Visit brilliant.org/TedEd/ to check out Brilliant’s 60+ courses in math, logic, science, and computer science. They feature storytelling, code-writing, interactive challenges, and plenty of puzzles for you to solve. And as an added bonus, the first 833 of you to use that link will receive 20% off the annual premium subscription fee.
@@bobonox4233 honestly, i believe that considering we can only use one disc throughout, our best shot in winning is to go with B, and hope that luck favours us in both the rounds, getting a four against A and C disc will do the trick, if C ends up in 1 instead of 5. The second way could be to pick A, in which case our odds of winning might just be a bit more better, considering B and C have greater odds of getting 2 or 1 respectively
@@bobonox4233 I believe A is slightly favored to win round 3. Multiply each disk's chance of winning in round 1 by their chance of winning round 2. Assuming your opponent will always choose their best option in round 1, odds of winning for each disk are: A (against C) = 51% B (against A) = 44% C (against B) = 22% Multiply by odds of winning round 2: A = 51% x 29% = 14.79% B= 44% x 33% = 14.52% C = 22% x 38% = 8.36% Therefore A is slightly favored over B in round 3!
I wanna spin the very best, like no one ever spun. To get a high level is my real test, no training to be number one. I will travel, across the land, spinning far and wide. Teach others to understand, the probability that lies! Diskymon! Gotta spin ‘em all! It’s the disk and me, Diskymon! I know which disk will win me! Diskymon! Math, is my best friend, where we spin and spin again! Diskymon! A spin so true, Diskymon! Our chances will pull us through! You teach me and I teach you, ran-dom-chance! Gotta spin ‘em all, Gotta spin ‘em all, Diskymon!
Yeah yeah yeahhhhhh. Every disk along the way, with luck and chance I will face. I will spin every day to claim my rightful place. Spin with me, the disk is right, there’s no better disk. Spin and spin, we’ll win the fight. It’s always been our dream! Diskymon! Gotta spin ‘em all! It’s disk and me, I know which disk will win me. Diskymon! Math, is my best friend, where we spin and spin again! Diskymon! Gotta spin ‘em all! A spin so true! Our chances will pull us through! You teach me and I teach you, Ran-dom-chance! Diskymon! Gotta spin ‘em all, gotta spin em’ all! Gotta spin em’ all! Gotta spin em’ all! Gotta spin em’ all! Yeah! Diskymon! Gotta spin ‘em all! It’s disk and me, I know which disk will win me. Diskymon! Math, is my best friend, where we spin and spin again! Diskymon! Gotta spin ‘em all! A spin so true! Our chances will pull us through! You teach me and I teach you, Ran-dom-chance! Diskymon! Gotta spin ‘em all, gotta spin em’ all, Diskymon!
All of the Pokémon references in this video: 0:01 Professor Oak telling you "Your very own tale of grand adventure is about to unfold." 0:10 Just the name "Diskymon" 0:29 Burgersaur is very similar to Bulbasaur, one of the three first starter Pokémon. 0:31 Churrozard is very similar to Charizard, the final evolution of Charmander, one of the three first starter Pokémon. 0:46 Wartortilla is very similar to Wartortle, the middle-stage evolution of Squirtle, one of the three first starter Pokémon. 1:01 "No matter what type it is" referencing the type system in Pokémon. 2:09 A reference to Jessie and James' famous line "Prepare for trouble! And make it double!" from the anime. 2:39 Being a Diskymon trainer is just like how you could be a Pokémon trainer. 4:47 A reference to "I wanna be the very best, that no one ever was!" from the theme song in the anime (I haven't seen that much of it so I don't know if it's still used in modern episodes). If I made some mistake or forgot something, let me know!
MORE REFERENCES! The crystals and eggs on the shelf at the start, are from the Fire Crystal Riddle and Egg Drop Riddle respectively. The Diskybowl Arena is reminiscent in design of the Wizard Dueling arena in the Wizard Standoff Riddle. (This one is more obscure, but I still count it) The art style and color scheme is that of Einstein's Riddle.
The bonus question is the nicest of all three: 1st round: There are two possibilities for any disk you pick. Because the opponent is going to pick according to your pick, we are going to consider only the lowest winning odds (the opponent will choose in order to minimize your odds). For A this is 51%. For B this is 44%. For C this is 38.22%. 2nd round: The video explains this well. Just considering the second round, the odds of A winning is 28.56%. For B this is 33.22%. For C this is 38.22%. 3rd round: In order to get the odds of A winning this round, one must multiply A1 and A2, which gives 14.57%. Multiplying B1 and B2 give 14.62%. C1 and C2 give 14.61%. We see that all three disks have incredibly similar odds, with B just barely coming out on top. But the point is that there really is no difference no matter which one you pick. Cool, innit?
Oh i get it now, by round 3 you mean the going through round 1 and round 2 in succession. Indeed this is pretty cool. The people coming up with these puzzles are amazing!
Indeed, but as B has a slight edge over the other two, you should go for that and stick with it. But, if it was that you have to face both of them in 1-on-1 matches, with all three of you locked to the same disk, followed by the battle royal with you still locked to the same disk, and you have to win all three matches, then B trumps out. This is because in the case where you have to fight both of the unchosen disks in succession, A has a 28.56% chance of coming out on top after both fights, B has a 27.18% chance and C has only an 18.73% chance. And, as Raen showed, A has the same odds to win the battle royal, B has a 33.22% chance to win and C has a 38.22% chance. Combining these, and A would have an 8.16% chance of winning all three fights, B would have a 9.03% chance and C would have a 7.16% chance of winning. Either way, those two rivals really wanted to rig the odds to be not in your favour.
I think you have to calculate every possibility so the chances add up to 100%. Your first opponent won't pick thinking only about beating you in round 1, their strategy should be the same as yours. This leaves Disk A with ~31% chance of winning, Disk B with ~35% and Disk C with ~33%. Now the odds for Disk B are better.
For anyone curious, the answer to the third round is disk B. There’s actually two ways to look at this: does your opponent always choose the best disk to beat you in round 1? Or do they calculate the best remaining disk to maximize their odds of winning both rounds? If we consider the first case, then the odds of winning are just multiplied: A: .51 * .2856 = 14.57% B: .44 * .3322 = 14.62% C: .3822 * .3822 = 14.61% So disk B is the best to keep for both rounds with a 14.62% chance of winning. Now consider if the opponent considers that they also have to keep their choice for the battle royale, so they pick based on those odds instead of just the best disk to beat you in the first round. I’ll save you the tedious calculations, but in every case the opponent still picks the same one and the results are the same.
@Carlos Soprano It's because if you choose Disk C in round 1 the best option to counter it is Disk B (51% chance of summoning level 1 to lose no matter what + ~11% (.22*.49) chance to lose even w the lvl 5 summon)
At first I calculated the mean result of each disk. Like that B was the best disk as it has a mean expected value of 3.32 which is more than 3 of disk A and 2.96 of disk C. I think the answer for the last riddle is B because with many tries you get the best expected value. But it could also be C if you really need to win each battleroyal and not get the most levels
MEANWHILE SOMEWHERE ELSE... Hey dude! Let’s play Diskemon! Ok! But first I need to get my Mintendo switch. *LATER...* Hmmm you have a 82% chance of getting a level 3 sturger...
For the bonus round the chances of winning the entire time: (Assuming your enemy makes the most intelligent decision to combat you each time. Also round one, disk A is best combated by disk C, not B like what was used in the video; I used the proper calculation to give you the worst odds per disc for each round. Also it's so close, that I had to skip rounding.) A: 14.5656% B: 14.6168% C: 14.607684% The winner is B, by a hair. It is fun to note that disk C has odds of winning 38.22% in both rounds.
@Carlos Soprano we have to choose first during the first riddle. If we were going second and our opponent picked C we would counter pick with B it's just A has no weak matchups so it's the best when you go first.
Can you explain a bit more please? The riddle is if you have to only pick ONE disc for BOTH rounds, which is the best disc to pick? So why does B have the best odds? Ty
Regarding the last question, it seem to me that the three options are surprisingly very close. I got 14,57% of a final win if choosing A throughout the all competition, 14,62% by choosing B and 14,61% by choosing C. Just, pick your favourite Diskymon and keep fingers crossed XP
@@blopszomb6813The easiest thing to see is that 2 different disks can win each event while there is no way you to be the winner if that were to happen. It is not a full explanation but you can see the total possible set of combinations is larger than the total set of combinations we are looking at .
@@blopszomb6813 You must understand that the sum of these %s are for choosing A1+A2, B1+B2 and C1+C2. Hence the rest of the % would be derived from A1+B2, A1+C2, B1+A2, B1+C2, C1+A2, and finally, C1+B2. Hence the % you see in the comment will not add up to 100%! Hope this makes sense!
References (I'll edit if I find more) 0:07 Elemental, Eggs. 0:10 Moldevort. 0:29 Kanto (the dreaded region) starters. 2:10 Prepare for trouble, and make it double. 4:47 The first opening lyrics. Overall: This video is a cease and desist bait for Nintendo.
ok ill uncliffhanger it for you Disk A: (probablity of winning part A) * (probablity of winning part B) = (probablity of being the very best) part 1 = if player 1 takes A p2 takes C so p1 has a 51% of winning part 2 = as mentioned in the vid, disk A has 28.56% chance of winning battle royal final probablity Disk A = 0.14.5656% Disk B part 1= p1 takes B p2 takes A, p1 has 44% of winning part 2 = 33.22% of winning BR final probablity Disk B = 14.6168% Disk C part 1 = p1 takes C p2 takes B, p1 has 38.22% of winning part 2 = also a 38.22% chance of winning BR final prob Disk C = 14.6077% The Best odds are with Disk B, but only by a sliver as they all round to 14.6%
@@833Rowan I don't agree with your calculation. For disc C round 1 the opponent will choose disc A giving you a 49% chance of winning. I agree with your round 2 calculation but then .3822*.49=18.7278% therefore disc C is best option.
For anyone wondering about the final question, you pick B! A has 51% (1st round) x 28.56% (2nd round) = 14.57% chance of winning both rounds. B has 44% (opponent would pick A round 1) x 33.22% = 14.62% chance of winning both rounds. C has 38.22% (opponent would pick B round 1) x 38.22% = 14.61% chance of winning both rounds. Fascinating how any one of the three disks can be made the best choice with such simple rule changes.
Edit: This is correct assuming that the opponent is logically good and will choose as you said in 1st round. Since that's not mentioned explicitly, we will have to assume that she has a 50% chance of selecting either of the other 2 disks. In round 1, opponent can pick any one of the 2 disks. A has 0.5×0.51+0.5×0.56=0.535 B wins over C 0.6178 of the times. Hence, if B is selected, 0.5×0.6178+0.5×0.44=0.5289 Similarly with C, 0.4361 Then the rest of the logic is similar as yours.
@@av.keshavshastry7785 Why is that so? The goal of the tournament is to win, so why not assume that the opponent is going to use the best possible move?
Had to scroll way down to find this, but no - THIS IS INCORRECT. They should pick C! The odds of C beating B is 51% x 78% = 39.78%. (It looks like you mistakenly jotted down 38.22% twice). This gives their overall odds to be 39.78% x 38.22% = 15.20%, making C the best chance of winning.
@@kumarsnehal1008 that is true also, i remember having some tangrowth that could kill my friend’s 200+hp mega ex in one hit(if you flip a heads, and because of it’s weakness)
@@zuriiiiiiiiiiiiiii 1 goomba and 2 badnicks. Cant add stuff with different labels unless combined into a third label. For example, the answer to this could be 3 video game enemies
Alex has reached Stage 3 of his Ted-Ed Riddle Cinematic Universe. References from his other riddles are constantly popping up in other riddles. (Crystals, Faberge Eggs, Moldevort and Drumbledrore)
Me: b Ted-ed: no its A but you can try again with different rules Me: b? Ted-ed: C, but you can try again again Me: B? Ted-ed: Doesn't say Me: LIFE ISN'T FAIR
@@MathematicallyAFox You probably might have a rounding error from using whole numbers; from what I've calculated, here's the full list: Choice A: In round 1, victory chances against B and C are 56% and 51% respectively; opponent chooses C to better their odds. In round 2, victory chances are 28.56%. Multiply together for a chance of 14.5656%. Choice B: In round 1, victory chances against A and C are 44% and 61.78% respectively; opponent chooses A to better their odds. In round 2, victory chances are 33.22%. Multiply together for a chance of 14.6168%. Choice C: In round 1, victory chances against A and B are 51% and 38.22% respectively; opponent chooses B to better their odds. In round 2, victory chances are 38.22%. Multiply together for a chance of 14.607684%. So, B comes out on top for its relative consistency across both rounds, but only very slightly.
I solved all 3 and it was quite fun! I solved by calculating losses instead which might've complicated things a bit but ended up basically doing the same thing. For round 1: If I pick A, opponent chooses C having better odds and I lose 49% time If I pick B, opponent choose A having better odds and I lose 56% time If I pick C, opponent chooses B having better odds and I lose 61.78% time So we pick A having least chances of a loss. For round 2: If I pick A, either opponents wins and I lose in (44+49)% = 93% time If I pick B and get a 2, I lose automatically... 56% time If I pick B and get a 4, I lose if disk C rolls a 5... 10.78% time So, on choosing B, I lose 66.78% time If I pick C and get a 1, I lose automatically... 51% time If I pick C and get a 5, I lose if disk B rolls a 6... 10.78% time So, on choosing C, I lose 61.78% time So we pick C having least chances of a loss. For round 3: You just check which disk lands you the best win chances. Disk A: (1-49) * (1-93) = 3.57% win chance Disk B: (1-56) * (1-66.78) = 14.61% win chance Disk C: (1-61.78) * (1-61.78) = 14.60% win chance So disk B barely gives us better odds against disk C.
"Now its time for round two, and while you've prepared for trouble, you didnt anticipate they'd make it double" team bracket invades the arena in the next tournament right?
Here’s an Interesting follow-up puzzle: You have acquired a Level 1 Wartortilla, a Level 2 Burgersaur, and a Level 3 Churrozard as your team. Your rival, on the other hand, has acquired a Level 1 Burgersaur, a Level 2 Churrozard, and a Level 3 Wartortilla. For this battle, you will each send out one Diskymon at a time, and the two will battle. The one with a higher level always wins, but there’s one catch: If both Diskymon are the same level, then the battle will be determined by type advantage (Churrozard will defeat Burgersaur, Wartortilla will defeat Churrozard, and Burgersaur will defeat Wartortilla). When a Diskymon is defeated, they will not be able to be used for the remainder of the battle, and the Trainer must use one of the other two, if they’re still in battle. When all three Diskymon are defeated, whoever remains wins overall. Can you think of a way to defeat your rival? What would be the optimal strategies for both you and your rival? Finally, how would things change if type advantage took precedence over level (for example, a Level 2 Churrozard would defeat a Level 3 Burgersaur)?
Well, if ur a master, the best way to win is to always make ur spin not random, as in train ur fingers to target specific areas while still spinning the arrow
For the bonus riddle its Disk B. Here is my reasoning: In single duels these are the winning odds: A against B = 56%, B against A = 44% A against C = 51%, C against A = 49% B against C = (51% + (22%x49%)) = 61.78% and C against B = 38.22% Winning Odds in combined duels: A winning = 56%x51% = 28.56% B winning = 22% + 22%x49% = 32.78% C winning = 49%x(100-22%) = 38.22% Now these are the winning odds for winning single duels and combined battles for each disk: A = 56%x51%x28.56% = 8.16% (Win against B and C and then win against both of them together) B = 44%x61.78%x32.78% = 8.91% (Same Logic) C = 38.22%x49%x38.22% = 7.16% Therefore, Disk B has the best odds of winning in a stretch of duels and a final fight. This makes sense as in the single Duels, B fairs better than C and fairs better than A in the combined battle.
For the last riddle the chances of each disk winning both competitions(Assuming on the first round your oponent picks the best avaliable disk) are A:14.5656% B:14.6168% C:14.607684% Therefore you should pick B, also the chances of each disk winning considering a 50/50 chane of picking are: A:15.2792% B:17.570058% C:16.67742% And considering its a way too accurate equipe rocket reference and they pick the worst possible disk on the first round: A:15.9936% B:20.2642% C:18.7278% So B is always best.
Agree with this results. My calculation is as followed: A > B = 0.56 A > C = 0.51 B > C = 0.22 + 0.22 * 0.51 = 0.6178 I. A = min(0.56, 0.51) = 0.51 B = min(1 - 0.56, 0.6178) = min(0.44, 0.6178) = 0.44 C = min(1 - 0.51, 1 - 0.6178) = min(0.49, 0.3822) = 0.3822 II. A = 0.56 * 0.51 = 0,2856 B = 0.22 + 0.22 * 0.51 = 0,3322 C = 0.49 * (1 - 0.22) = 0,3822 III A = I.A * II.A = 0,145656 B = I.B * II.B = 0,146168 C = I.C * II.C = 0,14607684
Bonus Question: Individual odds of winning both rounds by choosing the same disk both times: A: 51% × (51% × 56%) = 14.5656% B: 44% × (22% + 22% × 51%) = 14.6168% C: 49% × 78% × (49% × 78%) = 14.607684% - B has the highest odds of winning both rounds
It’s fun how the answer to the last riddle is always B, but if you assume your opponent will choose the best disk in the first fight it’s only better by a very slight margin, whereas if your opponent chooses randomly it becomes much more significant.
![ลองสุ่มไอโฟน 990 โกงมั้ย? [ โกงมั้ยครับ ep.101 ] | DOM](http://i.ytimg.com/vi/bfsWe0j492k/mqdefault.jpg)
Visit brilliant.org/TedEd/ to check out Brilliant’s 60+ courses in math, logic, science, and computer science. They feature storytelling, code-writing, interactive challenges, and plenty of puzzles for you to solve. And as an added bonus, the first 833 of you to use that link will receive 20% off the annual premium subscription fee.
I managed to figure this riddle quite quickly... This is the fifth Ted ed riddle I've successfully completed so far.
@@bobonox4233 honestly, i believe that considering we can only use one disc throughout, our best shot in winning is to go with B, and hope that luck favours us in both the rounds, getting a four against A and C disc will do the trick, if C ends up in 1 instead of 5. The second way could be to pick A, in which case our odds of winning might just be a bit more better, considering B and C have greater odds of getting 2 or 1 respectively
@@bobonox4233 I believe A is slightly favored to win round 3.
Multiply each disk's chance of winning in round 1 by their chance of winning round 2.
Assuming your opponent will always choose their best option in round 1, odds of winning for each disk are:
A (against C) = 51%
B (against A) = 44%
C (against B) = 22%
Multiply by odds of winning round 2:
A = 51% x 29% = 14.79%
B= 44% x 33% = 14.52%
C = 22% x 38% = 8.36%
Therefore A is slightly favored over B in round 3!
FINALLY Ted-Ed, an easier riddle that we can solve in *higher probability*
@@aaronhodgin7007 You made a small mistake, the odds of C winning against B are 38.22%, not 22%, so C edges out B slightly, but gets beaten by A
"while you expected trouble
You didn't think it'd be double" is the best sentence I've heard in a teded riddle
I hope you know that's a pokemon reference
Why else would he comment
You don’t see the reference? This whole ep is referring pokemons all over
the clash
I might not be that into Pokémon but I still love team rocket
Whoever wrote this definitely had a LOT of fun with the names
😂
true.
I had fun listening them 😅
Or discovered that Bugsnax is a demo of pokemon mayo and catsup
So many Pokémon references
2:01 that was actually a level four churrozard so the burgersaur should've lost there
Anything can be achieved through sheer willpower
It was power of friendship
@@deusvult4872 knowing pokemon it probably was tbh
@@jam4536 basically how pikachu gets through all the opponents in most of the Pokémon series.
Criticals....
1:00 "the higher level...will win..no matter what type it is."
Those who memorize the type chart: Surprised Pizza-chu face
Avert your eyes (or preferably not) here comes Gardenwar
I choose you, Pizza-chu!
at least here ash ketchup can win
finnally i got to use this joke
LOL 😂
psydish noises agreeing
the 3rd riddle doesnt matter because the other 2 guys will probably be disqualified for tying up the referee
Those two are the jessie and james of the series and the worlds officer jenny will be arriving soon.
@@lanluhylrean1122 tru
And they could have just stole all three discs....
@@PyroGothNerd well ... I guess
@@lanluhylrean1122 yeah
I wanna spin the very best, like no one ever spun. To get a high level is my real test, no training to be number one. I will travel, across the land, spinning far and wide. Teach others to understand, the probability that lies! Diskymon! Gotta spin ‘em all! It’s the disk and me, Diskymon! I know which disk will win me! Diskymon! Math, is my best friend, where we spin and spin again! Diskymon! A spin so true, Diskymon! Our chances will pull us through! You teach me and I teach you, ran-dom-chance! Gotta spin ‘em all, Gotta spin ‘em all, Diskymon!
This is one of the best comments I have ever seen.
Underrated
Good one
Yeah yeah yeahhhhhh. Every disk along the way, with luck and chance I will face. I will spin every day to claim my rightful place. Spin with me, the disk is right, there’s no better disk. Spin and spin, we’ll win the fight. It’s always been our dream! Diskymon! Gotta spin ‘em all! It’s disk and me, I know which disk will win me. Diskymon! Math, is my best friend, where we spin and spin again! Diskymon! Gotta spin ‘em all! A spin so true! Our chances will pull us through! You teach me and I teach you, Ran-dom-chance! Diskymon! Gotta spin ‘em all, gotta spin em’ all!
Gotta spin em’ all!
Gotta spin em’ all!
Gotta spin em’ all!
Yeah!
Diskymon! Gotta spin ‘em all! It’s disk and me, I know which disk will win me. Diskymon! Math, is my best friend, where we spin and spin again! Diskymon! Gotta spin ‘em all! A spin so true! Our chances will pull us through! You teach me and I teach you, Ran-dom-chance! Diskymon! Gotta spin ‘em all, gotta spin em’ all, Diskymon!
@@someone8581 That is not even remotely how Pov memes are used.
"You prepared for trouble, but you didn't anticipate they would make it double"
I hate this video and that's why I love it
I don't get it.
@@catoctober8005 Every frame of this video is a pokemon reference
*very best trophy appears on screen*
My brain: I wanna be the the very best
You’ll have to protect the world from devastation, and unite all people within our nation!
And denounce the world of truth and love and send it to the skies above.
All of the Pokémon references in this video:
0:01 Professor Oak telling you "Your very own tale of grand adventure is about to unfold."
0:10 Just the name "Diskymon"
0:29 Burgersaur is very similar to Bulbasaur, one of the three first starter Pokémon.
0:31 Churrozard is very similar to Charizard, the final evolution of Charmander, one of the three first starter Pokémon.
0:46 Wartortilla is very similar to Wartortle, the middle-stage evolution of Squirtle, one of the three first starter Pokémon.
1:01 "No matter what type it is" referencing the type system in Pokémon.
2:09 A reference to Jessie and James' famous line "Prepare for trouble! And make it double!" from the anime.
2:39 Being a Diskymon trainer is just like how you could be a Pokémon trainer.
4:47 A reference to "I wanna be the very best, that no one ever was!" from the theme song in the anime (I haven't seen that much of it so I don't know if it's still used in modern episodes).
If I made some mistake or forgot something, let me know!
you forgot diskymon master
Level based power system, that every Pomemon game has.
Hey that’s moldevort on the 0:10 paper he’s a teded riddle character
Bulbasaur, Wartortle, & Charizard.
Burgersaur, Wartotilla, & Churrozard.
Remember, Kids;
Pokemon & their Masters stick together!
You teach me, & I'll teach you!
POKEMON!!!
This might be a stretch, but the fact that the two opponents captured the referee may be a reference; Team Rocket often capture people in the show.
MORE REFERENCES!
The crystals and eggs on the shelf at the start, are from the Fire Crystal Riddle and Egg Drop Riddle respectively.
The Diskybowl Arena is reminiscent in design of the Wizard Dueling arena in the Wizard Standoff Riddle.
(This one is more obscure, but I still count it) The art style and color scheme is that of Einstein's Riddle.
Also Moldevort was in the Diskymon letter.
@@Dr._is_sleepy The inter-riddle references are getting out of hand.
OMG a TED Ex riddle cinematic universe
i saw them all. Fun fact the eggs are also Featured in the crystal riddle in the background
Yeah
The bonus question is the nicest of all three:
1st round: There are two possibilities for any disk you pick. Because the opponent is going to pick according to your pick, we are going to consider only the lowest winning odds (the opponent will choose in order to minimize your odds). For A this is 51%. For B this is 44%. For C this is 38.22%.
2nd round: The video explains this well. Just considering the second round, the odds of A winning is 28.56%. For B this is 33.22%. For C this is 38.22%.
3rd round: In order to get the odds of A winning this round, one must multiply A1 and A2, which gives 14.57%. Multiplying B1 and B2 give 14.62%. C1 and C2 give 14.61%.
We see that all three disks have incredibly similar odds, with B just barely coming out on top.
But the point is that there really is no difference no matter which one you pick. Cool, innit?
Where did round 3 come from?
We're just supposed to carry out the 2 rounds in succession with the same disk right?
Oh i get it now, by round 3 you mean the going through round 1 and round 2 in succession. Indeed this is pretty cool. The people coming up with these puzzles are amazing!
Indeed, but as B has a slight edge over the other two, you should go for that and stick with it. But, if it was that you have to face both of them in 1-on-1 matches, with all three of you locked to the same disk, followed by the battle royal with you still locked to the same disk, and you have to win all three matches, then B trumps out.
This is because in the case where you have to fight both of the unchosen disks in succession, A has a 28.56% chance of coming out on top after both fights, B has a 27.18% chance and C has only an 18.73% chance. And, as Raen showed, A has the same odds to win the battle royal, B has a 33.22% chance to win and C has a 38.22% chance.
Combining these, and A would have an 8.16% chance of winning all three fights, B would have a 9.03% chance and C would have a 7.16% chance of winning.
Either way, those two rivals really wanted to rig the odds to be not in your favour.
I think you have to calculate every possibility so the chances add up to 100%. Your first opponent won't pick thinking only about beating you in round 1, their strategy should be the same as yours. This leaves Disk A with ~31% chance of winning, Disk B with ~35% and Disk C with ~33%. Now the odds for Disk B are better.
'YOU JUST WON BY LUCK!"
"Y...yes....that's literally how everyone wins this game..."
😂
For anyone curious, the answer to the third round is disk B.
There’s actually two ways to look at this: does your opponent always choose the best disk to beat you in round 1? Or do they calculate the best remaining disk to maximize their odds of winning both rounds?
If we consider the first case, then the odds of winning are just multiplied:
A: .51 * .2856 = 14.57%
B: .44 * .3322 = 14.62%
C: .3822 * .3822 = 14.61%
So disk B is the best to keep for both rounds with a 14.62% chance of winning.
Now consider if the opponent considers that they also have to keep their choice for the battle royale, so they pick based on those odds instead of just the best disk to beat you in the first round. I’ll save you the tedious calculations, but in every case the opponent still picks the same one and the results are the same.
@Carlos Soprano It's because if you choose Disk C in round 1 the best option to counter it is Disk B (51% chance of summoning level 1 to lose no matter what + ~11% (.22*.49) chance to lose even w the lvl 5 summon)
At first I calculated the mean result of each disk. Like that B was the best disk as it has a mean expected value of 3.32 which is more than 3 of disk A and 2.96 of disk C. I think the answer for the last riddle is B because with many tries you get the best expected value. But it could also be C if you really need to win each battleroyal and not get the most levels
It's common sense lol
Respect for actually calculating that
I would choose B only because it was not the better options before, so it must be now 😂😂
i can't be the only one who's impressed with how the narrator manages to say every single reference without missing a beat
I am because, as per my above comment, it lacked essential information.
MEANWHILE SOMEWHERE ELSE...
Hey dude! Let’s play Diskemon!
Ok! But first I need to get my Mintendo switch.
*LATER...*
Hmmm you have a 82% chance of getting a level 3 sturger...
@@sebxiou-lifestyle4465 you do not have a comment above this one?
I'm just impressed about the fact that he didn't burst out laughing every single time he made a reference
@@Alexander-km2jo I mean he could just retry
Noone talking about the crystals and eggs from other riddles
S a y w h a t?
@@joongminshin2769 0:08
the invitation's signed by moldevort as well!
hmmm I commented about this
Wow I didn't even notice! Thanks for pointing it out, now I enjoy this video even more :)
I love the fire crystal riddle reference at the beginning. And I also love the constant pokemon references.
There was also the egg riddle reference
Me on christmas: I can’t wait to play my Diskydisk on my Sintendo Ditch
cant wait to play pybercunk in my zbox
I'm a huge fan of Fame Greek, developers of the Diskydisk video games, myself.
Dude you should get duigi's zansion tree on the Sintendo Ditch.
@@Dxuser5 That honestly not a good one
I can't wait to play tinecraft on my zbox for chirstmas
ted-ed: did you get the answer?
me: ulu but actually ozo...
I get it!
So ulu is yes and ozo is no?
@@MoreToFind1 well ulu but actually ozo
Aaaaaah, I agree
This does not answer there question
For the bonus round the chances of winning the entire time: (Assuming your enemy makes the most intelligent decision to combat you each time. Also round one, disk A is best combated by disk C, not B like what was used in the video; I used the proper calculation to give you the worst odds per disc for each round. Also it's so close, that I had to skip rounding.)
A: 14.5656%
B: 14.6168%
C: 14.607684%
The winner is B, by a hair. It is fun to note that disk C has odds of winning 38.22% in both rounds.
@Carlos Soprano They will choose B, because 1 loses to everything.
@Carlos Soprano Basically, A is best response to B and B is best response to C and C is best response to A.
@Carlos Soprano we have to choose first during the first riddle. If we were going second and our opponent picked C we would counter pick with B it's just A has no weak matchups so it's the best when you go first.
@@carlossoprano571 0.3822
Can you explain a bit more please?
The riddle is if you have to only pick ONE disc for BOTH rounds, which is the best disc to pick?
So why does B have the best odds?
Ty
Nobody:
Ted Ed: *PUNS*
@Danielle Anderson if Ted Ed liked this, it means *PUNS*
@Danielle Anderson pretty sure they're portmanteaus
Thats what they do!😂 sometimes puns make a great TH-cam channel
"while you expected trouble
You didn't think it'd be double" is the best sentence I've heard in a teded riddle
*YES!!!*
Regarding the last question, it seem to me that the three options are surprisingly very close.
I got 14,57% of a final win if choosing A throughout the all competition, 14,62% by choosing B and 14,61% by choosing C.
Just, pick your favourite Diskymon and keep fingers crossed XP
I get the same.. Choose B.
I was always a Wartortilla kid myself
That’s just wrong that adds up to 42%, where’s the other 58?
@@blopszomb6813The easiest thing to see is that 2 different disks can win each event while there is no way you to be the winner if that were to happen. It is not a full explanation but you can see the total possible set of combinations is larger than the total set of combinations we are looking at
.
@@blopszomb6813 You must understand that the sum of these %s are for choosing A1+A2, B1+B2 and C1+C2. Hence the rest of the % would be derived from A1+B2, A1+C2, B1+A2, B1+C2, C1+A2, and finally, C1+B2. Hence the % you see in the comment will not add up to 100%!
Hope this makes sense!
Funfact: Every Ted-Ed riddle has atleast one comment that says “confirm you have green eyes”
How else are you going to summon the Green-Eyes Chicken Quesadilladragon?
Confirm if the guy Who say it has green eyes, then let they in peace
Ozo.
Ever other normal persons dream:- Crush confessing their love for you
My dreams:- DISKY DISC
RIGHT!?!
BURGERSAUR DEFEATS THE CHURROZARD!!
wat?!!?
@Sophia Guild it prob got bad luck
th-cam.com/video/E2OW_plGc5Q/w-d-xo.html
"You'd prepared for trouble, but didn't anticipate they would make it double" LOL
And that’s why they have 12 million subscribers. And deserve a whole lot more😃😃
To protect the world from devastation!
@@jorgemtzb9359 "To unite all people with in our nation" 😉
@@jorgemtzb9359 to unite the people within our nation!
@Mighel do PUBG James!
And while you prepared for trouble. You didn't anticipate it being double *chef's kiss*
0:11 Moldevort?? THE MOLDEVORT we trapped in a chess board? He escaped!!!!
That’s what I said
Of course they were going to tie up the referee
WHERE IS THE CHESSBOARD???
Man the TCU (Ted-ed Cinematic Universe) goes deep
0:09 Reference to fire crystal and egg drop riddle
YEP
How many riddles you want to put in a single video?
Ted Ed: yes
ozo
Ozo
Ozo
Or is it ulu?
@amit kumar uzu, best of both worlds.
I love that Wartortilla at 0:46. It's like a burrito upgraded to a B-52
Burrito 52
Blastoisuritto
I'll keep this in mind next time I enter the finals of a monster fighting and summoning competition.
References (I'll edit if I find more)
0:07 Elemental, Eggs.
0:10 Moldevort.
0:29 Kanto (the dreaded region) starters.
2:10 Prepare for trouble, and make it double.
4:47 The first opening lyrics.
Overall: This video is a cease and desist bait for Nintendo.
2:10 i love how he mad a team rocket reference
This whole thing is Pokemon reference with math and disks
@@connorbeighley6981 Math is half of Pokémon
"Becoming that which no one ever was" how is nobody talking about this lol
i don't get why we should be talking about this part
@pycl oohhh bro idk how i didnt realize that 😭 thanks
And that bit saying "and while you've prepared for trouble, you didn't anticipate they'd make it double"
this whole video is gold
@@Berilia and the eggs from the egg drop riddle, and the crystals from the crystal riddle.
@@claudeh.8145 have you watched Pokemon?🤔
You can't end on a cliffhanger like this!
ok ill uncliffhanger it for you
Disk A: (probablity of winning part A) * (probablity of winning part B) = (probablity of being the very best)
part 1 = if player 1 takes A p2 takes C so p1 has a 51% of winning
part 2 = as mentioned in the vid, disk A has 28.56% chance of winning battle royal
final probablity Disk A = 0.14.5656%
Disk B
part 1= p1 takes B p2 takes A, p1 has 44% of winning
part 2 = 33.22% of winning BR
final probablity Disk B = 14.6168%
Disk C
part 1 = p1 takes C p2 takes B, p1 has 38.22% of winning
part 2 = also a 38.22% chance of winning BR
final prob Disk C = 14.6077%
The Best odds are with Disk B, but only by a sliver as they all round to 14.6%
@@833Rowan damn thats why i thought it could be B or C it added up to almost the same in my head thanks for clearing it up :)
@@dimusoikatherz526youre correct. my bad that's a typo but at least the calculations are still correct.
@@dimusoikatherz526 *losing
@@833Rowan I don't agree with your calculation. For disc C round 1 the opponent will choose disc A giving you a 49% chance of winning. I agree with your round 2 calculation but then .3822*.49=18.7278% therefore disc C is best option.
“You prepared for trouble, but didn’t anticipate they would make it double”
I love this video
For anyone wondering about the final question, you pick B!
A has 51% (1st round) x 28.56% (2nd round) = 14.57% chance of winning both rounds.
B has 44% (opponent would pick A round 1) x 33.22% = 14.62% chance of winning both rounds.
C has 38.22% (opponent would pick B round 1) x 38.22% = 14.61% chance of winning both rounds.
Fascinating how any one of the three disks can be made the best choice with such simple rule changes.
Edit:
This is correct assuming that the opponent is logically good and will choose as you said in 1st round.
Since that's not mentioned explicitly, we will have to assume that she has a 50% chance of selecting either of the other 2 disks.
In round 1, opponent can pick any one of the 2 disks.
A has 0.5×0.51+0.5×0.56=0.535
B wins over C 0.6178 of the times.
Hence, if B is selected, 0.5×0.6178+0.5×0.44=0.5289
Similarly with C, 0.4361
Then the rest of the logic is similar as yours.
I chose B for every round here
@@av.keshavshastry7785 Why is that so? The goal of the tournament is to win, so why not assume that the opponent is going to use the best possible move?
Either way, the odds are definitely stacked against you if you cannot switch.
Had to scroll way down to find this, but no - THIS IS INCORRECT.
They should pick C!
The odds of C beating B is 51% x 78% = 39.78%. (It looks like you mistakenly jotted down 38.22% twice).
This gives their overall odds to be 39.78% x 38.22% = 15.20%, making C the best chance of winning.
The logic games are awesome, but props to whoever makes the scenarios and names, they obviously love what they do and do it right
0:10 moldevort
Wait, I thought he was stuck in that game?!
lol, Professor Oak quotes?
I would have gone with: "This is my grandson. He's been your rival since you were a baby. …Erm, what is his name again?"
Shocking, isn't it?
@@carltonleboss no, i think his name was Gary
@@San-lh8us LMAO
*names my rival Kix*
Wow, never imagined there would be a Ted-Ed opening quote by prof. Oak, that too from the games and not the anime! Yay!
Next week on Ted-Ed:
Can You Solve The Card Game Riddle
"My grandfather's deck has no pathetic cards" -Yugi Muto
"I'll kill myself if you don't forfeit this children card game Yugi!"
-Seto Kaiba probably.
me: calculates probabilities
ted ed: before you start calculating probabilities
"Which disk should you choose?" Me, an intellectual: "Ozo"
nice reference
VG player?
Probably, "Ulu"
No I think olu
HAHAHAHHAHAHAHAAHAHHAAAHAHAA
The quote at the beginning was from pokemon and the animals like
Churrozard was like Charzard.
I AM SO HAPPYY
Are we not gonna talk about how TED-Ed is now putting Easter eggs (literally, and also Easter crystals and signs)?
Yup
Let's apriciate that ted Ed REALLY think to make these riddles
0:07 the crystal riddle reference omg-
"The higher level Diskymon will always win, no matter what type it is." I mean that's how I play normally, sooooo.
I did but it doesn't work in the discy dis.. I mean pokemon league
@@kumarsnehal1008 as proven in season 1, episode 9, the school of hard knocks.
@@imanaxolotl7999 oh yeah
The one with that school
I was referring to the games
@@kumarsnehal1008 that is true also, i remember having some tangrowth that could kill my friend’s 200+hp mega ex in one hit(if you flip a heads, and because of it’s weakness)
I have a tendency to end up just using levels even when I try to take type into account-
"Your very own tale of grand adventure is about to unfold"
-Professor Oak
Yes
proffesor oat
mom can we have pokemon?
mom:we have pokemon at home
pokemon at home:
As soon as I heard the word “Diskymon” I knew this would be a favorite.
Ted-Ed: Before you start calculating probabilities-
Me: *screeches and throws spreadsheet out the window*
but it helps for the final answer
00:05 love how the elemental magic and egg drop riddle is there
This is like those teachers trying to make their lessons include games so that the students would be more interested
NoW KIds WHAts 1 GooMBa plus 2 BADnICKS
@@zuriiiiiiiiiiiiiii 1 goomba and 2 badnicks. Cant add stuff with different labels unless combined into a third label. For example, the answer to this could be 3 video game enemies
Atleast they are trying
Well yes, that's exactly what this is
Yeah except its actually working here
0:02 you just made us respect you more🙇♂️
Alex has reached Stage 3 of his Ted-Ed Riddle Cinematic Universe. References from his other riddles are constantly popping up in other riddles. (Crystals, Faberge Eggs, Moldevort and Drumbledrore)
is nobody gonna talk about how MoldeVort signed the invitation letter thingy?
oh god every riddle is in the same universe!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I just relised
it's from Moldevort
The lore is growing...
@@benjaminschmitt1562 _dramatic music_
"You anticipated the trouble, but prepare for double"
Lol. Team rocket reference.
The *whole thing* is a Pokémon reference.
*y e s*
forget the fact that si actually deduced this intuitively, this script was pure perfection!! amazing job!!
Ted Ed: can you solve the riddle?
Me: probably not, but I’ll have fun watching the animation.
ah yes, the referee explaining the rules to a game at the finals😂
Shhhh, let's say it's for the audience to understand 😂
Considering it went from 1-on-1 to a battle royale, I wouldn't be surprised if the rules changed a bit each round.
th-cam.com/video/E2OW_plGc5Q/w-d-xo.html
th-cam.com/video/E2OW_plGc5Q/w-d-xo.html
It's as silly as the champion being surprised that you know about super-effective moves!
Leon would be proud of this referee.
This is the first ever Ted Ed Riddle I actually got
How to win:
Step one: know the difference between ozo and ulu
If I asked you to stop making obnoxious references to past riddles, would you say "ozo"?
Wait that's illegal
@Henry Amazeing step two: confirm you have green eyes.
Wait which riddle video was this from?
@@torazely ozo
Me: b
Ted-ed: no its A but you can try again with different rules
Me: b?
Ted-ed: C, but you can try again again
Me: B?
Ted-ed: Doesn't say
Me: LIFE ISN'T FAIR
Last one is C
@@MathematicallyAFox rlly?
@@MathematicallyAFox how do u know
@@suitcaseofsmarts Just multiply the odds for the two matches and you get the percent.
@@MathematicallyAFox
You probably might have a rounding error from using whole numbers; from what I've calculated, here's the full list:
Choice A:
In round 1, victory chances against B and C are 56% and 51% respectively; opponent chooses C to better their odds.
In round 2, victory chances are 28.56%.
Multiply together for a chance of 14.5656%.
Choice B:
In round 1, victory chances against A and C are 44% and 61.78% respectively; opponent chooses A to better their odds.
In round 2, victory chances are 33.22%.
Multiply together for a chance of 14.6168%.
Choice C:
In round 1, victory chances against A and B are 51% and 38.22% respectively; opponent chooses B to better their odds.
In round 2, victory chances are 38.22%.
Multiply together for a chance of 14.607684%.
So, B comes out on top for its relative consistency across both rounds, but only very slightly.
2:05 that was a level 4 !!!! Churrozard should’ve won
The tournament was rigged, all for the profit of the Big Disky.
One, two oops you messed up!
How to win diskeymon league:
Step 1: confirm you have green eyes
Step 2: Ask the champion to leave
Ask opponents if they reply ozo if u select disk 2
@@BhavinRughaniwould prefer ulu
ozo
Step 4: Divi up the gold so everyone is happy.
"Whoever summons the higher level diskimon wins"
Me: " I will cut you into a certain shape of triangle. "
Well ozo, but actually ulu
*WHAT ARE YOU DOING? WE ONLY HAVE 17 MINUTES TO CROSS THE BRIDGE BEFORE THE ZOMBIES COME*
Bruh 17? We have to lower 25 levels of water and be the ones to take it to zero.
Don’t worry, we just have to confirm the highest floor the egg won’t break on and we’ll be okay
make sure it has green eyes
I like the callbacks to other riddles, like the eggs and elemental crystals on the shelves at the beginning
1:20 Ted ed: "Before you start counting probabilities..."
Me: Oh no...
the same 😂
keep the disky-disks away from those crystals, they would become even worse.
@Sonal Mohrir obviously yes
@@Noname-67 man scratch sucks *BIG TIME*
@@unholycrusader69 r u d e
@@catandcomparator bruh ur so late
MY ANSWER FOR QUESTION 4:46, B
best = b
*good enough*
My brain: Why is this my favourite teded video
Also my brain: *P U N S V E R Y G O O D*
*c h u r r o z a r d*
Bruh this is literally the first riddle from Ted-Ed that I actually found out before he explained it.
*My brain is expanding*
As a Pokemon fan, I absolutely love this riddle.
I read it as "digimon" lol
Also btw halfway in I literally forgot this was a riddle, the pokemon names and drawings are too funny
Ted ed is the tool I use to sharpen my brain over online class.
I solved all 3 and it was quite fun! I solved by calculating losses instead which might've complicated things a bit but ended up basically doing the same thing.
For round 1:
If I pick A, opponent chooses C having better odds and I lose 49% time
If I pick B, opponent choose A having better odds and I lose 56% time
If I pick C, opponent chooses B having better odds and I lose 61.78% time
So we pick A having least chances of a loss.
For round 2:
If I pick A, either opponents wins and I lose in (44+49)% = 93% time
If I pick B and get a 2, I lose automatically... 56% time
If I pick B and get a 4, I lose if disk C rolls a 5... 10.78% time
So, on choosing B, I lose 66.78% time
If I pick C and get a 1, I lose automatically... 51% time
If I pick C and get a 5, I lose if disk B rolls a 6... 10.78% time
So, on choosing C, I lose 61.78% time
So we pick C having least chances of a loss.
For round 3:
You just check which disk lands you the best win chances.
Disk A: (1-49) * (1-93) = 3.57% win chance
Disk B: (1-56) * (1-66.78) = 14.61% win chance
Disk C: (1-61.78) * (1-61.78) = 14.60% win chance
So disk B barely gives us better odds against disk C.
I can’t blame you editor, it’s almost time for lunch :)
Your very own tale of grand adventure is about to unfold
- *Professor Oak, in Pokémon*
Being a competitive pokemon player myself this riddle was actually surprisingly in line with how you treat probability in comp pokemon
Me: "I got his, here we go"
TedEd: "There are three disky-disks"
Me: "Ok, I'm out"
i got the first one
"Now its time for round two, and while you've prepared for trouble, you didnt anticipate they'd make it double" team bracket invades the arena in the next tournament right?
0:13 Those are the disks from the crystal riddle where some students ate them up!
0:13 the priceless eggs and the fire water earth and air crystal
So we're all going to leave that poor referee chained for the eternity as if it's nothing?
#freethereferee
this video taught me a probability concept better than any I've ever had before thanks guys!!!
Here’s an Interesting follow-up puzzle:
You have acquired a Level 1 Wartortilla, a Level 2 Burgersaur, and a Level 3 Churrozard as your team. Your rival, on the other hand, has acquired a Level 1 Burgersaur, a Level 2 Churrozard, and a Level 3 Wartortilla. For this battle, you will each send out one Diskymon at a time, and the two will battle. The one with a higher level always wins, but there’s one catch: If both Diskymon are the same level, then the battle will be determined by type advantage (Churrozard will defeat Burgersaur, Wartortilla will defeat Churrozard, and Burgersaur will defeat Wartortilla). When a Diskymon is defeated, they will not be able to be used for the remainder of the battle, and the Trainer must use one of the other two, if they’re still in battle. When all three Diskymon are defeated, whoever remains wins overall.
Can you think of a way to defeat your rival? What would be the optimal strategies for both you and your rival? Finally, how would things change if type advantage took precedence over level (for example, a Level 2 Churrozard would defeat a Level 3 Burgersaur)?
You can't win the first one because your opponent has an immediate trump card. It just sends out the Level 3 Wartortilla 3 times and you are wiped.
I was about to say that died off quickly because of the lack of comments, but the riddle was there for over a year
Nobody:
TED-Ed: WHaT's PoKEMoN!?
Even a year later and the writing in this video still makes me come back too it
The namer definitely was eating while working!Funny and surprising video by Ted-Ed as always Thanks so much!
When you realize this is just Pokémon but food:
“Pokémon: Gotta *Vore* Them All!”
NO
You mean pocket foodsters?
NO
th-cam.com/video/E2OW_plGc5Q/w-d-xo.html
*EAT*
Well, if ur a master, the best way to win is to always make ur spin not random, as in train ur fingers to target specific areas while still spinning the arrow
Everyone: getting prepared for solving ted ed riddle
Me: laughing at her hairstyle
00:18
th-cam.com/video/E2OW_plGc5Q/w-d-xo.html
th-cam.com/video/E2OW_plGc5Q/w-d-xo.html
That chourozard was a level 4 on the first batle
For the bonus riddle its Disk B.
Here is my reasoning:
In single duels these are the winning odds:
A against B = 56%, B against A = 44%
A against C = 51%, C against A = 49%
B against C = (51% + (22%x49%)) = 61.78% and C against B = 38.22%
Winning Odds in combined duels:
A winning = 56%x51% = 28.56%
B winning = 22% + 22%x49% = 32.78%
C winning = 49%x(100-22%) = 38.22%
Now these are the winning odds for winning single duels and combined battles for each disk:
A = 56%x51%x28.56% = 8.16% (Win against B and C and then win against both of them together)
B = 44%x61.78%x32.78% = 8.91% (Same Logic)
C = 38.22%x49%x38.22% = 7.16%
Therefore, Disk B has the best odds of winning in a stretch of duels and a final fight. This makes sense as in the single Duels, B fairs better than C and fairs better than A in the combined battle.
For the last riddle the chances of each disk winning both competitions(Assuming on the first round your oponent picks the best avaliable disk) are A:14.5656%
B:14.6168%
C:14.607684%
Therefore you should pick B, also the chances of each disk winning considering a 50/50 chane of picking are:
A:15.2792%
B:17.570058%
C:16.67742%
And considering its a way too accurate equipe rocket reference and they pick the worst possible disk on the first round:
A:15.9936%
B:20.2642%
C:18.7278%
So B is always best.
Agree with this results. My calculation is as followed:
A > B = 0.56
A > C = 0.51
B > C = 0.22 + 0.22 * 0.51 = 0.6178
I.
A = min(0.56, 0.51) = 0.51
B = min(1 - 0.56, 0.6178) = min(0.44, 0.6178) = 0.44
C = min(1 - 0.51, 1 - 0.6178) = min(0.49, 0.3822) = 0.3822
II.
A = 0.56 * 0.51 = 0,2856
B = 0.22 + 0.22 * 0.51 = 0,3322
C = 0.49 * (1 - 0.22) = 0,3822
III
A = I.A * II.A = 0,145656
B = I.B * II.B = 0,146168
C = I.C * II.C = 0,14607684
i mean, B IS the first letter in best
The trophy with the words "The Very Best" absolutely slayed me
Bonus Question:
Individual odds of winning both rounds by choosing the same disk both times:
A: 51% × (51% × 56%) = 14.5656%
B: 44% × (22% + 22% × 51%) = 14.6168%
C: 49% × 78% × (49% × 78%) = 14.607684%
- B has the highest odds of winning both rounds
"Prepare for trouble"
"Make it double"
"Battle royale"
Ah, I see you having fun with Sun and Moon :D
Also for the last one:
Use your Z-move bro!
Cleverly designed riddle. Awesome video as always 👌
I manage to solve them, good job with the names!
It’s fun how the answer to the last riddle is always B, but if you assume your opponent will choose the best disk in the first fight it’s only better by a very slight margin, whereas if your opponent chooses randomly it becomes much more significant.