Moving on to a Maths Puzzle

[ACE2M]

7.8%.

[thetank]

I googled for hints.....


Consider the generating function
f(x) = (1 + x)4(1 + x2)4...(1 + x13)4 = a0 + a1x + a2x2 ... + a364x364.

Each exponent in the generating function represents a total score, and the corresponding coefficient represents the number of ways of obtaining that score.

Well that's just giving it away ffs.

[TightEnd]

you are close enough


the workings, long form, are reproduced below for all those who have been pestering me about it. Especially Poppet.


There are 2 to the power 52 equally likely configurations of face up/face down.  We seek the number of configurations for which the sum of face up card values is divisible by 13.

Consider the generating function
f(x) = (1 + x)4(1 + x2)4...(1 + x13)4 = a0 + a1x + a2x2 ... + a364x364.(1)

There is a bijection between each card configuration and a contribution to the corresponding term in the generating function.  Each exponent in the generating function represents a total score; the corresponding coefficient represents the number of ways of obtaining that score.

Hence we seek the sum S = a0 + a13 + ... + a364.

Express S in terms of f(1), f(w), ... , f(w12)
Let w be a (complex) primitive 13th root of unity.  Then w13 = 1 and 1 + w + w2 + ... + w12 = 0.  Consider

f(1)  = a0 +  a1 +  a2 + ... +  a13 +  a14 + ... +  a363 +  a364
f(w)  = a0 +  a1w +  a2w2 + ... +  a13 +  a14w + ... +  a363w12 +  a364
f(w2)  = a0 +  a1w2 +  a2w4 + ... +  a13 +  a14w2 + ... +  a363w11 +  a364
f(w3)  = a0 +  a1w3 +  a2w6 + ... +  a13 +  a14w3 + ... +  a363w10 +  a364
 ...
f(w12)  = a0 +  a1w12 +  a2w11 + ... +  a13 +  a14w12 + ... +  a363w +  a364

Since w is a primitive root of unity, the set of values {wk, (wk)2, ... , (wk)12} is a permutation of {w, w2, ... , w12}, for any integer k not divisible by 13.
Therefore, adding these 13 equations, we obtain
f(1) + f(w) + ... + f(w12) = 13(a0 + a13 + ... + a364).
Hence S = [f(1) + f(w) + ... + f(w12)]/13.

Evaluate f(1), f(w), ... , f(w12)
Clearly, from (1), f(1) = 252.
Also, f(w) = [(1 + w)(1 + w2)...(1 + w13)]4.(2)

Consider g(x) = x13 − 1 = (x − w)(x − w2)...(x − w13).
Then g(−1) = −2 = (−1 − w)(−1 − w2)...(−1 − w13).
Hence (1 + w)(1 + w2)...(1 + w13) = 2.

Again, since w is a primitive root of unity, the terms of f(w2), ... , f(w12), will simply be a permutation of those for f(w), in (2).
Hence f(w) = f(w2) = ... = f(w12) = 2.

Conclusion

Putting the above results together, we obtain S = (252 + 12·24)/13.

Therefore, the probability that the total value is divisible by 13 is S/252 =

 
0.0769
 

Remarks

The full expansion of the generating function is given below.  So, for example, the mode is total value = 182, which can occur in 62256518307724 different ways.

f(x) = 1 + 4x + 10x2 + 24x3 + 51x4 + 100x5 + 190x6 + 344x7 + 601x8 + 1024x9 + 1702x10 + 2768x11 + 4422x12 + 6948x13 + 10748x14 + 16404x15 + 24722x16 + 36816x17 + 54242x18 + 79112x19 + 114285x20 + 163644x21 + 232364x22 + 327324x23 + 457652x24 + 635320x25 + 875970x26 + 1199972x27 + 1633653x28 + 2210864x29 + 2975012x30 + 3981404x31 + 5300170x32 + 7019992x33 + 9252384x34 + 12136984x35 + 15848122x36 + 20602388x37 + 26667864x38 + 34375320x39 + 44131176x40 + 56433032x41 + 71888218x42 + 91235276x43 + 115369242x44 + 145371580x45 + 182544588x46 + 228451436x47 + 284963083x48 + 354311768x49 + 439152692x50 + 542635472x51 + 668485014x52 + 821093960x53 + 1005628534x54 + 1228147584x55 + 1495737359x56 + 1816664180x57 + 2200545074x58 + 2658538920x59 + 3203561115x60 + 3850521524x61 + 4616588568x62 + 5521483156x63 + 6587801685x64 + 7841371392x65 + 9311642126x66 + 11032113124x67 + 13040798405x68 + 15380734776x69 + 18100530454x70 + 21254957700x71 + 24905593127x72 + 29121503120x73 + 33979976422x74 + 39567307824x75 + 45979628488x76 + 53323783760x77 + 61718262076x78 + 71294167920x79 + 82196238282x80 + 94583905196x81 + 108632394234x82 + 124533856716x83 + 142498536263x84 + 162755956512x85 + 185556125278x86 + 211170753288x87 + 239894471656x88 + 272046039176x89 + 307969536456x90 + 348035526772x91 + 392642171257x92 + 442216294116x93 + 497214373326x94 + 558123441452x95 + 625461891139x96 + 699780156504x97 + 781661253340x98 + 871721171276x99 + 970609086769x100 + 1079007378496x101 + 1197631437798x102 + 1327229243384x103 + 1468580679735x104 + 1622496595508x105 + 1789817571030x106 + 1971412375504x107 + 2168176115226x108 + 2381028044032x109 + 2610909019874x110 + 2858778616460x111 + 3125611864711x112 + 3412395614952x113 + 3720124536976x114 + 4049796740152x115 + 4402409013015x116 + 4778951709128x117 + 5180403273004x118 + 5607724412712x119 + 6061851960721x120 + 6543692427032x121 + 7054115261412x122 + 7593945881104x123 + 8163958479348x124 + 8764868641560x125 + 9397325841272x126 + 10061905840124x127 + 10759103030662x128 + 11489322805332x129 + 12252873985920x130 + 13049961360604x131 + 13880678420106x132 + 14745000336692x133 + 15642777234404x134 + 16573727850160x135 + 17537433631546x136 + 18533333318564x137 + 19560718110862x138 + 20618727464160x139 + 21706345555304x140 + 22822398515948x141 + 23965552467364x142 + 25134312386636x143 + 26327021892068x144 + 27541863967292x145 + 28776862637474x146 + 30029885667328x147 + 31298648284803x148 + 32580717918316x149 + 33873519998414x150 + 35174344804916x151 + 36480355320421x152 + 37788596117900x153 + 39096003239480x154 + 40399414996720x155 + 41695583699166x156 + 42981188239084x157 + 44252847439018x158 + 45507134142632x159 + 46740589955049x160 + 47949740515536x161 + 49131111260522x162 + 50281243567912x163 + 51396711143755x164 + 52474136595712x165 + 53510208073162x166 + 54501695821760x167 + 55445468589999x168 + 56338509765548x169 + 57177933082296x170 + 57960997841452x171 + 58685123525124x172 + 59347903653156x173 + 59947118833482x174 + 60480748900460x175 + 60946984006739x176 + 61344234635400x177 + 61671140451814x178 + 61926577881548x179 + 62109666406075x180 + 62219773526440x181 + 62256518307724x182 + 62219773526440x183 + 62109666406075x184 + 61926577881548x185 + 61671140451814x186 + 61344234635400x187 + 60946984006739x188 + 60480748900460x189 + 59947118833482x190 + 59347903653156x191 + 58685123525124x192 + 57960997841452x193 + 57177933082296x194 + 56338509765548x195 + 55445468589999x196 + 54501695821760x197 + 53510208073162x198 + 52474136595712x199 + 51396711143755x200 + 50281243567912x201 + 49131111260522x202 + 47949740515536x203 + 46740589955049x204 + 45507134142632x205 + 44252847439018x206 + 42981188239084x207 + 41695583699166x208 + 40399414996720x209 + 39096003239480x210 + 37788596117900x211 + 36480355320421x212 + 35174344804916x213 + 33873519998414x214 + 32580717918316x215 + 31298648284803x216 + 30029885667328x217 + 28776862637474x218 + 27541863967292x219 + 26327021892068x220 + 25134312386636x221 + 23965552467364x222 + 22822398515948x223 + 21706345555304x224 + 20618727464160x225 + 19560718110862x226 + 18533333318564x227 + 17537433631546x228 + 16573727850160x229 + 15642777234404x230 + 14745000336692x231 + 13880678420106x232 + 13049961360604x233 + 12252873985920x234 + 11489322805332x235 + 10759103030662x236 + 10061905840124x237 + 9397325841272x238 + 8764868641560x239 + 8163958479348x240 + 7593945881104x241 + 7054115261412x242 + 6543692427032x243 + 6061851960721x244 + 5607724412712x245 + 5180403273004x246 + 4778951709128x247 + 4402409013015x248 + 4049796740152x249 + 3720124536976x250 + 3412395614952x251 + 3125611864711x252 + 2858778616460x253 + 2610909019874x254 + 2381028044032x255 + 2168176115226x256 + 1971412375504x257 + 1789817571030x258 + 1622496595508x259 + 1468580679735x260 + 1327229243384x261 + 1197631437798x262 + 1079007378496x263 + 970609086769x264 + 871721171276x265 + 781661253340x266 + 699780156504x267 + 625461891139x268 + 558123441452x269 + 497214373326x270 + 442216294116x271 + 392642171257x272 + 348035526772x273 + 307969536456x274 + 272046039176x275 + 239894471656x276 + 211170753288x277 + 185556125278x278 + 162755956512x279 + 142498536263x280 + 124533856716x281 + 108632394234x282 + 94583905196x283 + 82196238282x284 + 71294167920x285 + 61718262076x286 + 53323783760x287 + 45979628488x288 + 39567307824x289 + 33979976422x290 + 29121503120x291 + 24905593127x292 + 21254957700x293 + 18100530454x294 + 15380734776x295 + 13040798405x296 + 11032113124x297 + 9311642126x298 + 7841371392x299 + 6587801685x300 + 5521483156x301 + 4616588568x302 + 3850521524x303 + 3203561115x304 + 2658538920x305 + 2200545074x306 + 1816664180x307 + 1495737359x308 + 1228147584x309 + 1005628534x310 + 821093960x311 + 668485014x312 + 542635472x313 + 439152692x314 + 354311768x315 + 284963083x316 + 228451436x317 + 182544588x318 + 145371580x319 + 115369242x320 + 91235276x321 + 71888218x322 + 56433032x323 + 44131176x324 + 34375320x325 + 26667864x326 + 20602388x327 + 15848122x328 + 12136984x329 + 9252384x330 + 7019992x331 + 5300170x332 + 3981404x333 + 2975012x334 + 2210864x335 + 1633653x336 + 1199972x337 + 875970x338 + 635320x339 + 457652x340 + 327324x341 + 232364x342 + 163644x343 + 114285x344 + 79112x345 + 54242x346 + 36816x347 + 24722x348 + 16404x349 + 10748x350 + 6948x351 + 4422x352 + 2768x353 + 1702x354 + 1024x355 + 601x356 + 344x357 + 190x358 + 100x359 + 51x360 + 24x361 + 10x362 + 4x363 + x364.


and to think, I spent years on this stuff and now I have to resort to copy and pasting from google. Shakes head.

[Royal Flush]

And to think boldie can do it in under 6 minutes!

[Graham C]

I was almost there, just on my last few +3203561115x304's!

[Moskvich]

Therefore, the probability that the total value is divisible by 13 is S/252 =


0.0769

I did 1/13 = 0.0769.

Please can I have a maths degree?

from Moskvich (aged 8 and a half)

[Rod Paradise]

Bugger - was going to say 1/4096.... but that's only the chance of perfect pairings (same no of Q's as A's, J's as 2's etc.

Didn't take into account 1xQ, 2xA's, 3x6's, 1x7 etc.

Head hurts now....

[thetank]

Moskovich was indeed, exactly right.

Sod all the explanation, 1 divided by 13, ship it to the lad.

[TightEnd]

Moskvich can have his degree but only if it is from a former (sniff, turns nose up) polytechnic...


you know, a pretend one...

[Wardonkey]

Doe the formula hold if you take out the kings?

With no kings your looking for a multiple of 12.

Probability = 1/12?

[AndrewT]

Therefore, the probability that the total value is divisible by 13 is S/252 =


0.0769

I did 1/13 = 0.0769.

Please can I have a maths degree?

from Moskvich (aged 8 and a half)

Er, yeah.

Correct my thinking if I'm wrong, but the maximum score is 364. Any value from 0 to 364 is equally likely. As 364 itself is a multiple of 13, the chances of any randomly selected number in the range being divisible by 13 is 1/13.

[TightEnd]

Doe the formula hold if you take out the kings?

With no kings your looking for a multiple of 12.

Probability = 1/12?

don't ask me, i'm just a good copy and paster

[Wardonkey]

Therefore, the probability that the total value is divisible by 13 is S/252 =


0.0769

I did 1/13 = 0.0769.

Please can I have a maths degree?

from Moskvich (aged 8 and a half)

Er, yeah.

Correct my thinking if I'm wrong, but the maximum score is 364. Any value from 0 to 364 is equally likely. As 364 itself is a multiple of 13, the chances of any randomly selected number in the range being divisible by 13 is 1/13.

I don't think they're equally likely, there is only one way to make zero and a loads and loads, to use a techincal term, to make 182.

[ACE2M]

Therefore, the probability that the total value is divisible by 13 is S/252 =


0.0769

I did 1/13 = 0.0769.

Please can I have a maths degree?

from Moskvich (aged 8 and a half)

Er, yeah.

Correct my thinking if I'm wrong, but the maximum score is 364. Any value from 0 to 364 is equally likely. As 364 itself is a multiple of 13, the chances of any randomly selected number in the range being divisible by 13 is 1/13.

I don't think they're equally likely, there is only one way to make zero and a loads and loads, to use a techincal term, to make 182.

i made the same mistake intially.

[Wardonkey]

Doe the formula hold if you take out the kings?

With no kings your looking for a multiple of 12.

Probability = 1/12?

don't ask me, i'm just a good copy and paster

I think it does I treid to solve the original problem by using a simpler dice scenario and then extrapolating, I had the 0.0769 or 1/13 thing at one point, but I'm technically challenged and didn't really know how I'd got there or if it was correct (I had some other numbers as well).