blonde poker forum

(Solution will be provided by Dr Richard Prew, B.A( hons) , Phd, ACIS, but only after you've had a good go!).

A standard pack of cards is thrown into the air in such a way that each card, independently, is equally likely to land face up or face down.  The total value of the cards which landed face up is then calculated.  (Card values are assigned as follows: Ace=1, 2=2, ... , 10=10, Jack=11, Queen=12, King=13.  There are no jokers.)

What is the probability that the total value is divisible by 13?


TAKE YOUR TIME.....

1

lol..you don't make it easy do you? PM sent...don't want to ruin it for everyone. less then 1 though.

i know, just taking an early punt. will work out correct answer.

 ;djinn; I'll leave this one to Wardonkey......

lol you worked it out in 6 minutes......hmmmmm

I'm a genius flushy didn't ya know? ;whistle;

He worked it out in less than one minute...fold  :D

Eeerrrrrm!!

Me 'ead 'urts.

Tightend, the probability of this happening is 50%, I will let others elaborate.

fascinating. it's somewhat less than 50% though.

0.078125 or 5/64.

that could well be wrong.

i'm at work here.

It could be google!

In the absence of any brain power, I'd have to go for 1 in 13. But I suspect that may be too easy.

eleventy

7.8%.

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.

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.

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

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

I did 1/13 = 0.0769.

Please can I have a maths degree?

from Moskvich (aged 8 and a half)

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....

Moskovich was indeed, exactly right.

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

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


you know, a pretend one...

Doe the formula hold if you take out the kings?

With no kings your looking for a multiple of 12.

Probability = 1/12?

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.

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

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.

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).

But like in craps, the probability of very small numbers and very big numbers being rolled will be symmetrical and very neatly cancel each other out.

Unlike in craps, the small number begins at zero. Easy when you know the answers.



Here's my favourite from that site.....

Q - Find the smallest natural number greater than 1 billion (10⁹) that has exactly 1000 positive divisors.  (The term divisor includes 1 and the number itself.  So, for example, 9 has three positive divisors.)

A - Of course, it is 1,060,290,000.


That guy should really write trivia for christmas crackers

Presumably that doesn't matter, as the distribution of scores will be a normal curve around the 182-point, no..? So the 182s will be balanced out by 181s, 183s etc, which aren't divisble by 13.

Isn't the problem with this that there's a chance of getting 0. So there are 365 possible scores, not 364. Playing to 13, there's a 1 in 4503599627370496 chance of getting 0. (I just did it, but the floor's a right mess now).

So actually the chance of the score being divisible by 13 is marginally less than 1 in 13. But the discrepancy will be higher the lower down you go. Do it with just one ace - the chance of the score being divisble by 1 isn't 1 in 1, cos now there's a 50% chance of the score being 0.

That's what I was saying too, only slower.

It would appear you all have.

So, given that they both give the same answer, is the correlation between the two answers merely a result of the fact that the top score is divisible by the required number, and that the unlikely scores either side of the centre cancel each other out?

Would the probability of the answer being exactly divisible by, say, 9, be far away from 1/9?

Book answer :   0.0769230769230802025049342773854732513427734375
Moskvich 1/13   0.076923076923076923076923076923077 (to a finite number of decimal places)

So aswell as being right, it looks like he's right about being wrong too.

What a guy  ;applause; ;applause;

yeah...way off the mark lol

Before i read the rest of the thread i just have to say LOL @ this.