What are the odds?

[Bill Wheeler]

I have recently performed the "Tom Hawbecker Triple Prediction" with good results for some friends and family. A forceful or dynamic personality may be able to keep a paying audience interested.

Part of the routine requires two spectators to deal their shuffled decks of cards until they both have a "match". My question then is:

What are the odds that two shuffled decks of cards will have at least one card matching? (That is, the same card in the same location in both decks).

Wouldn't the odds improve if both decks were in the same order?

PS: The trick is in "The Magic of Francis Carlyle" the description is only a skosh better than those found in "Moe and his Miracles" which is probably why I missed it for so long.

[Tommy]

"What are the odds that two shuffled decks of cards will have at least one card matching? (That is, the same card in the same location in both decks)."

Bill, I do not know but I would guess it's close to certain. Lets put it this way: If the Ace of Spades is 3rd in one deck it is 51 to 1 that it is 3rd in the other deck. As it can only be in one of 52 places. To put it another way it has about a 2% chance of being in the same spot, a bit less. It is the same for all the other cards. Adding up each cards chances gives about 99% chance. I would think. However, to say that the Ace of Spades will be 3rd in one deck and it will be 3rd in the other is 52x52 = 2704 = 2703 to 1.

The Endless Chain principle, I think would help your chance, I think. All this is just guess work. Place two Spade Royal Flushes on top of each deck, both in order. Give each deck two riffle shuffles you are bound to get a hit in the in the first few cards.

I am not familiar with the tricks you mention, sorry if none of this helps.

[Guest]

1-1/e = 63.21%

Derangement

[Guest]

The odds are better than you think if you are only talking about a one card match. Who knows real odds for two cards...three cards...etc.?

James Washer...
not meant as an affront...could you please explain a little more your formula? I'm a math buzz and would love an explanation...that's really cool, man.

[Guest]

Yup...I remember that in a book on betting. You bet that a card will match and you'll win over the long haul.
Steve V

[Rafael Benatar]

Yeah, the Washer formula looks great. Even better if it holds true. Without attempting to get to the bottom of this, here is a hint of how complex the precise calculation can get:

Say that after the shuffling, the top card of deck A is the Ace of Spades. Given that, the chances that the top card of deck B matches should 1/52 alright. Say you got the 2H in deck B, which doesn't match.

The trouble starts when you are going to calculate the odds of the 2nd card matching. You have 51 cards left in each deck, but only 50 of them have a chance. In other words: The AS of deck B and the 2H of deck A don't have a chance of matching, and if only one of those cards turns up 2nd from the top, that position cannot have a match. There is even the possibility of both cards ending up unmatched at the 2nd position or that none of them are there (the most likely case, of course), all of which needs to be accounted for. And take it from there.

[Q. Kumber]

Originally posted by Bill Wheeler:
I have recently performed the "Tom Hawbecker Triple Prediction" with good results for some friends and family. A forceful or dynamic personality may be able to keep a paying audience interested.

Check out Ted Lesley's book PARAMIRACLES for ideas on how to make a potentially boring counting trick riveting. Look up his handling for PREMONITION. I saw Ted perform it in his full evening show in Berlin in 1994. The audience was truly riveted.

[Guest]

Instead of thinking in terms of cards (Ace of spades, King of clubs, etc.) think of the deck in terms of the numbers 1-52. If we order the deck {1, 2, 3, 4, ..., 52} we simply need to count how many of the 52! combinations are derangements of {1, 2, 3, ..., 52}.

Let's simply first. Consider a 3 card deck {1, 2, 3}. The derangements are {2, 3, 1} and {3, 1, 2}. In other words, if one of the decks was ordered {1, 2, 3} then the other deck must be ordered {2, 3, 1} or {3, 1, 2} in order for no matches to be made. Therefore in this case the probability of no matches would be 2/3! = 1/3. And the probability of at least one match = 1 - probability of no matches = 1 - 1/3 = 2/3.

The number of derangements for n elements (denoted !n) = the nearest integer of (n!/e). Let n = 3, then !3 = nearest integer of (3!/e) = 2.

For large n, the nearest integer part becomes insignificant and !n can be approximated using n!/e.

So (if you are still awake) the number of derangements for a 52 card deck = 52!/e. The probability of no matches = (52!/e) / 52! = 1/e. Thus, the probability of at least one match = 1 - 1/e.

Does that make any sense?
James Washer <- Applied Mathematics Major in College :)

[DChung]

James is right on the money. In fact, I remember this problem was first posed to me (in a slightly different form) in my first college probability course.

The exact question was: Suppose 7 people with 7 different hats are at a party. If they each randomly take a hat when they leave, what are the odds that at least one person leaves with his proper hat?

Even with just 7 people, the odds are ridiculously close to 1-1/e. So when he means large n, 52 is well more than large enough.

Cooper, be careful about adding probabilities. It just doesn't work like that. Otherwise the odds of a card being red, diamond, or a heart (which is the same as the prob of a card being red) = 1/2 + 1/4 + 1/4 = 1. Not likely. One must take the inclusion-exclusion principle into account.

The human mind has particularly bad intuition regarding probability. That's how we're wired. So one has to be very careful about these things. When people (educated or not) guess, they're generally way off.

Rafael, these things generally get less complicated if you look at them the right way. However, finding the right way is not always so easy.

Cheers,
Derrick

[Guest]

Originally posted by Derrick Chung:
The human mind has particularly bad intuition regarding probability.

I used to find that very useful (financially) when playing backgammon.

Some people would believe that the probability of throwing a double (with two dice) was the same as the probability of throwing any other combination. And, of course, it isn't.

I'm told that people used to believe that if you tossed two coins, it was twice as likely that they'd land showing identical faces as differing faces. The argument being that they could land head-head, tail-tail, or head-tail. The same error of reasoning as for the backgammon case.

The odds against you "knowing" the thought-of card(s) seem much greater for Twisted Sisters than for B'wave. But they're not.

Dave

[Rafael Benatar]

Thank you guys. Brilliant. So I guess the derangement, when applied to cards, is the number of ways the deck can be cut without altering its cyclic sequence, right?

[DChung]

That's not right, Rafael. One should not be thinking in terms of sequences and cuts. It's sort of hard to explain but I'll give it a try.

From the mathworld site that James referenced: "Derangements are permutations without fixed points."

In the context of playing cards, a derangement is an ordering of the cards such that no card is where it is supposed to be (i.e. where they should be in New Deck Order, say).

In other words, a derangement corresponds to not having any matched cards while you deal through both decks.

So what Bill wanted to know was the probability that a given deck ordering is NOT a derangement.

If this still doesn't make any sense, I'll be happy to try again.

Derrick

[Bill Wheeler]

Many thanks for all the responses. Clearly the work behind this was a lot more complicated than my class in finite mathematics could handle. Anyway, its good to know that I will only have to use an out 36.79% of the time.

Quentin: Thank you for the reference, I'll be checking that out.

[Guest]

Thank you Mr. Washer. You keep right on applying those mathamatics. I got every bit of it.
Back to school for some, eh?

[Pete McCabe]

Not to be difficult, but I think another good question to consider is what are the odds that this process of selecting a card will be meaningful, interesting, and/or entertaining to the audience?

[Q. Kumber]

Originally posted by Pete McCabe:
Not to be difficult, but I think another good question to consider is what are the odds that this process of selecting a card will be meaningful, interesting, and/or entertaining to the audience?

I would not present this as part of a regular set, rather as an encore or something 'special' and framed as 'something extraordinary'.

With such a framework it should be devastating.

[magicfish]

In Paramiracles" Ted Lesley sought the expertise of Toronto's Tom Ransom for this very same information.