# Small Risk, HUGE Reward Magic Trick

NEW SCAM SCHOOL CHANNEL!! http://www.youtube.com/user/scamschool?feature=watch

If you’re comfortable with taking a little risk for a chance at a HUGE reward, you’re going to love this magic trick. It’s a card trick that lets you scam free drinks and more from almost anyone.

Funny how this was made when netflix was so small that they advertised on scam school lol

Evreryone i tried it onit didnt work

Hey brushwood….. Would you please put the openers, tweeners, and closers from scam school on separate playlists or on different pages on your site? PLEASE?! NAVIGATING IS DIFFICULT!!!! Thanks a billion….

I love the girl at 8:09

got around 82.42951653 percent of course i used a calculator and yes i looked for an equation to solve this online and solved it within ten minutes because i am awesomeness

I wrote a quick console app in C# to run simulations and I get about 46.5% success.

Or… 47.2% if you consider the deck to be circular (the front and back cards are next to each other)

I've only ran it for a few seconds but it seems to converge and hold a steady value very quickly (doesn't take long to get to 1 million iterations)

Source code:

using System;

namespace Card_Shuffler

{

public class Shuffler

{

private int chosen1;

private int chosen2;

private int[] deck;

private int[] newDeck;

private Random rng;

private long iterations = 0;

private int success = 0;

private int success2 = 0;

public Shuffler()

{

rng = new Random();

createNewDeck();

int i = 1;

while (true)

{

initDeck();

pickTwoCards();

checkDeck();

if (i == 100000)

{

Console.WriteLine("Status report " + iterations.ToString());

Console.WriteLine(" – Normal success " + Math.Round(((decimal)success * 100 / (decimal)iterations), 2).ToString() + "% : " + success.ToString());

Console.WriteLine(" – Circular success " + Math.Round(((decimal)success2 * 100 / (decimal)iterations), 2).ToString() + "% : " + success2.ToString());

i = 0;

}

i++;

}

}

private void createNewDeck()

{

newDeck = new int[52];

for (int i = 0; i < 4; i++) //Four suits

{

for (int j = 0; j < 13; j++) //13 values

{

newDeck[i * 13 + j] = j;

}

}

}

private void initDeck()

{

deck = new int[52];

int[] tempDeck = new int[52];

for (int i = 0; i < 52; i++)

{

tempDeck[i] = newDeck[i];

}

int[] tempDeck2;

int tempPosition;

for (int i = 0; i < 52; i++)

{

tempPosition = rng.Next(0, tempDeck.Length);

deck[i] = tempDeck[tempPosition];

tempDeck2 = new int[tempDeck.Length – 1];

for (int j = 0; j < tempDeck.Length; j++)

{

if (j == tempPosition)

{

continue;

}

int targetPosition;

if (j > tempPosition)

{

targetPosition = j – 1;

}

else

{

targetPosition = j;

}

tempDeck2[targetPosition] = tempDeck[j];

}

tempDeck = tempDeck2;

}

}

private void pickTwoCards()

{

chosen1 = rng.Next(0, 13);

chosen2 = rng.Next(0, 13);

}

private void checkDeck()

{

iterations++;

bool s1 = false;

bool s2 = false;

if ((deck[0] == chosen1 && deck[51] == chosen2) || (deck[0] == chosen2 && deck[51] == chosen1))

{

s2 = true;

}

for (int i = 0; i < 51; i++)

{

if ((deck[i] == chosen1 && deck[i + 1] == chosen2) || (deck[i] == chosen2 && deck[i + 1] == chosen1))

{

s1 = true;

break;

}

}

if (s1 == true)

{

success++;

success2++;

}

else

{

if (s2 == true)

{

success2++;

}

}

}

}

}

p(2, 4)=77% roughly,as in I rounded up

The second trick works a little over 60% of the time.

There are 52! (52 factorial) ways that a deck of cards can be shuffled. We need to calculate the number of ways 2 cards of two particular ranks could be next to each other and divide it by this total to find the probability of this occurring. There are 4 cards of one rank, and 4 cards of another, which means 16 combinations of the two in 2 different orders, meaning 32 ways two ranks could be next to each other. Those two cards could be in 51 different positions in relation to the rest of the deck. The other 50 cards can be in 50! (50 factorial) variations of order. This gives us the following equation:

32 * 51 * 50! / 52! = 32 / 52 = 8 / 13 ~ 61.5%

Just tried is 6 times with a fresh shuffle each time, only got it right once so yeah, na :P

the odds of the second one would be about 76%

OMG He be advertising DA best show on DA planet my life is complete.

I take a picture of the 4 of hearts and sent it to them and it makes it even better

watch at 8:49 : imma get drinks.

sign says i dont give a shit

I actually use the pick a card trick on my phone, I just send the photo of the 5 cards tell them to pick and what not then send more photos each with a card missing until sending the 4 of hearts. Worked like… I would agree with 80% of the time like Mr. Brushwood said lol. And if it doesn't work then I am all like… well I guess magic doesn't translate into the digital era, need you here in person next time.

I picked the 7 of clubs because it's the only one that isn't symmetrical if turned upside down.

Avatar: The Last Airbender is the best! I have the complete DVD collection now :D

lol 8:25 Amazingly enough and i dont give a shit

Here's my calculation/estimation. Let me know if I'm doing something wrong. Let's say I pick Q's and J's. A decent percentage of the time, the Q's will all be separated from each other with at least 2 cards between each, and neither Q as the top or bottom card of the deck. In this scenario, there are 8 cards touching a Q (the maximum possible). For the first J's position, there are 40 "safe spots" out of 48 total spots. 40/48 odds of not touching. Second Jack's safe odds: 39/47. Third: 38/46. Fourth: 37/45. 40*39*38*37/48*47*46*45 = 0.47. So in this scenario, there is a 47% chance of none touching, and a 53% chance that at least one Q is touching a J. But all scenarios that do not fit this description would have to have a Q on the top or bottom of the deck (losing one touching spot, gaining one safe spot), a pair of Q's with only 1 card separating them (losing one touching spot, gaining one safe spot), or two Q's touching (losing TWO touching spots, gaining TWO safe spots). In the extreme case, if all four Q's are in order at one end of the deck, you lose two touching spots for each of the three consecutive pairs of Q's (2*3 = 6 touching spots lost), and you lose one more touching spot by having one Q on the end of the deck. A total of 7 touching spots are lost, leaving only 1. Obviously, this stands to reason as you can easily imagine visually that there would be only 1 touching spot in this scenario. So, the way I'm seeing it… BEST case scenario is 53% of touching, and all the exceptions to that case will be LESS than 53%. I would estimate that this would put it at least down to 50% if not several percentage points lower. Brian says it is "well over" 50%. So, someone tell me, what is wrong with my reasoning?

I saw this first on numberphile.. Believe it or not the second trick the odds are NOT over 50% its actually like 48%. Math guy actually calculated the odds.

The second trick isn't a viable scam. The actual probability (even though people are saying 60+%) is only around 53%. That's only slightly better odds than a coin toss. And to all the people getting 60+%, you're doing your math wrong because you're not accounting for a LOT of factors in probability.