Response: Scam School Small Risk, HUGE Reward Magic Trick

[MrJack2449]

Okay you have way to much time on your hands >.> you are smart i'll give you that, but he's a MAGICIAN.

[eneses93]

if a method works for a deck with four cards type A, four cards type B and some other 44 cards then it should also work for a deck with four cards type A, four cards type B and a single cad type C (9 cards total) but in this deck we can get the number of possible ways that have at least one pair since the only ways to have no pairs are AAAACBBBB and BBBBCAAAA. that makes a total of 1152 wrong ways and 9! - 1152 = 361 728. now if your method is right you should get the same number :)

[eneses93]

@eneses93 I don't have an answer yet, just decided to show why some previously posted solutions are wrong. I did try to use a method from one of Henry Dudeney's riddles. It was in the book "536 Puzzles and Curious Problems" - 447. The magisterial bench.

I thought it could be altered so that it is applicable for this problem but I haven't managed to do that so far and now I really doubt that this is the way to do it.

[SealFormulaMaster]

My math skills are horrid but, this is what I figure. If you divide 4 (The amount of each card) by 13 (The number of cards in each suit) for 30.0 (After multiplying by 100), then take that number and divide again by 52 (The total amount of cards in the deck) for 59.1 (Again, after multiplying by 100), or 59%. Again, Im bad at math but, at least from my understanding, you can separate each solution to find the next solution in the problem. Correct me if Im wrong please.

[acdc10133]

Ok, 52 spaces, so total number of ways to shuffle deck is 52! Now to find number of ways where card type A and card type B are next to each other. So let's put a card from the 4 A's and a card from the 4 B's in two of the 52 spaces so they are next to each other. There are 2 * (4 * 4) * 51 ways to do this. After this it doesn't matter how we place the remaining cards we will satisfy the condition. So we get 2 * (4 * 4) * 51 * 50!. 2 * (4 * 4) * 51! / 52! = 32/52 = .62. That's what I got...

[tubelarr]

@acdc10133 Darn, I was so close but I think I made a mistake and you got it right. I had [2(4*4)50!]/52! At first I thought it wasn't neccesary to count the position of the 2 cards in the deck but you are right, that needs to be counted also. well done :)

[Loomis1979]

The probability there is a Jack next to a 3 given that the 3s have 8 DISTINCT neighbors is actually only about 53%. You can't just multiply 4/48*8 -- that's double counting. Calculate the probability of Jacks not being next to a 3 and subtract from 1, so you get 1 - (44/48)*(43/47)...(37/41) = 53%.

8 neighbors is the max possible, so 53% is an upper bound. But, the probability of having 8 distinct neighbors is only 50.14%. With 7 neighbors, there is 48% of having a pair. The total is < 50%.

[derekmcgoldrick]

Time's 1 ... what an idiot ;)

[knightsfly5]

kewl british accent 

[bdavis1218]

Has anyone pointed out that Brian never said the shuffle trick was "way over 50%"? The trick with the 4 cards laid out and a person selecting one was "way over 50%, like 80%" whereas the card trick involving the shuffled deck and two values he said was about 50%, so this video really just reiterated what Brian said without ever paying attention to what he was referencing in the video. I like it as background noise too but this was wasted potential.

[singingbanana]

@bdavis1218 He said "way over 50%" at 7 minutes 12 seconds. Hence the need to respond.

[bdavis1218]

@singingbanana I stand corrected. I was going off at the beginning of the video he emphasized the other trick as well over 50. Huh.

[HaslamCorp]

Ok, I see it now. If the first 3 has a 3 next to it, then the second 3 automatically looses one of its "possibilities" - Thanks! Do you get many new subscribers that go back to your earliest posts and plod through them? I would guess that there are a lot us like that.

[singingbanana]

@HaslamCorp I don't know, but I've been reading your comments. Thanks! :)

[HaslamCorp]

Threes and Jacks - My logic went: There's four 3s in the deck. There can be a card on either side of each, so that's 8 slots to fill. Possibilities for each slot are the other 51 cards and the top or bottom of the deck or 53 possible. (Would symmetry make it 52?)  Success is one of the four Jacks. So, I figured 8 times 4/53 or 32/53 or about 60%. BUT, my pascal program I used to test it out came up with your success rate: just over 48%. Where was my logic wrong?

[singingbanana]

@HaslamCorp You considered _3_ _3_ _3_ _3_ but you haven't considered combinations like _3333_ or _333_ _3_ etc.

[IhaveWoodforSheep]

What I did was to consider all the possible distinct *pairs* of cards, without caring for suit. You have 13 choices for the first, and 12 for the second (we'll count doubles later), but this overcounts (that is, (2,3) and (3,2) are both counted), so we divide by 2, then add 13 for doubles. Then we have 91 distinct pairs possible. A deck of 52 cards has 51 pairs when considered in order. 51/91=.56, or 56% chance of occurrence. It may be lower, considering multiple instances of the same pair.

[PhenixReviewz]

@IhaveWoodforSheep dont try, u cant beat him :P

[IR335]

"Infact, I'm just a tiny, tiny man." LMAO!

[MrBoingz]

this guy is ANTI-SOCIAL....try hitting the bar hahaha

[singingbanana]

@MrBoingz Pardon?

[rakib567]

you!!!!!1 i have seen you at my school!!! :D yesss aston manor school???

wow!! you have a greate youtube channel :D

[singingbanana]

@rakib567 Hello again :)

[gmsherry1953]

This is my non-mathematician's intuition (which is generally wrong). The 4 cards of one value will have 8 cards next to them, which could be any of the 12 remaining values, so there's about an 2/3 chance of success. Cards of a chosen value at the top or bottom, or next to each other, or two successes in the same shuffle, all decrease that chance of success. (4 cards of a chosen value all at the top of the deck make the chance 1 in 12.) I'm surprised it takes it down to less than 50-50, though.

[jackfree619]

are u paying out the brushwood

[singingbanana]

@jackfree619 I don't know what that means.

[TheFlyingTiger129]

Oh, wait, that's me!

LOL!

[TheFlyingTiger129]

How many possible positions is it for a deck of cards? I came up with more then ten undecillions (one undecillion has 67 zeroes), but I think that's a little bit to much...

[orion285]

80,658,175,170,943,878,571,660­,636,856,403,766,975,289,505,4­40,883,277,824,000,000,000,000­

[TheFlyingTiger129]

Just ignore my last comment!!!!

I mixed everything up... I just wrote everything wrong! I think I actually came up with the right answere.

[sgtforge21]

james are you sure there's 32 ways that don't get a kq combo? because i got 64

[singingbanana]

You're right, I don't know how I missed that. 64.

[sgtforge21]

never mind. i actually got 80

[singingbanana]

There's always something. Proves you're the first one to fact check me.

[sgtforge21]

don't count on that yet. i'm still counting. actually, can i email you my results so you can check?

[singingbanana]

No you are right, it's KKJQJQ KKJJQQ KJKJQQ KJQQJK QQJKJK QQJJKK QJQJKK QJKKJQ You can times each of those by 8. Somehow I missed the ones beginning KK and QQ. I think every video I make has one mistake in it, but this is particularly disappointing. One day I'll rerecord them all!

[singingbanana]

Fortunately it doesn't change my argument in any way.

[sgtforge21]

because there are 16 that start with the jacks

[singingbanana]

Ah,

JKKJQQ

JQQJKK

I obviously did this quickly.

[sgtforge21]

lol. i actually wrote out every combination. how would i go about sending you them

[singingbanana]

No need, this is it. That's 80 not 64. I obviously did not spend enough time checking.

[sgtforge21]

i know what you mean. that's why i get c's in math

[singingbanana]

That's annoying because it took 10 seconds to fix once you pointed it out. But that's the problem when making these by yourself.

[sgtforge21]

lol. hey man for being on your own you still make really cool vids so don't put yourself down.

[TheFlyingTiger129]

I also wonder if you could make a video about "The tower of Hanoi"?

[sgtforge21]

there are exactly 80,658,175,170,943,878,571,6 60,636,856,403,766,975,289,505 ,440,883,277,824,000,000,000,0­00 ways to shuffle a deck of 52 cards

[Jenai1997]

u may be a tiny man but ur brain is HUGE!!!

[TiagoMLF22]

my head hurts :(

[mastermagic100]

ill bet you are a little man lol

no just kiddin but your tooooooo smart

[TheAlakazams]

I love probablity maths :) great job explaining!

[BAMitzBRADERS]

lol cool

[stuartstudios]

thats great!!

[Jenai1997]

What happens if the jokers r included in the deck.....that means thr will be 54 cards what is the probability then??

[Haguatchi]

nice cards.

[plstrom]

dude thank god for people like you, seriously, i could NEVER do things like this,not even if my life depended on it...

i've already subscribed.. ; ) congrats

[gC222SA]

youramazingly simple,cool cardz,cya lill man!

[2tonearmy1192]

cool you got a green screen!

[kid29a]

Great video. You're my favorite youtube channel at the moment. I wrote a little PHP script simulating 10000 repeats of this experiment and I got about 0.6 probability. When pair has to be in given order it's around 0.3. But of course it can be (but I doubt it is) biased by shuffling and randomizing functions implementations.

[DAWNCB]

you are great!!!

[Maper555]

ur a college math teacher right? lol nice thanks for the propability catch up didnt know how to do it (forgot? lol)

[TheUltimateNihilist]

You changed your presentation. Nice. Keep up the good maths.

[just4fun3O1]

you look like your in your 20's...how old are you?

[Strijdparel]

Wow, that was just cool, also good camera work. Your funny!

[dragonrid204]

i seriously wish my teachers in Highscool are like this but sadly i dont even understand my math i'm in right now :( (this is a comment to all of ur video's that you do!)

[RickyAustinR]

tell me about it lol,

i hate how my teachers, show me how to solve a weird equation. then do different ones 50 times, over and over. and never show me how to put it to use in a story problem (switch is all that really matters)

[yullsjenkins]

i love scam school. its so much fun doing all of the scams and tricks on other people

[singingbanana]

Agreed.

[ortega24024]

Nice one again.

[singingbanana]

Thanks :)

[xKargatx]

So you would be better of betting there wouldn't be those two cards beside eachother! Onley nowone would bett for a Drink :)

[singingbanana]

True.

[mrcoments83]

Do you teach math to anyone besides youtube???!

[singingbanana]

Are you looking for tuition? I'm afraid I'm too busy!

[mismag822]

80 million trillion trillion trillion trillion trillion?

Their should be a lottery predicting the order of 52 random cards. The Carryover Jackpot would be pretty nice.

[mismag822]

Funny, and Genius ... To My Favorites!!!

I once gave you this problem and you came up with a number over 50%. Everything I've ever read has said the odds are anywhere between 50%-80%. How confident are you that the probability are less than 50%?

1- Approximately how many 0s are there in the number of possible combinations of 52 cards?

2- What Is the a name for a number that high?

I've read that there are more stars in the universe than there are possible combinations of 52 cards.

[singingbanana]

Yup, it was mismag822 who gave me this problem - ages ago too, but I wanted to make a good effort on the video.

I'm confident with 48%, I've just discovered on the Scam School forums several were getting 48.6% so I would go with that.

Combinations of 52 cards is an 8 followed by 67 zeros. I don't know if there is a name for that, no one really bothers after a 'trillion', which is twelve zeros. So this would be 80 million trillion trillion trillion trillion trillion.

[zyzzyzus]

According to Wolfram Alpha it is 80 unvigintillion.

[xExodeusx]

Well to work that out you do 52! (Factorial) then of course compare that to the number of stars..

[sebbbaastian]

i tried it 10 times and it worked 5 times XD

[singingbanana]

That proves it ;)

[Wewuvwabbitz]

Hehe I like how you make it funny. :D

[Error081688]

Concerning the intro though,

I like the idea, but the music selection surprised me. If I had to pick a soundtrack that I think of for your videos, it would be the song on "My Juggly Afternoon". Just my 2 cents.

[Error081688]

Brilliant video! haha, wow. Nice backround and setup, several camera angles and a few takes, and even an intro! Excellent quality.

[PolarBearMath]

<3

I wish I was as awesome as you MrBanana :(

[victoriscool1]

hes just a tiny tiny man hahaha! lol!

greetings from holland jim

[singingbanana]

Hello Victor, hello Holland.

[victoriscool1]

if you want you can look and use some of my card tricks they dont really look gtood cuz its dark and i was pretty young there butt still if you use it can you leave a bit credit! thank you i love your videos i really like how you explaine the monty hall problem! thank you for that! greetings from victor from...well holland!

[ViktorBluhme]

Great editing! And nice video! Awesome!

[DeathRazorSk8ing]

Very interesting, I saw this trick from Marcofrezza...it freaked me out.

[Skope111]

You are sooo clever, I love your little quips =D. Nice explanation by the way.

[thepianoaddict]

you look creepy when i stare at the ending pic to long :S

[singingbanana]

Don't do that then! There's meant to be an annotation covering me up.

[thepianoaddict]

haha ok :P don't worry i'll still watch your vids :P

[Kaiserks3338]

I shared this video to 125meadowbrook

[singingbanana]

Hey, thanks.

[v0lv0xx]

o.O...

he must be hiding a second brain...

[TyYann]

They say "A scientific impossiblility. We're gonna defy all logic and learn how to break the laws of probability."

When you hear that, you know what to expect.

[rossmccluskie]

This video response says 'win'.

[Jenai1997]

2nd!

[link211211]

1st