CraigD Posted February 10, 2013 Report Posted February 10, 2013 I was watching a science fiction anime series (Space Brothers – not recommended by me for any but the most hard-core and eclectic spaceflight fans, so slowly paces that NASA withdrew its approval of it on essentially the grounds that “real astronaut training isn’t this complicated and boring”), in which 5 people choose 2 from their numbers by playing a single round of rock-paper-scissors. It worked in one round, 2 throwing paper, 3 stone, giving 2 winners. The narrator expressed wonder and satisfaction, having expected that many rounds would be needed. I wondered how likely such an outcome is, so wrote this little MUMPS program to calculate its probability, assuming the 5 player follow a completely random strategy: n (XMRPSN,R) s A=R k R s R=A,N=+R,M=$p(A,",",2) s:'M M=3 f A=0:1:M**N-1 x XMRPSN(1,1) s R(C)=$g(R(C))+1 ;XMRPSN(1): returns R(#nonlosers)=count given R=#player,#plays s C=0 f B=1:1:N s AB=A\(M**B/M)#M,F=0 x XMRPSN(1,1,1) s:'F C=C+1 ;XMRPSN(1,1) f D=1:1:N s F=A\(M**D/M)#M-AB#M#2 q:F ;XMRPSN(1,1,1) S R=5 x XMRPSN(1) gives R(2)/(3**5)= 30/243 =~ 12.3%, showing that the depicted outcome is lucky, but not extraordinarily so.A look at these counts of the number of players not eliminated in a single round of play for different numbers of players N, N=2 0 6 3 N=3 6 9 9 3 N=4 36 12 18 12 3 N=5 150 15 30 30 15 3 N=6 540 18 45 60 45 18 3 N=7 1806 21 63 105 105 63 21 3 N=8 5796 24 84 168 210 168 84 24 3 Showed something that surprised me. Ignoring the first column (which shows the count of plays in which 0 players avoid elimination) and dividing by a common factor of 4, it’s a variation on Pascal’s triangle, where the cell-setting rule is the usual [math][x,y] = [x-1,y-1]+[x,y-1][/math], and 1 of the 2 edge cell-setting rules is the usual [math][y,y]=1[/math],but the other is, rather than the usual [math][1,y]=1[/math], [math][1,y]=y[/math] 1 2 1 3 3 1 4 6 4 1 5 10 10 5 1 6 15 20 15 6 1 7 21 35 35 21 7 1 ... Folk who socialize with the right sort of folk, or just watch the popular TV show about folk who do The Big Bang Theory, know that the rock/paper/scissor game can be played, and remain optimal-strategied with a evenly weighted random choice of play, for not just 3, but 5 possible plays. A bit of thinking (or a quick read of wikipedia) reveals that it is so for any properly handled odd number of possible plays. Adding a second argument to the above program shows the counts for some odd number of possible plays M >3 N=2,M=5 0 20 5 N=3,M=5 30 60 30 5 N=4,M=5 300 160 120 40 5 N=5,M=5 2070 400 400 200 50 5 N=6,M=5 12300 960 1200 800 300 60 5 N=7,M=5 67830 2240 3360 2800 1400 420 70 5 N=8,M=5 359100 5120 8960 8960 5600 2240 560 80 5 N=9,M=5 1857270 11520 23040 26880 20160 10080 3360 720 90 5 N=10,M=5 9475500 25600 57600 76800 67200 40320 16800 4800 900 100 5 N=2,M=7 0 42 7 N=3,M=7 84 189 63 7 N=4,M=7 1176 756 378 84 7 N=5,M=7 11340 2835 1890 630 105 7 N=6,M=7 94080 10206 8505 3780 945 126 7 N=7,M=7 724164 35721 35721 19845 6615 1323 147 7 N=2,M=9 0 72 9 N=3,M=9 180 432 108 9 N=4,M=9 3240 2304 864 144 9 N=5,M=9 40140 11520 5760 1440 180 9 N=6,M=9 427680 55296 34560 11520 2160 216 9 Their corresponding Pascall triangle cell-filling rules, however, are more complicated: [math][x,y] = [x-1,y-1] +k[x,y-1], \, [1,y] = f(y), \, [x,y] = 1[/math] where [math]f(y) = \frac{f(y-1)ky}{y -1}, k = \frac{M -1}{2}[/math] From [imath]f(y)[/imath] for M=3, we can see a trivial but interesting identity, [math]n+1 = \prod_{a=1}^n \frac{2(a+1)}{a}[/math] Sketching [imath]f(y)[/imath] it for M=5 gives[imath]1 \cdot \frac{4}{1} = 4[/imath] [imath]4 \cdot \frac{6}{2} = 12[/imath] [imath]12 \cdot \frac{8}{3} = 32[/imath] [imath]32 \cdot \frac{10}{4} = 80[/imath] [imath]80 \cdot \frac{12}{5} = 192[/imath] [imath]192 \cdot \frac{14}{6} = 448[/imath] [imath]448 \cdot \frac{16}{7} = 1024[/imath]... I’ve a vague hunch there’s some deep number theoretical profundity buried in these coincidences. Quote
tetrahedron Posted March 3, 2013 Report Posted March 3, 2013 There is always some profundity to be had- the question is at what cost to you in time and in sanity (with or without spacer). As for me, after months of lean pickings I finally found ANOTHER Pascal-related nuclear numerical relation. And I'm probably now certifiable! Jess Tauber Quote
Turtle Posted March 5, 2013 Report Posted March 5, 2013 ...[math]n+1 = \prod_{a=1}^n \frac{2(a+1)}{a}[/math] Sketching [imath]f(y)[/imath] it for M=5 gives[imath]1 \cdot \frac{4}{1} = 4[/imath] [imath]4 \cdot \frac{6}{2} = 12[/imath] [imath]12 \cdot \frac{8}{3} = 32[/imath] [imath]32 \cdot \frac{10}{4} = 80[/imath] [imath]80 \cdot \frac{12}{5} = 192[/imath] [imath]192 \cdot \frac{14}{6} = 448[/imath] [imath]448 \cdot \frac{16}{7} = 1024[/imath]... I’ve a vague hunch there’s some deep number theoretical profundity buried in these coincidences. i got some profundity on my shoes once & so i started going barefoot. :D since you first posted i have been trying to place your values there, {4 12 32 80 192 448 1024...}, in some context i grok, but no lok. but wait!!... circumstances prompted me today to throw them at the Wolfram Alpha Engine and among other things it gave an=2n(n+1) . not sure if that's the same as your above notation or not. help? :thumbs_up no help? :thumbs_do the engine gives this continuation: {4 12 32 80 192 448 1024 2304 5120 11264 24576 53248 114688 ...} 4 12 32 80 192 448 @Wolfram Alpha Quote
CraigD Posted March 5, 2013 Author Report Posted March 5, 2013 since you first posted i have been trying to place your values there, {4 12 32 80 192 448 1024...}, in some context i grok, but no lok. but wait!!... circumstances prompted me today to throw them at the Wolfram Alpha Engine and among other things it gave an=2n(n+1) . not sure if that's the same as your above notation or not. help? :thumbs_up no help? :thumbs_do the engine gives this continuation: {4 12 32 80 192 448 1024 2304 5120 11264 24576 53248 114688 ...}That’s it, for the 5-play case (AKA rock/paper/scissor/lizard/Spock :)) I noticed this shortly after my first post, without (I swear ;)) consulting the oracle of Wolfram, but didn’t post it – a good thing, as you likely wouldn’t have had the searching and groping for grocking fun you did if I had, Turtle. The recursive[math]f(y) = \frac{ky}{y -1}f(y-1), k = \frac{M -1}{2}[/math]where M is the number of plays, which must be odd.can be rewritten as the non-recursive[math]f(y) = yk^{y-1}[/math] Either give sequences like this:1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28, ...1,4,12,32,80,192,448,1024,2304,5120,11264,24576,53248,114688,245760,524288, ...1,6,27,108,405,1458,5103,17496,59049,196830,649539,2125764,6908733,22320522, ...1,8,48,256,1280,6144,28672,131072,589824,2621440,11534336,50331648,218103808, ...1,10,75,500,3125,18750,109375,625000,3515625,19531250,107421875,585937500, ...1,12,108,864,6480,46656,326592,2239488,15116544,100776960,665127936,4353564672, ...1,14,147,1372,12005,100842,823543,6588344,51883209,403536070,3107227739, ...1,16,192,2048,20480,196608,1835008,16777216,150994944,1342177280,11811160064, ...1,18,243,2916,32805,354294,3720087,38263752,387420489,3874204890,38354628411, ...1,20,300,4000,50000,600000,7000000,80000000,900000000,10000000000,110000000000, ...1,22,363,5324,73205,966306,12400927,155897368,1929229929,23579476910, ...1,24,432,6912,103680,1492992,20901888,286654464,3869835264,51597803520, ...1,26,507,8788,142805,2227758,33787663,501988136,7341576489,106044993730, ...1,28,588,10976,192080,3226944,52706752,843308032,13282101504,206610467840, ...1,30,675,13500,253125,4556250,79734375,1366875000,23066015625,384433593750, ...... The recursive formula seems prettier and more fitting, though, given that the other terms use recursive, Pascal-esque triangle generating formulae for all the terms except the 0 and 1 winner ones. What amazes me the most about this is how these fairly simple formulae exactly describe the probabilities of outcome of a randomly (and optimally) played, fair, simple, well-known game (and its more obscure variants), and their practical value in showing how large a gang of people can play such games and expect to get a useful outcome (ie: something other than everybody loses to someone, 0 winners). Quote
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