Stanley really deserved a million dollars with all of the effort he put into coming up with the correct answers on these super tough questions that I don't believe anyone else in the cubing community but him could have aced. Great job Stanley!!
I didn’t realize the 2x2x50 was made by him, but I knew the other two. That was one of the easiest questions for me, because I’d seen videos on all three.
*How to solve the final question for a human:* Section 1: Prerequisites You have to know some approximate values of logarithms. Very rough values. And you have to know Stirling's approximation (which I'm disappointed Stanley didn't know/use) and how to change log base (and also basic properties of the logarithm, which I'm not writing here). ln(2)=0.7 ln(3)=1.1 Log(e)=0.43 ln(n!)=n*ln(n)-n (This is Stirling's approximation) Log(x)=Log(e)*ln(x) (This is *exact* ) All of these equalities are approximate. Furthermore Stirling's formula is more accurate the bigger n is. So don't use for 4!, but for n>=10 it's fair game. Optional prerequisites: knowing the rough number of combinations on a 2x2 (3.7e+6) and on a 3x3 (4.3e+19). Also for intuition purposes 1/0.43=2.3 is helpful for immediately knowing if the end result is gonna be bigger or smaller than 100. Section 2: 5x5 As Stanley correctly pointed out you can factor out the combinations on a 3x3, because midges and corners form a 3x3. The combinatorics itself is simple: there are 24 wings, 24 +-centers (in 6 sets of 4 identical ones) and 24 x-centers (also in 6 sets of 4 identical ones). They can be permuted in all possible ways (no permutation parity) and they do not have an orientation. Thus we need to calculate (24!)^3/(4!)^12. Let's calculate the natural log of this: ln(24)=ln(2^3*3)=3ln(2)+ln(3)=3*0.7+1.1=3.2=ln(4!) ln(24!)=24*ln(24)-24=24*(ln(24)-1)=24*2.2=52.8 ln[(24!)^3/(4!)^12)]=3ln(24!)-12ln(4!)=3*52.8-12*3.2=158.6-38.4=120.2 Now, even without doing further detailed computations you can see that we will never reach 10^100: this is because 120.2/2.3 is certainly gonna be less than 60, and adding 19 isn't gonna bring the total to 100. Going ahead with the calculation anyways though we need to convert the natural log to a base 10 log: discarding that 0.2 at the end of the previous result (because seriously, who cares?) Log[(24!)^3/(4!)^12)]=Log(e)*ln[(24!)^3/(4!)^12)]=0.43*120=12*4.3=51.6, call it 52. Adding 19 gives 71. Wiki says 74. This may be attributed to errors stacking up, especially with Stirling's formula and that multiplication by 12. Section 3: 6x6 We can fix the DBL corner, and pretend that the corners form a 2x2, thus we can just focus on centers and wings, and worry about the factor of 10^6 later. The combinatorics is done exactly like before: we need to compute (24!)^6/(4!)^24. We already did most of the hard work before. ln[(24!)^6/(4!)^24]=6ln(24!)-24ln(4!)=6*52.8-24*3.2=316.8-76.8=240 Immediately if we divide by 2.3 we will get something that's bigger than 100. And we haven't even added the 2x2 contribution! The answer is already locked in. Log[(24!)^6/(4!)^24]=Log(e)*ln[(24!)^6/(4!)^24]=0.43*240=103.2 Adding 6 we get 109. Wiki says 116. Appendix A: 2x2 Refer to a multitude of resources for the combinatorics. The final result is 7!*3^6. 7! can be computed by hand: it's only slightly tedious. It's 5040. 3^6 can be computed by hand or with logarithms. We'll do it by hand this time. 3^6=9^3=9^2*9=81*9=729. Thus the final number is 729*5040 which is about 3.7 milion. Appendix B: 3x3 Again the combinatorics is well known. We need to compute 12!*8!*3^7*2^10. ln(12)=ln(2^2*3)=2ln(2)+ln(3)=2*0.7+1.1=2.5 ln(12!)=12*(ln(12)-1)=12*1.5=6*3=18 ln(12!*3^7*2^10)=ln(12!)+3ln(3)*10ln(2)=2.5+7*1.1+10*0.7=18+7.7+7=32.7 This time we'll divide by 2.3. Compute 2.3*15: it's 23*1.5=23+23/2=23+11.5=34.5. Subtracting 2.3 gives 2.3=14=34.5-2.3=32.2, thus Log(12!*3^7*2^10) is about 14. We didn't include the factor of 8!=40320, which means that the number of combinations is about 10^18 (actual number is 4.3*10^9, not bad for our rough estimate). Takeaway: Stanely bad lmao.
Congrats for Stanley for making history and how you find answers for a question takes a lot of work but it helps you find the answers and that is what matters and DG what music and time the music is happening in is perfect and makes the show more interesting.
I think it was the same for all of this season, but in the original season it was actually: Which of these was never the 3x3 single world record? A: 8.15 B: 7.03 C: 6.65 D: 6.24 (I ended up changing the times to make it slightly easier and I used this as the $4,000 question in episode 1 of this season. As it turns out it still had some remnants of a $500,000 question since the contestants found it pretty difficult!)
True, good point! I didn't explicitly state it but that question was referring to a traditional mechanism where center pieces/caps can't be removed. I would've clarified if there was any confusion about that by the player, but Stanley seemed to get where the question was coming from :)
![Who Wants To Be A Millionaire: Cubing Edition [S2E2]](http://i.ytimg.com/vi/2JeypqD8q3E/mqdefault.jpg)
![Who Wants To Be A Millionaire: Cubing Edition [S2E2]](/img/tr.png)
![Pointers in C / C++ [Full Course]](http://i.ytimg.com/vi/zuegQmMdy8M/mqdefault.jpg)
Hope y'all enjoy this one :)
The next episode will be livestreamed tomorrow (Sunday, July 18) on Twitch at 4-5PM EDT! www.twitch.tv/dgcubeslive
Stanley wasting a lifeline for fun at 500000 is so savage.
Thanks Stanley and DG that was so much fun to watch!
Hey Tingman, glad you enjoyed it! :)
Stanley really deserved a million dollars with all of the effort he put into coming up with the correct answers on these super tough questions that I don't believe anyone else in the cubing community but him could have aced. Great job Stanley!!
Aww thank you
I'm still waiting for Vlog #3
@@natecubed Soon come
What's interesting is the 1x2x111, 2x2x50, and 33x33x33 were all made by the same person - Gregoire Pfennig.
I didn't even realize that until this comment! What an inspirational designer :)
I didn’t realize the 2x2x50 was made by him, but I knew the other two. That was one of the easiest questions for me, because I’d seen videos on all three.
Stanley Big brained everything.
Congrats to him because that was insane lol
Getting a heart from dg cubes is when you know life is good lol
*How to solve the final question for a human:*
Section 1: Prerequisites
You have to know some approximate values of logarithms. Very rough values. And you have to know Stirling's approximation (which I'm disappointed Stanley didn't know/use) and how to change log base (and also basic properties of the logarithm, which I'm not writing here).
ln(2)=0.7
ln(3)=1.1
Log(e)=0.43
ln(n!)=n*ln(n)-n (This is Stirling's approximation)
Log(x)=Log(e)*ln(x) (This is *exact* )
All of these equalities are approximate. Furthermore Stirling's formula is more accurate the bigger n is. So don't use for 4!, but for n>=10 it's fair game.
Optional prerequisites: knowing the rough number of combinations on a 2x2 (3.7e+6) and on a 3x3 (4.3e+19). Also for intuition purposes 1/0.43=2.3 is helpful for immediately knowing if the end result is gonna be bigger or smaller than 100.
Section 2: 5x5
As Stanley correctly pointed out you can factor out the combinations on a 3x3, because midges and corners form a 3x3. The combinatorics itself is simple: there are 24 wings, 24 +-centers (in 6 sets of 4 identical ones) and 24 x-centers (also in 6 sets of 4 identical ones). They can be permuted in all possible ways (no permutation parity) and they do not have an orientation. Thus we need to calculate (24!)^3/(4!)^12. Let's calculate the natural log of this:
ln(24)=ln(2^3*3)=3ln(2)+ln(3)=3*0.7+1.1=3.2=ln(4!)
ln(24!)=24*ln(24)-24=24*(ln(24)-1)=24*2.2=52.8
ln[(24!)^3/(4!)^12)]=3ln(24!)-12ln(4!)=3*52.8-12*3.2=158.6-38.4=120.2
Now, even without doing further detailed computations you can see that we will never reach 10^100: this is because 120.2/2.3 is certainly gonna be less than 60, and adding 19 isn't gonna bring the total to 100.
Going ahead with the calculation anyways though we need to convert the natural log to a base 10 log: discarding that 0.2 at the end of the previous result (because seriously, who cares?) Log[(24!)^3/(4!)^12)]=Log(e)*ln[(24!)^3/(4!)^12)]=0.43*120=12*4.3=51.6, call it 52. Adding 19 gives 71. Wiki says 74. This may be attributed to errors stacking up, especially with Stirling's formula and that multiplication by 12.
Section 3: 6x6
We can fix the DBL corner, and pretend that the corners form a 2x2, thus we can just focus on centers and wings, and worry about the factor of 10^6 later. The combinatorics is done exactly like before: we need to compute (24!)^6/(4!)^24. We already did most of the hard work before.
ln[(24!)^6/(4!)^24]=6ln(24!)-24ln(4!)=6*52.8-24*3.2=316.8-76.8=240
Immediately if we divide by 2.3 we will get something that's bigger than 100. And we haven't even added the 2x2 contribution! The answer is already locked in.
Log[(24!)^6/(4!)^24]=Log(e)*ln[(24!)^6/(4!)^24]=0.43*240=103.2
Adding 6 we get 109. Wiki says 116.
Appendix A: 2x2
Refer to a multitude of resources for the combinatorics. The final result is 7!*3^6. 7! can be computed by hand: it's only slightly tedious. It's 5040. 3^6 can be computed by hand or with logarithms. We'll do it by hand this time. 3^6=9^3=9^2*9=81*9=729. Thus the final number is 729*5040 which is about 3.7 milion.
Appendix B: 3x3
Again the combinatorics is well known. We need to compute 12!*8!*3^7*2^10.
ln(12)=ln(2^2*3)=2ln(2)+ln(3)=2*0.7+1.1=2.5
ln(12!)=12*(ln(12)-1)=12*1.5=6*3=18
ln(12!*3^7*2^10)=ln(12!)+3ln(3)*10ln(2)=2.5+7*1.1+10*0.7=18+7.7+7=32.7
This time we'll divide by 2.3. Compute 2.3*15: it's 23*1.5=23+23/2=23+11.5=34.5. Subtracting 2.3 gives 2.3=14=34.5-2.3=32.2, thus Log(12!*3^7*2^10) is about 14. We didn't include the factor of 8!=40320, which means that the number of combinations is about 10^18 (actual number is 4.3*10^9, not bad for our rough estimate).
Takeaway: Stanely bad lmao.
@Dan He Then just read the takeaway.
Lol the way Stanley used his expertise to make this so much more interesting is so entertaining
Lets go, Stanley is so mathematical, that last question tho
Congrats to Stanley! Really fun watch :D
Hey Krish!! Glad you enjoyed :)
Congrats for Stanley for making history and how you find answers for a question takes a lot of work but it helps you find the answers and that is what matters and DG what music and time the music is happening in is perfect and makes the show more interesting.
Yessss, Stanley's a legend!! And thank you so much, I put a lot of effort into these shows so seeing comments like these really makes my day :)
Do one with the cubing historian!!
yes!
@@putian_ye Doesn't mean they're not active elsewhere
@@putian_ye they are, trust me
Can't believe I knew the answer to the million dollar question : )
Yoooo that's awesome :)
Completely blown away 😮
SOMEONE REMIND THIS MAN TO DO PYRA WALKTHROUGHS 2021 THE STREAK CANT BE BROKEN
OOOOOO I see a beard growing 👀👀
Are you going to make a season 3?
Was the $500,000 question also the same during every episode? No one besides scalpel reached that question so I assume that
I think it was the same for all of this season, but in the original season it was actually:
Which of these was never the 3x3 single world record?
A: 8.15
B: 7.03
C: 6.65
D: 6.24
(I ended up changing the times to make it slightly easier and I used this as the $4,000 question in episode 1 of this season. As it turns out it still had some remnants of a $500,000 question since the contestants found it pretty difficult!)
great show, very entertaining. kudos to the showmaster and the superbrain. :-)
The only thing I am better at than scalpel is chess, shogi and go. No chance in Cubing and Cubing facts.
Excited for 1st August episode
what's yr chess elo
My chess elo is around 2100
I'm only 1300
It's has dropped to 2050
But I'm already a millionaire... 😉
:O
@@DGCubes jkjk lol
Your alive?!?!
@@QwerkyCuber yeah
@@Kewbix yaeaeaeahhhhh!!!!!
Impressive memory
If you are assembling a 3x3 you are also assembling the center pieces meaning that the probability should be lower than 1/12.
True, good point! I didn't explicitly state it but that question was referring to a traditional mechanism where center pieces/caps can't be removed. I would've clarified if there was any confusion about that by the player, but Stanley seemed to get where the question was coming from :)
2:00 Stickered or stickerless
I'd love to see Jay McNeill on this, I feel like he'd be quite good
my mans actually did it
39 minutes of mental math test
AYOOOOOOOOOOOOOOOOOOOOOOOO
WOOOOOO
@@DGCubes BIG HYPE
I keep thinking I’m gonna see Rufus Dinkleton :(
lol
Is the prize money real?
4th
First
Legend states that DG will not pin this comment. :)