Response: Scam School Small Risk, HUGE Reward Magic Trick

My response to the magic trick described on Scam School. A trick based on probability, I work out the exact odds. See the original Scam School video here http://www.youtube.com/watch?v=B7MOlKQCPmE

See the discussion on the Scam School forum here http://revision3.com/forum/showthread.php?t=28018 They were getting a figure of 48.6%, like I said my video isn't the final word, so I would go with that.

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For more about probability, risk and uncertainty, have a look at the channel of my colleague David Spiegelhalter http://youtube.com/undunc

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[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.

[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.

[IR335]

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

[singingbanana]

@MrBoingz Pardon?

[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]

@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.

[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 :)

[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 :)

[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...

[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