Board Game Designers Forum - Comments for "#!@*?# Math"
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Comments for "#!@*?# Math"en#!@*?# Math
https://www.bgdf.com/forum/archive/archive-game-creation/game-design/math#comment-60831
<p></p><p>Thanks again guys; this is exactly the info. I needed. Much appreciated.</p>
Thu, 27 Jul 2006 15:38:42 +0000Carloscomment 60831 at https://www.bgdf.com#!@*?# Math
https://www.bgdf.com/forum/archive/archive-game-creation/game-design/math#comment-60830
<p></p><p><div class="bb-quote"><b>Carlos wrote:</b><blockquote class="bb-quote-body"><br />
So LSJ, the reason why I limited my question to column A was because the other columns (except G) are much less likely to allow a winning condition in the first hand--or so I thought. After seeing your analysis, I'm wondering if the possibility of getting an immediate winning hand in the other columns, especially B and F, might warrant their own statistical investigation.<br />
</blockquote></div></p>
<p>B has a probability of 1.28% of being instantly winable. Once every 78 draws or so.</p>
Wed, 26 Jul 2006 23:54:37 +0000Sebastiancomment 60830 at https://www.bgdf.com#!@*?# Math
https://www.bgdf.com/forum/archive/archive-game-creation/game-design/math#comment-60829
<p></p><p><div class="bb-quote"><b>Carlos wrote:</b><blockquote class="bb-quote-body">Epigone (btw, that's a sad moniker! you are not an epigonist, but a creator of original games! :) )</blockquote></div></p>
<p>After reading Sebastian's and Epigone's post that puts the chance at getting 7 or more points for the A column in the first column in the first draw of the game at 6%, I think I need to shrink the hand size. That just seems too lop-sidedly high.<br />
I use it in the sense of "standing on the shoulders of giants". It's good enough for Newton. :)</p>
<p>For reference, the probs for different hand sizes are:<br />
2: 0.06%<br />
3: 0.23%<br />
4: 0.53%<br />
5: 0.99%<br />
6: 1.63%<br />
7: 2.48%<br />
8: 3.55%<br />
9: 4.86%<br />
10: 6.4%<br />
11: 8.18%</p>
<p>Could the 4x cards just be too overpowered? They're easily several times as good as the variety-pack cards, since you should be able to get four times the effect from them for only a small versatility hit. Imagine a game where one player gets 5+ of the 4x cards and 5+ of the 3x cards throughout the game... this should happen uncomfortably often, does it lead to a pretty much guaranteed win?</p>
Wed, 26 Jul 2006 20:23:13 +0000Epigonecomment 60829 at https://www.bgdf.com#!@*?# Math
https://www.bgdf.com/forum/archive/archive-game-creation/game-design/math#comment-60828
<p></p><p>Thanks for helping me on this everyone.</p>
<p>Epigone (btw, that's a sad moniker! you are not an epigonist, but a creator of original games! :) ) and LSJ, what Emphyrio was alluding to in his post about varying lengths of columns was this less-than-lucid part of my post:</p>
<p><div class="bb-quote"><b>Carlos wrote:</b><blockquote class="bb-quote-body">The first column has 8 squares that must be traversed, the starting square plus seven others; the 2nd 10; the 3rd 12; the 4th 14; the 5th 12; the 6th 10; and the 7th 8. </blockquote></div></p>
<p>So to put in more clearly, let's assign each of the seven rows a letter, A-G; I'll put the length of each row next to each letter. Remember, the length of each includes the starting square:</p>
<p>A=8<br />
B=10<br />
C=12<br />
D=14<br />
E=12<br />
F=10<br />
G=8</p>
<p>So LSJ, the reason why I limited my question to column A was because the other columns (except G) are much less likely to allow a winning condition in the first hand--or so I thought. After seeing your analysis, I'm wondering if the possibility of getting an immediate winning hand in the other columns, especially B and F, might warrant their own statistical investigation.</p>
<p>But there are other factors in play. Every card has a total of four points on it, represented by the mix of symbols for the seven suits. So, for instance, there is an A A A A card, but there is also a B D E G card. And so the more cards that are used toward the conquering of one column, the less that are available to defend against your opponent's assaults. That's why I was initially more concerned with the outer columns and the 4-point and 3-point cards: there's little you can do with those cards except play them on the appropriate column. For cards with a mix of symbols, though, you have to choose which column you should play it, and choosing one will leave you more vulnerable on another.</p>
<p>After reading Sebastian's and Epigone's post that puts the chance at getting 7 or more points for the A column in the first column in the first draw of the game at 6%, I think I need to shrink the hand size. That just seems too lop-sidedly high.</p>
<p>Once more, thanks to all of you math wizzes for helping me and my math-addled brain out!</p>
<p>--Carlos</p>
Wed, 26 Jul 2006 18:58:44 +0000Carloscomment 60828 at https://www.bgdf.com#!@*?# Math
https://www.bgdf.com/forum/archive/archive-game-creation/game-design/math#comment-60827
<p></p><p>If you want to analyze it for any color (possibly more than one at once), rather than just a single color, the solution is about 3 of every 10 hands. </p>
<p>Analysis follows:</p>
<p>Success means getting a single-color sum of 7 or more in a hand of 10.</p>
<p>Probability = #successful hands / # hands</p>
<p># hands is the easy part: C(56,10) = 35607051480<br />
That's number of ways to choose 10 things from 56 things.<br />
C(n,k) = n!/(k! * (n-k)!)</p>
<p># successful hands is the tricky part, especially since a hand that fails in getting to 7 in A may get to 7 in B.</p>
<p>First, calculate the prob of NOT getting a match given a set # of cards of a single color. For instance, if you get 3 cards of color A, what is the prob of not getting a 7 sum in A?</p>
<p>FailColor(N) = prob of missing 7 with N cards of a single color.</p>
<p>FC(8) = 0<br />
FC(7) = 0<br />
FC(6) = 0<br />
FC(5) = 2 / C(8,5) = 2/56 = 1/28<br />
FC(4) = (#1111 + #2111 + #2211 + #3111) / C(8,4)<br />
= (1 + 8 + 6 + 4) / 70 = 19/70<br />
FC(3) = 1 - ((#431 + #432 + #421 + #422) / C(8,3))<br />
= 1 - ((4 + 2 + 8 + 1) / 56) = 41/56<br />
FC(2) = 1 - (#43 / C(8,2)) = 1 - (1/28) = 27/28<br />
FC(1) = 1<br />
FC(0) = 1</p>
<p>Find out how likely it is to get exactly a particular pattern of color (sorting each hand by largest group of color to smallest group). For instance, to get a hand that looks like AAAAAAAABB, you have to pick a color to be A (7 choices) and a color to be B (6 choices). For A, you have to pick all 8 cards of that color (1 way to do that). For B, you have to pick 2 of the 8 = C(8,2) = 28 ways. So 7*6*28 = 1176 hands that match the pattern.</p>
<p>With this hand, you have FC(8)*FC(2) = 0*(27/28) = 0 probability of missing, or a prob 1 of success. Which means all 1176 hands of AAAAAAAABB are successful.</p>
<p>Shortcutting, we notice that all AAAAAAAAXX, AAAAAAAXXX, and AAAAAAXXXX hands are successful (with X being any color but A, to avoid overcounting), from the table of FC(N) values. The number of such hands is 7*C(8,8)*C(48,4) + 7*C(8,7)*C(48,3) + 7*C(8,6)*C(48,2) = 7*1*194580 + 7*8*17296 + 7*28*1128 = 2551724 (which includes the 1176 above).</p>
<p>#AAAAABBBBB = C(7,2)*C(8,5)*C(8,5) = 65856 (Note we use C(7,2) and not 7*6, since it A=red,B=green is the same hand as A=green,B=red).<br />
Of these, FC(5)*FC(5) fail, so 1-FC(5)*FC(5) = 1-(1/28)(1/28) = 783/784 succeed. 65856 * 783/784 = 65772 successful hands</p>
<p>For AAAAABBBBC, we have 7*C(8,5)*6*C(8,4)*5*C(8,1) * (1-FC(5)*FC(4)*FC(1)) = 6521760 successful hands.</p>
<p>Successful hands in patterns:<br />
AAAAAA??XX= 2551724<br />
AAAAABBBBB = 65772<br />
AAAAABBBBC = 6521760<br />
AAAAABBBCC = 17974740<br />
AAAAABBBCD = 82091520<br />
AAAAABBCCD = 142618560<br />
AAAAABBCDE = 325570560<br />
AAAAABCDEF = 61931520<br />
AAAABBBBCC = 13382565<br />
AAAABBBBCD = 82355616<br />
AAAABBBCCC = 19696005<br />
AAAABBBCCD = 596245440<br />
AAAABBBCDE = 675440640<br />
AAAABBCCDD = 162772820<br />
AAAABBCCDE = 206786160<br />
AAAABBCDEF = 1244651520<br />
AAAABCDEFG = 93585408<br />
AAABBBCCCD = 119498400<br />
AAABBBCCDD = 258966750<br />
AAABBBCCDE = 213801840<br />
AAABBBCDEF = 625766400<br />
AAABBCCDDE = 945972160<br />
AAABBCCDEF = 3013785600<br />
AAABBCDEFG = 634454016<br />
AABBCCDDEE = 60090681<br />
AABBCCDDEF = 559204800<br />
AABBCCDEFG = 325283840</p>
<p>Total successful hands: 10491066817</p>
<p>Odds of getting such a hand: 10491066817/35607051480 = 29.46%</p>
Wed, 26 Jul 2006 18:13:14 +0000LSJcomment 60827 at https://www.bgdf.com#!@*?# Math
https://www.bgdf.com/forum/archive/archive-game-creation/game-design/math#comment-60826
<p></p><p><div class="bb-quote"><b>Emphyrio wrote:</b><blockquote class="bb-quote-body">Since you have two columns with 8 spaces, the upper bound is actually about twice as much. Then you also have to figure the chances you might get cards adding up to 9 for columns B and F, and so on... By this time your upper bound is becoming less and less useful.</blockquote></div></p>
<p>Another alternative would be to write a program to enumerate all 3.5 billion possible hands (the binomial coefficient of 56, 10) and count how many of them fulfill the criteria. If I get a chance I'll try it.<br />
It looks like you have information we don't... 9 for columns B and F, etc.?</p>
<p>For just column A you don't need to enumerate all of those hands, just... 60. And then multiply by the probabilities of getting those hands. I get the same result Sebastian got, 2279313410/35607051480 = 6.4%.</p>
Wed, 26 Jul 2006 17:56:52 +0000Epigonecomment 60826 at https://www.bgdf.comRe: #!@*?# Math
https://www.bgdf.com/forum/archive/archive-game-creation/game-design/math#comment-60825
<p></p><p><div class="bb-quote"><b>Carlos wrote:</b><blockquote class="bb-quote-body">Is there too great a probability that in YOUR FIRST HAND you will draw enough cards FOR COLUMN A that add up to a value of 7, thereby guaranteeing you a victory on that row?</blockquote></div></p>
<p>There is a 6% chance that this will occur. That is, with properly shuffled cards, it will come up every fiveteenth or sixteenth game.</p>
Wed, 26 Jul 2006 13:27:45 +0000Sebastiancomment 60825 at https://www.bgdf.com#!@*?# Math
https://www.bgdf.com/forum/archive/archive-game-creation/game-design/math#comment-60824
<p></p><p><div class="bb-quote"><b>Emphyrio wrote:</b><blockquote class="bb-quote-body"><br />
Another alternative would be to write a program to enumerate all 3.5 billion possible hands (the binomial coefficient of 56, 10) and count how many of them fulfill the criteria. If I get a chance I'll try it.</blockquote></div></p>
<p>I love you.</p>
<p>I mean ... thanks! I'd be really grateful! :)</p>
<p>Thanks to for the responses so far.</p>
Wed, 26 Jul 2006 12:47:05 +0000Carloscomment 60824 at https://www.bgdf.com#!@*?# Math
https://www.bgdf.com/forum/archive/archive-game-creation/game-design/math#comment-60823
<p></p><p>I'm not a probability expert, but here's how I'd analyze the problem:</p>
<p>You have a 10 in 56 chance of drawing A4, and if you do, there is a 9 in 55 chance of drawing A3. So your chance of drawing A4 + A3 = (10*9)/(56*55) = about 2.9%.</p>
<p>Some are a bit trickier. For example, A4 + A2 + A1 you have 10/56 chance to draw A4, but since there are two A2 cards, your chance of drawing at least one is 1 - chance of not drawing either, which is 1 - (53/55 * 52/54 * 51/53 * ... * 45/47) = 30.3%. I'm sure there's an easier way to compute this. Then your chance of drawing at least one A1 is 1 - (50/54 * 49/53 * 48/52 * ... * 43/47) = 48.4%. So your total chance of drawing A4 + A2 + A1 = 10/56 * .303 * .484 = 2.6%.</p>
<p>Finally, you need to combine all these separate probabilities. These are not independent probabilities, so strictly speaking you can't just add them up. However, they're small enough that you can get an upper bound on the chances by adding them. Since you have two columns with 8 spaces, the upper bound is actually about twice as much. Then you also have to figure the chances you might get cards adding up to 9 for columns B and F, and so on... By this time your upper bound is becoming less and less useful.</p>
<p>Another alternative would be to write a program to enumerate all 3.5 billion possible hands (the binomial coefficient of 56, 10) and count how many of them fulfill the criteria. If I get a chance I'll try it.</p>
Wed, 26 Jul 2006 06:14:47 +0000Emphyriocomment 60823 at https://www.bgdf.com#!@*?# Math
https://www.bgdf.com/forum/archive/archive-game-creation/game-design/math#comment-60822
<p></p><p>I'm not sure about the math for the answer (in haste I have a guess, below), but if it turns out you do (or even if you don't), you could vary the values for each suit. Like for suit A you could have one 3, two 2's and five 1's or something.</p>
<p>Here's a guess...</p>
<p>So there are 15 total points in each color (4+3+2+2+1+1+1+1), and 7 colors, so there are 15*7=105 points total.</p>
<p>In your opening draw you will maybe have an average of 10/56 of those points in hand, or 18 to 19 points.</p>
<p>Since the cards are evenly distributed over the suits, then on the average you should have say 19/7=2 or 3 points per color.</p>
<p>Now, that's an average, so how often will you have significantly more than that in one color and less in another, and how oftn will that color be suit A? I don't know.</p>
<p>Finally, you could add a rule that says players dealt 5 or more points in suit A must declare as much, reveal their hand, and re-draw a new hand. It's wonky, but it might work if it only happens once in a while.</p>
Wed, 26 Jul 2006 05:35:28 +0000sedjtrollcomment 60822 at https://www.bgdf.com