Divide 100 coins into 2 Piles with the Same # of Heads

rip
.

[@Smona]

okay but licking the coins is still a valid strategy: while your fingers might not be sensitive enough to discern whether a coin side is heads or tails the tongue can, with a little training. Of course the maths solution is easier here, so using that would be faster, but also less funny

[@dumbidiot1119]

If they are American pennies (or maybe quarters, Idr) finger touch can be used by itself

[@meezivanhoutendijk3591]

Your a sick man and i love you

[@Dr.1.]

Ahahah

[@YeprilesteR]

Yummy

[@littlefish9825]

Thanks for not only mathematically proving it, but also giving an intuitive answer for *why* it works that way. Fantastic vids

[@sorgan7136]

This is a very difficult problem to explain. You did it masterfully. Well done!

[@MathNerd1729]

I read the title and knew what puzzle this was referring to! Still, the additional perspective at the end was awesome! :)

[@agargamer6759]

Solving this puzzle is one thing, but how do people come up with this?!

[@Kurushimi1729]

For those who have trouble seeing how they could have figured this out, here's a good strategy. Try changing the problem a little to make it easier. What if there had only been one coin instead of 30? I think in this case it's easier to see that picking a random coin and flipping it would give the desired result.

[@theaureliasys6362]

Aaaargh. This is a generalized version of a card trick. Split a 32 card deck in half (16 and 16), the number of black cards in one pile will equal the number of red cards in the other. Didn't know you could generalize, but of course you can. Clever.

[@BleachWizz]

2:00 - ok, thanks for showing me a new hack for mathematics... we usually try to find a common variable to make subistitutions, well... * sigh *

[@pipo3686]

I don't understand this is basic system

[@PacoCotero1221]

you deserve to be a BIG channel some day. one of my best discoveries in years!!!! woo hooo

[@sweatyhands9830]

Man, you already habe 7k subs! You're growing fast, do what you're doing

[@randompersonasdf]

let the number of heads in pile A (30 coins) = x, then the number of heads in pile B (70 coins) = 30 - x. If you flip all of the pile A you get 30 - x heads in pile A = 30 - x heads in pile B

[@robindosanjos9672]

Michael Stevens from Vsauce also explained this phenomenon very well with two decks of cards. The number of red cards in one deck always equals the number of black cards in the other: m.kzbin.info/www/bejne/hZPMon2wm950ea8 😃

[@pranjalsrivastava4260]

Great work!

[@arxalier2956]

Indeed a marvellous video and solution

[@suricate2964]

Great video !

[@HA7DN]

I think I read a variation of this in a book about Rodolfo (a Hungarian magican).

[@Leo-if5tn]

Marvelous!

[@theothetorch8016]

Death by maths.

[@chair1694]

brilliant

[@Dr.1.]

You are literally me

[@Dr.1.]

Imagine someone imagining the solution

[@sdspivey]

Red is dead, no more videos.

[@rigoluna1491]

Neat

[@incredulity]

helio!

[@rohinv2752]

hello!

[@Fantastish29]

I've got a solution for the case without given amount of tails: divide each coin in half, and turn one half upside down. This will break the coins, but this will guarantee that we will have equal amount of tails and heads

[@Sir_Isaac_Newton_]

ain't no way

[@helloiamenergyman]

:)

[@ieatgarbage8771]

Separate into 2 piles and flip 1 pile over completely. H1+T2=30 + + T1+H2=70 = = S1 S2 H2+T2=S2 So H1+T2=S2 But H1+T2=30 So S2=30 Separate the coins into piles of 70 and 30. Flip every coin in the pile with 30

[@brahmandsaraswat867]

Good, great, Now do that with 60 heads not 30,

[@immigrantmammoth5162]

Simple: Separate the coins into a pile of 60 and a pile of 40, then flip the pile of 60 coins.

[@brahmandsaraswat867]

@@immigrantmammoth5162 hey, great observation, I was here flipping the 40 coins.. 🤦, Thank you for replying.

[@alimahdi15]

🥲youre the bestt