 08 Mar 2017 21:51:02 +0000Daggazcomment 89825 at https://www.bgdf.comIm too tired to explain this
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89824
<p>Im too tired to explain this clearly, so I will go to your example.</p>
<p>Given<br />
6 3 3</p>
<p>6 6 Both of these would be counted by my method, assuming you want a 6</p>
<p>9 3<br />
9 3 Only one of these would be counted, this is a degenerate result. This is degenerate whether you want a 9, or a 3, because in either situation, you can only take one set of either one single die, or one pair of die. NOTE that you do count the nine for one outcome (I want a nine), and you do count the 3 for another (I want a three), but you do NOT count two nines, or two threes. </p>
<p>The difference is that in the first situation, even though there are two numbers that are equal, you get them by different methods. Whereas in the second example, the outcome is duplicated, and this happens because you pick the same method twice, using the duplicated primary dice. Duplicated outcomes are known in combinatorics as "degenerate" and you dont count them twice unless there is a specific reason to do so (one such reason comes up in the physics of electron energy levels, but thats another topic).</p>
<p>The odds I give are for a single theater showing up in a given cast of three dice by either method (single or sum) allowed. If you want the odds of the second theater, GIVEN that the first theater is chosen, then it gets yet again more complicated and at this point I really dont feel like running that out. But it's also besides the point. In the design of the tracks as OP describes, he will only be concerned with the primary chance of any one theater showing up. The secondary statistic is simply the primary statistic divided amongst the allowable neighbors.</p>
Wed, 08 Mar 2017 21:36:57 +0000Daggazcomment 89824 at https://www.bgdf.comOops
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89823
<p><div class="quote-msg"><div class="quote-author"><em>Daggaz</em> wrote:</div>Hey Frank, </p>
<p>Im not sure without looking at your sheet and understanding your method where it went wrong, but the total percentages have to add up to 100% as a first thing.</div><br />
There were two outcomes per state, but I only divided by the 216 states. That made everything sum up to 200%.</p>
<p>* hangs head in shame *</p>
<p>It wasn't clear to me what to make of, for example, 4 and 2+2. If the player would be allowed to move that track's pawns <em>twice</em>, double-counting is perfectly appropriate. If the player is prohibited from picking the same track twice, then the second 2~6 should be excluded.</p>
Wed, 08 Mar 2017 21:00:28 +0000FrankMcomment 89823 at https://www.bgdf.comDaggaz wrote:I discard
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89822
<p><div class="quote-msg"><div class="quote-author"><em>Daggaz</em> wrote:</div>I discard anything that has already been counted, because the rules do not give you extra points for rolling your value twice. You can only use it once. In other words, if a given permutation has multiple paths to getting the desired outcome (many of them do), you only count that permutation once.</div>Where did he say that? I was under the impression that if you rolled for example 6, 3, 3. That you could turn that into:<br />
6 and 6, 9 and 3 and another 9 and 3.</p>
<p>Regarding the 6 and 6, they both can be used, stated by his rules. Right? So that counts twice. But....</p>
<p>Are you saying that this 9 and 3, which can be created twice, is counted only once? Because, then I finally get where you where at.</p>
<p>I'll look into my table once more. Since I got many of those doubles indeed. I am going to (dis)count them, literly. So please be patient on this one :)</p>
Wed, 08 Mar 2017 20:50:03 +0000X3Mcomment 89822 at https://www.bgdf.comI discard anything that has
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89821
<p>I discard anything that has already been counted, because the rules do not give you extra points for rolling your value twice. You can only use it once. In other words, if a given permutation has multiple paths to getting the desired outcome (many of them do), you only count that permutation once.</p>
<p>Regarding 1296, you dont get this number because not all values have a choice. You can ONLY roll a 12 with two dice. You can ONLY roll a 1 with one. So you dont get to double the entire spread.</p>
Wed, 08 Mar 2017 20:11:57 +0000Daggazcomment 89821 at https://www.bgdf.comIm not sure without looking
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89816
<p>Hey Frank, </p>
<p>Im not sure without looking at your sheet and understanding your method where it went wrong, but the total percentages have to add up to 100% as a first thing.</p>
<p>Another issue, which is why I went through each matrix individually, is that the overlap changes depending on the value. Some values are easy to hit because there are lots of combinations, and you can choose to either hold or add. 1 is tricky because you simply have to roll it outright (notice that my counting method returns the same number as my calculation : 91), 7 and higher you have to achieve with two dice. </p>
<p>If you count it out by hand, you can see how this overlap evolves as you mark the rows and columns and look for the extra hits. </p>
<p>Again with respect to 1: notice how I have 36, 11 11 11 11 11. These are the base odds for rolling a value outright with one dice. That matches what you did. But as soon as you move to 2 or higher, this changes. These odds underlie the results for 2 through 6, but no longer contribute to 7 and higher (with the caveat that the 36 moves across the row as the desired number grows).</p>
Wed, 08 Mar 2017 20:05:58 +0000Daggazcomment 89816 at https://www.bgdf.comNo need to become aggresive
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89820
<p>Just trying to understand the way how you think. That is all. I can learn from that. Either way.</p>
<p>When looking at that last table, I got that impression. It looks a bit vague to me. 1 to 12 by 1 to 6.</p>
<p>I have written out for the numbers that I got, just to be sure. A list of 216 (rolling 6*6*6). Also looking at the 3 arrangements (A,B+C/B,A+C/C,A+B). Which gives a total of 648. By the rules that NewbieDesigner has given, thus using both value's. We get a total of 1296.<br />
And I was simply counting each value, 1 to 12. Really, counting.</p>
<p>When using permutations, you are discarding some numbers. This is what I don't understand. When AND where do you discard, what you have rolled? I already offered to email the excel. You can point out where I went wrong?</p>
<p>You don't have to believe me. Lets agree that we disagree.</p>
Wed, 08 Mar 2017 20:04:56 +0000X3Mcomment 89820 at https://www.bgdf.comNo problem. It was a fun
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89819
<p>No problem. It was a fun problem and now I will have to try not to solve it explicitly (math) and instead finish my manuscript. Good luck on your game and don't be afraid to ask questions.</p>
Wed, 08 Mar 2017 20:01:01 +0000Daggazcomment 89819 at https://www.bgdf.comDude... you don't honestly
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89813
<p>X3M, your response is rather condescending, which is made all the worse by the fact that you clearly don't understand what I did. I explained the method earlier in this thread and you didnt seem to have a problem with it, so you must have lost track. I will explain it explicitely now.</p>
<p>FIRST I take 3 dice, A B C. I put two of them A and B into a summation matrix, and the third C I hold on the outside. PLEASE NOTE: This does NOT mean that I add the first two, the matrix is simply a tool to keep track of the possible summation values. Its a matrix after all, by itself it does not represent a singular quantity.</p>
<p>THEN I iterate the third dice. First 1. Then 2. Then 3...etc. up to 6</p>
<p>THEN for EACH iteration, I look at the wanted values, 1 through 12, and count the possibilities using the matrix. ALL of the possibilities. </p>
<p>EXAMPLE</p>
<p>IF you have a 3 on the third dice, and you need to get a final value of 4, you can either roll a 4 on either of dice A or B (mark the 4 column and the 4 row), OR you can roll a 1 on either dice A or B which you add to dice C (mark the 1 column and 1 row), OR you can roll A = 2 B = 2 for one more hit. Add all of these possibilities together, making sure not to count anything twice. You get 21. Again, if this is not clear, this includes ALL possible combinations of the three dice which result in 4. Now look at the chart I posted. Third Dice = 3, Wanted Value = 4, Returns 21</p>
<p><div class="quote-msg"><div class="quote-author"><em>X3M</em> wrote:</div>The first 2 dice are added up. They can't count as a 1.</div><br />
Dude... you don't honestly think I added two dice to get a one do you? </p>
<p><div class="quote-msg"><div class="quote-author"><em>X3M</em> wrote:</div>Adding up asks for 36 possibilities. Then you are stuck with 36 rolls on 1 dice. That is how I get my 6 times a 1.</div> Im not sure how 6 * 1 has anything to do with 36, you are going to need to explain this leap of logic a bit more clearly. Especially the part about 36 rolls on 1 dice, because that never happens. Not even equivalently.</p>
<p><div class="quote-msg"><div class="quote-author"><em>X3M</em> wrote:</div>When everyone doubts:<br />
I got a table in excel with all the numbers. (including the list of FrankM). Can I email it to you, so that you can take a look?</div></p>
<p>Now you are not only rude, but you are directly contradicting yourself. If you are so sure of your numbers, why have you not posted them (I did), and why are you asking FrankM to check your work privately in the same message?</p>
<p>If I sound pissed off, it's because I spent time in my busy schedule today to solve this problem instead of working, and while I welcome legitimate criticism quite openly, yours is anything but. And to top it off you are simply wrong.</p>
Wed, 08 Mar 2017 19:58:03 +0000Daggazcomment 89813 at https://www.bgdf.comThanks for confirming.
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89818
<p>Thanks for confirming.</p>
Wed, 08 Mar 2017 19:57:42 +0000NewbieDesignercomment 89818 at https://www.bgdf.comNewbieDesigner wrote:Thanks
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89817
<p><div class="quote-msg"><div class="quote-author"><em>NewbieDesigner</em> wrote:</div>Thanks Daggaz. Another question regarding the percentages in case I am missing something obvious or asking a new question. Based on my movie question above about rolling three dice (locking two) and choosing two theatres based on the results. Are your most recent results saying that I will be able to choose theatre six 14.51% of the time I roll the 3 dice (for comparison sake, I would be able to choose theatre six 14% of the time with just rolling two dice)? </p>
<p>That strikes me as low intuitively based on the 3 dice selection so I'm assuming I'm misunderstanding something (or perhaps it is that low) or asking a different question.</div></p>
<p>The results say that you will get to choose theater six 14.51% of the time if you roll three dice.</p>
<p>If that seems low to you, remember that you are also adding lots of possibilities to the other results as well, and the final probability is the number of results you want divided by the total number of possibilities. </p>
<p>Look at the entire distribution from 1 to 12 and you can see that it is very skewed. You added a new value 1, and the distribution is no longer centered perfectly on 7.</p>
Wed, 08 Mar 2017 19:53:49 +0000Daggazcomment 89817 at https://www.bgdf.comWhere my table came from
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89814
<p>I had counted everything from a 216-row exhaustive list of outcomes using Excel, though the results make sense this way: for A there are six possibilities and each one spans a full set of 36 outcomes for the other two dice... and for B+C there are 36 permutations (11 distinct sums) and each one spans a full set of 6 outcomes for the other die. That's why everything in the first column is a multiple of 36 and everything in the second column is a multiple of 6.</p>
Wed, 08 Mar 2017 19:33:20 +0000FrankMcomment 89814 at https://www.bgdf.comDaggaz
The first 2 dice are
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89808
<p>Daggaz<br />
The first 2 dice are added up. They can't count as a 1.<br />
Adding up asks for 36 possibilities. Then you are stuck with 36 rolls on 1 dice. That is how I get my 6 times a 1.</p>
<p>Please keep in mind that both value's are used at the same time. Thus A and B+C or B and A+C or C and A+B.</p>
<p>When everyone doubts:<br />
I got a table in excel with all the numbers. (including the list of FrankM). Can I email it to you, so that you can take a look?</p>
Wed, 08 Mar 2017 17:56:01 +0000X3Mcomment 89808 at https://www.bgdf.comThanks Daggaz. Another
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89801
<p>Thanks Daggaz. Another question regarding the percentages in case I am missing something obvious or asking a new question. Based on my movie question above about rolling three dice (locking two) and choosing two theatres based on the results. Are your most recent results saying that I will be able to choose theatre six 14.51% of the time I roll the 3 dice (for comparison sake, I would be able to choose theatre six 14% of the time with just rolling two dice)? </p>
<p>That strikes me as low intuitively based on the 3 dice selection so I'm assuming I'm misunderstanding something (or perhaps it is that low) or asking a different question.</p>
Wed, 08 Mar 2017 15:36:25 +0000NewbieDesignercomment 89801 at https://www.bgdf.comThe sub-permutation of 3
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89797
<p>The sub-permutation of 3 comes from the fact that you can pick either dice A, B, or C as the single die. Or {(A :(B+C)), (B:(A+C)), (C:(A+B))} </p>
<p>So you have 6*6*6*3 permutations at this point. But its not done there. </p>
<p>You have the final set of values you are trying to achieve, and you make a choice at this point if you want value X (the single die) or value Y (the summed dice) to achieve this final value. </p>
<p>The final set of values is 1 through 12. But as I mentioned, only (2,3,4,5,6) can be made by either a single die or a summed pair of dice. That leaves seven final values which cannot have a choice, and one of them can only be made with one die, while the other six must be made with the sum. </p>
<p>If every value had a legitimate choice between the two, then 6*6*6*3*2 = 1296 would be the maximum limit, but they dont, so the real number will be lower than this. I am tempted to say it is :</p>
<p>6*6*6*3*2*(5/12) + 6*6*6*3*(7/12) = 540 + 378 = 918. </p>
<p>But just looking at that it seems like there could be a mistake because of the complex overlap of the sub-sets. This just looks too easy. So lets actually count these. I drew out the summation matrix for 2d6, then laid transparent plastic strips over the rows and columns that could give me the target value with or without the third die, and then included any additional counts found on the matrix not covered by these strips, which are summations not involving the third die. This was done iterating six times for the possible values of the third die, see the top row.</p>
<p>.....01 02 03 04 05 06 ..... (this is the value of the third die)<br />
01 : 36 11 11 11 11 11 : 091 : 08.69%<br />
02 : 20 36 12 12 12 12 : 104 : 09.93%<br />
03 : 20 20 36 13 13 13 : 115 : 10.98%<br />
04 : 21 22 21 36 14 14 : 128 : 12.23%<br />
05 : 22 22 22 22 36 15 : 139 : 13.28%<br />
06 : 23 23 24 23 23 36 : 152 : 14.51%<br />
07 : 15 15 15 15 15 15 : 090 : 08.60%<br />
08 : 05 14 14 15 14 14 : 076 : 07.26%<br />
09 : 04 04 13 13 13 13 : 060 : 05.73%<br />
10 : 03 03 03 12 13 12 : 046 : 04.39%<br />
11 : 02 02 02 02 11 11 : 030 : 02.87%<br />
12 : 01 01 01 01 01 11 : 016 : 01.53%</p>
<p>value we want : the number of ways to get the wanted value given the third die : total number of ways to get this value: percent chance</p>
<p>Total permutations : 1047<br />
Sum of percentages : exactly 100% (amazing considering I rounded)</p>
<p>PS: If your probabilities sum to over 100%, you know immediately you have a problem. Total probability MUST be equal to 100%. If it is over, you have overcounted somewhere. If it is under, you are missing some outcomes or undervalued something. </p>
<p>I think this puts the nail in the coffin. Both regarding the answer, and the point that you really should work these out the long way rather than trying to calculate the answer, if you do not know exactly what you are doing and why you are doing it (I certainly didnt).</p>
Wed, 08 Mar 2017 14:02:31 +0000Daggazcomment 89797 at https://www.bgdf.comThanks for the additional
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89763
<p>Thanks for the additional info everyone.</p>
Tue, 07 Mar 2017 18:35:37 +0000NewbieDesignercomment 89763 at https://www.bgdf.comFrankM is right, regarding
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89762
<p>FrankM is right, regarding the chances (a total of 200%). I went with the minimal result. And forgot that fact, that the player is going to use 2 value's per roll.</p>
<p>Either way, the proportions seem to be the same.</p>
<p>@Daggaz<br />
I don't get how you got those permutations? Please elaborate.</p>
Tue, 07 Mar 2017 18:29:10 +0000X3Mcomment 89762 at https://www.bgdf.comNot sure I understand the question
https://www.bgdf.com/forum/game-creation/mechanics/dice-probability-question-cant-stop#comment-89756
<p>It sounds like the question is roll three dice, and you would like the distribution of outcomes from a single die and the sum of the other two.</p>
<p>There are <em>four</em> variables here: A, B and C are dice (where Nature chooses a value of 1 to 6 for each) and D (the player's decision of which die to stand alone). This gives you 6x6x6x3 = 648 total states.</p>
<p>But if you want the raw probabilities without any situation-specific strategic thinking by the player, as mentioned above you can simplify this by always picking A as the stand-alone die.</p>
<table>
<thead>
<tr>
<th>Result</th>
<th>A</th>
<th>B+C</th>
<th>Either</th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td>36</td>
<td>00</td>
<td>36 / 216 = 16.67%</td>
</tr>
<tr>
<td>2</td>
<td>36</td>
<td>06</td>
<td>42 / 216 = 19.44%</td>
</tr>
<tr>
<td>3</td>
<td>36</td>
<td>12</td>
<td>48 / 216 = 22.22%</td>
</tr>
<tr>
<td>4</td>
<td>36</td>
<td>18</td>
<td>54 / 216 = 25.00%</td>
</tr>
<tr>
<td>5</td>
<td>36</td>
<td>24</td>
<td>60 / 216 = 27.78%</td>
</tr>
<tr>
<td>6</td>
<td>36</td>
<td>30</td>
<td>66 / 216 = 30.56%</td>
</tr>
<tr>
<td>7</td>
<td>00</td>
<td>36</td>
<td>36 / 216 = 16.67%</td>
</tr>
<tr>
<td>8</td>
<td>00</td>
<td>30</td>
<td>30 / 216 = 13.89%</td>
</tr>
<tr>
<td>9</td>
<td>00</td>
<td>24</td>
<td>24 / 216 = 11.11%</td>
</tr>
<tr>
<td>10</td>
<td>00</td>
<td>18</td>
<td>18 / 216 = 08.33%</td>
</tr>
<tr>
<td>11</td>
<td>00</td>
<td>12</td>
<td>12 / 216 = 05.56%</td>
</tr>
<tr>
<td>12</td>
<td>00</td>
<td>06</td>
<td>06 / 216 = 02.78%</td>
</tr>
</tbody>
</table>
<p>If the plan is to make these track length proportional to the probability, you could divide the number of outcomes by 6. But whatever factor you use, the "6" track is always going to be 11 times the length of the "12" track.</p>
Tue, 07 Mar 2017 17:27:17 +0000FrankMcomment 89756 at https://www.bgdf.com