The eight corner cubes of the black cube formed in
Paint it black (the cubes with half their faces already painted black) are assembled randomly into a smaller cube, with the outside of the new cube once again being painted black.
The eight cubes are then, again, disassembled and rebuilt randomly.
What is the probability that the outside of this cube is again completely black?
Bonus: What would the probability be if the smaller cube of eight cubes were a random selection of any eight cubes from the original black cube instead of only the eight corner cubes?
There are 77 configurations of what original type pieces wind up in the final 2x2x2 cube, such as 6 vertex pieces, 1 center piece, 1 edge piece and no face pieces. If v, c, e and f represent the number of vertex, center, edge and face pieces respectively, the probability of any given such configuration is 8!*12!*6!*C(8,v)*C(8-v,c)*C(8-v-c,e)/((8-v)!*(12-e)!*(6-f)!*(27!/19!))
We must get the conditional probabilities for each type of cube to eventually show all black faces, and then for each such configuration, raise each conditional probability to the power of the number of such cubes in the configuration, to get the overall probability that that configuration will produce an all-black outer surface, and then multiply by the probability that the configuration will in fact be that, as defined in the preceding paragraph.
We've already shown that the probability of an originally corner piece eventually showing all black is 27/64. What about the others?
Originally edge:
There are three positional edges at which the double-black edge can be aligned so that one extra face gets painted black, making three faces black. Another three positional edges hide the already-painted faces completely in the intermediate stage, so that five faces come out black. In the remaining 6 cases, one black face is outside and two white, which become black, making 4 black altogether. The table:
faces prob. prob of showing all black product
3 1/4 1/8 1/32
4 1/2 1/4 1/8
5 1/4 1/2 1/8
The sum of the products is 9/32.
Originally face:
The probability is 1/2 that the black face will be buried and the cube then gets a total of 4 black faces, and 1/2 that its black face merely gets a second coat of black, along with two other faces getting their first coat. The table:
faces prob. prob of showing all black product
3 1/2 1/8 1/16
4 1/2 1/4 1/8
The sum of these products is 3/16.
If a cube was originally in the center, it will definitely have 3 painted faces going into the final phase, leaving a probability of 1/8 of showing all black at the end.
The following program incorporates these probabilities:
10 VProb=27//64:EProb=9//32:FProb=3//16:CProb=1//8
100 for NoVert=0 to 8
200 if NoVert=8 then CLim=0:else CLim=1
300 for NoCent=0 to CLim
400 ELim=8-NoVert-NoCent
500 for NoEdge=0 to ELim
600 NoFace=8-NoVert-NoCent-NoEdge
700
800 if NoFace<=6 then
810 :print NoVert;NoCent;NoEdge;NoFace;
900 :PConfig=!(8)//!(8-NoVert)*!(12)//!(12-NoEdge)*!(6)//!(6-NoFace)
1000 :PConfig=PConfig//(!(27)//!(19))
1100 :PConfig=PConfig*combi(8,NoVert)*combi(8-NoVert,NoCent)*combi(8-NoVert-NoCent,NoEdge)
1200 :Tp=Tp+PConfig
1300 :print PConfig;
1400 :Ct=Ct+1
1500 :TProb=TProb+VProb^NoVert*EProb^NoEdge*FProb^NoFace*CProb^NoCent*PConfig:print VProb^NoVert*EProb^NoEdge*FProb^NoFace*CProb^NoCent
1600
1700 next NoEdge
1800 next NoCent
1900 next NoVert
2000 print Tp,Ct:print TProb:print TProb/1,1/TProb
It finds a probability for the second case, of 247324600785879/4628855992006737920, or
0.0000534310424029106 which is 1 in 18715.7119724380656633462
The particular probabilities making this up are:
conditional
# of pieces prob of prob of success
vtx ctr edge face this config given this config
0 0 2 6 2/67275 59049/17179869184
0 0 3 5 8/13455 177147/34359738368
0 0 4 4 1/299 531441/68719476736
0 0 5 3 32/4485 1594323/137438953472
0 0 6 2 28/4485 4782969/274877906944
0 0 7 1 16/7475 14348907/549755813888
0 0 8 0 1/4485 43046721/1099511627776
0 1 1 6 4/740025 6561/4294967296
0 1 2 5 4/22425 19683/8589934592
0 1 3 4 4/2691 59049/17179869184
0 1 4 3 4/897 177147/34359738368
0 1 5 2 8/1495 531441/68719476736
0 1 6 1 56/22425 1594323/137438953472
0 1 7 0 8/22425 4782969/274877906944
1 0 1 6 32/740025 177147/34359738368
1 0 2 5 32/22425 531441/68719476736
1 0 3 4 32/2691 1594323/137438953472
1 0 4 3 32/897 4782969/274877906944
1 0 5 2 64/1495 14348907/549755813888
1 0 6 1 448/22425 43046721/1099511627776
1 0 7 0 64/22425 129140163/2199023255552
1 1 0 6 8/2220075 19683/8589934592
1 1 1 5 64/246675 59049/17179869184
1 1 2 4 16/4485 177147/34359738368
1 1 3 3 128/8073 531441/68719476736
1 1 4 2 8/299 1594323/137438953472
1 1 5 1 128/7475 4782969/274877906944
1 1 6 0 224/67275 14348907/549755813888
2 0 0 6 28/2220075 531441/68719476736
2 0 1 5 224/246675 1594323/137438953472
2 0 2 4 56/4485 4782969/274877906944
2 0 3 3 448/8073 14348907/549755813888
2 0 4 2 28/299 43046721/1099511627776
2 0 5 1 448/7475 129140163/2199023255552
2 0 6 0 784/67275 387420489/4398046511104
2 1 0 5 56/740025 177147/34359738368
2 1 1 4 112/49335 531441/68719476736
2 1 2 3 224/13455 1594323/137438953472
2 1 3 2 112/2691 4782969/274877906944
2 1 4 1 56/1495 14348907/549755813888
2 1 5 0 224/22425 43046721/1099511627776
3 0 0 5 112/740025 4782969/274877906944
3 0 1 4 224/49335 14348907/549755813888
3 0 2 3 448/13455 43046721/1099511627776
3 0 3 2 224/2691 129140163/2199023255552
3 0 4 1 112/1495 387420489/4398046511104
3 0 5 0 448/22425 1162261467/8796093022208
3 1 0 4 56/148005 1594323/137438953472
3 1 1 3 896/148005 4782969/274877906944
3 1 2 2 112/4485 14348907/549755813888
3 1 3 1 448/13455 43046721/1099511627776
3 1 4 0 56/4485 129140163/2199023255552
4 0 0 4 14/29601 43046721/1099511627776
4 0 1 3 224/29601 129140163/2199023255552
4 0 2 2 28/897 387420489/4398046511104
4 0 3 1 112/2691 1162261467/8796093022208
4 0 4 0 14/897 3486784401/17592186044416
4 1 0 3 56/88803 14348907/549755813888
4 1 1 2 56/9867 43046721/1099511627776
4 1 2 1 56/4485 129140163/2199023255552
4 1 3 0 56/8073 387420489/4398046511104
5 0 0 3 224/444015 387420489/4398046511104
5 0 1 2 224/49335 1162261467/8796093022208
5 0 2 1 224/22425 3486784401/17592186044416
5 0 3 0 224/40365 10460353203/35184372088832
5 1 0 2 56/148005 129140163/2199023255552
5 1 1 1 448/246675 387420489/4398046511104
5 1 2 0 112/67275 1162261467/8796093022208
6 0 0 2 28/148005 3486784401/17592186044416
6 0 1 1 224/246675 10460353203/35184372088832
6 0 2 0 56/67275 31381059609/70368744177664
6 1 0 1 56/740025 1162261467/8796093022208
6 1 1 0 112/740025 3486784401/17592186044416
7 0 0 1 16/740025 31381059609/70368744177664
7 0 1 0 32/740025 94143178827/140737488355328
7 1 0 0 8/2220075 10460353203/35184372088832
8 0 0 0 1/2220075 282429536481/281474976710656
For example, that last line shows the 282429536481/281474976710656 probability, given all 8 cubes are original corner cubes, but also the only 1/2220075 probability that all 8 would be corner cubes (8!(27-8)!/27!), so the product of these two fractions is the contribution to the overall probability.
A simulation program on this one is as follows
RANDOMIZE TIMER
DO
REDIM cube(8, 6)
REDIM had(27)
FOR c = 1 TO 8
DO
r = INT(RND(1) * 27 + 1)
LOOP UNTIL had(r) = 0
had(r) = 1
SELECT CASE r
CASE 1 TO 8
blk = 3
CASE 9 TO 20
blk = 2
CASE 21 TO 26
blk = 1
CASE ELSE
blk = 0
END SELECT
FOR f = 1 TO blk
cube(c, f) = 1
NEXT
NEXT
FOR c = 1 TO 8
GOSUB chooseFace
cube(c, f1) = 1
cube(c, f2) = 1
cube(c, f3) = 1
NEXT
good = 1
FOR c = 1 TO 8
GOSUB chooseFace
IF cube(c, f1) = 0 THEN good = 0: EXIT FOR
IF cube(c, f2) = 0 THEN good = 0: EXIT FOR
IF cube(c, f3) = 0 THEN good = 0: EXIT FOR
NEXT
totGood = totGood + good: ct = ct + 1
PRINT totGood; ct; totGood / ct;
IF totGood > 0 THEN PRINT ct / totGood: ELSE PRINT
LOOP
END
chooseFace:
f1 = INT(RND(1) * 6 + 1)
DO
f2 = INT(RND(1) * 6 + 1)
LOOP WHILE f2 = f1 OR f2 + f1 = 7
DO
f3 = INT(RND(1) * 6 + 1)
LOOP WHILE f3 = f1 OR f3 + f1 = 7 OR f3 = f2 OR f3 + f2 = 7
RETURN
which gives the result as .0000971585E-05 or 1/25178.86 after 1082631 trials were run, for 43 successes.
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Posted by Charlie
on 2007-04-18 09:35:40 |
bonus solution
