1

Take a standard pack of playing card, no jokers and no magic shop gimmicks. The 'Magician' invites you to name 2 cards, but without their suit; eg "two" and "queen" (ie but not "two of hearts" and "queen of spades") and says that he can make them 'stick together'.

From the pack face down, he turns over the first card off the top and then second... and third... and keeps going. Almost every time (and that's the point of the question!) - almost every time, long before he's got to over the last card, he'll have dealt a "two" immediately followed by a "queen" or vice-versa. There they are, 'stuck together'.

The audience is impressed and thinks there is sleight of hand but no, he hands them the pack. They shuffle it and the 'magician' names 2 cards which they turn over themselves. It works! Someone says that the magician must have already known of 2 cards that would come up; so the magician asks the volunteer to choose anyone else to name 2 cards and they deal the cards out themselves a second time. It works again and the magician isn't even involved!

Sometimes, if it doesn't work, the magician blames the 'novice' magician and berates them for not concentrating. Worst case, the 2 cards are normally only separated from being "stuck together" by one stray card. "Try again!" says the magician - and it works! This delights children and confounds adults!

So, clearly down to probability and not magic but it almost always works. Question is... what is the probability of it working on any given 'deal'?

Thanks Mike

MikeG
  • 11
  • 1
    I know I've seen this problem. I don't remember if it was here or on youtube... let me see if I can remember the argument and/or find a link. – JMoravitz Nov 14 '20 at 14:59

2 Answers2

1

Let us assume without loss of generality that the cards selected are Kings and Queens.

Consider the event $K_{\spadesuit}$ that the king of spades is adjacent somewhere in the deck to a some queen. Similarly define $K_{\heartsuit},K_{\clubsuit}$ and $K_{\diamondsuit}$.

For the purposes of estimation, let us assume these are independent events and also let us assume that the deck "wraps around" so the top card is said to be adjacent also to the bottom card. I know they are not, but it will help simplify calculations considerably and cause the numbers to be off by a relatively insignificant amount.

We try to find $\Pr(K_\spadesuit \cup \dots \cup K_{\diamondsuit})$ which expands by incusion-exclusion as $\Pr(K_\spadesuit)+\dots+\Pr(K_\diamondsuit) - \Pr(K_\spadesuit\cap K_{\heartsuit}) - \dots - \Pr(K_{\clubsuit}\cap K_{\diamondsuit}) + \Pr(K_\spadesuit \cap K_\heartsuit \cap K_{\clubsuit}) +\dots - \Pr(K_{\spadesuit}\cap \dots \cap K_{\heartsuit})$

Now, $\Pr(K_{\spadesuit}) = \frac{8}{51}-\frac{12}{51\cdot 50}\approx 0.152$

$\Pr(K_{\spadesuit}\cap K_{\heartsuit})$ then by our simplifications/assumptions is about $0.152^2$ and so on...

This gives our estimated probability as being about $4\cdot 0.152 - 6\cdot 0.152^2 + 4\cdot 0.152^3 - 0.152^4\approx 0.4828$, an astounding (to some) almost half of the time. If we were to include allowing being separated by a card, that raises $\Pr(K_{\spadesuit})$ up to $\approx 0.286$ bringing the total to closer to $0.7401\dots$

Getting an exact value for this seems like a lot of work. Approximations like the one used here should get close enough (I expect within $0.05$ of the true value).

JMoravitz
  • 79,518
0

That’s extraordinary! Thank you. I wouldn’t have known where to start! So - half the time and, with a little patter to explain how the cards came super close every other turn in 4, that is perhaps why it seems so successful. I don’t imagine that most people intuitively think it would be that likely - and why it is so easy to amaze and believe some mystical forces are at work! Of course, the mathematician is is the real magician here - but thank you for sharing the secret.

Mike

MikeG
  • 11