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The Magic Cafe Forum Index » » Magical equations » » An old bar bet - how can we apply it to an effect? (0 Likes) Printer Friendly Version

Nir Dahan
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Take a deck of cards.
Shuffle a few times.

Deal 3 cards face down on the table, peek at the cards, and turn over 2 of the same color (either 2 reds or 2 blacks.) It is always possible to turn over 2 of the same color.
What are the odds of the other (face down) card being of the opposite color to the two face up cards?
The answer is a bit counter-intuitive and is 3/4.
(Just list all the cases and check for yourself.)

The question is, “how to apply it in an effect?”
I thought of doing it as a sort of telephone effect with the entire deck.

n
Ben Blau
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I can't seem to recall where I read this, but there is another interesting statistical challenge that is secretly in the performer's favor:

Have the spectator shuffle a deck, and cut it into three piles. The bet is simply that there will either be a jack, a four, or an ace on top of one or more of the piles.

If the deck is repeatedly shuffled and cut as described, the performer will win more often than he will lose. (Though there would seem to be abundantly more ways to lose and only a few ways to win).

The thing about using this in a magic effect is simply that the statistical advantage won't become impressively noticeable until several rounds are completed.

Ben Blau
Nir Dahan
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Ben,

I use this old bet (from one of will Dexter books I believe) as a phone effect.
I call it "The Moroccan Poker."
With you having 3 cards and the spectator the rest the chances are around 65% in your favor.
With 4 cards for you the chances for your winnings go as high as 94%.
To calculate the exact probability is not an easy matter, so I wrote some computer simulations and converged to these values.
Anyway, back to the effect. I present it as the way they play poker in morocco, and ask the guy on the other side of the line to do all the shuffling dealing and so on.
I usually use 4 cards for me just to play it on the safe side.

nir
Larry Barnowsky
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Is that really so counterintuitive? Of the 8 outcomes, 2 have all the colors the same. So, 6/8 times the other color will not match the like pair. That gives you the probability of 3/4 as you stated. I think it is interesting but not something that would be the basis of a baffling magic effect.
landmark
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Nir,

I'm not clear on what the effect is for what you describe as Moroccan Poker. Are you talking about Ben's three pile bet? If so, the correct odds of getting at least one King, Queen or Jack on top of one of three piles is 55%. The odds of getting an Ace, King, Queen, or Jack on one of the three piles is 68%. I don't want to bore anyone with the details, but if you or anyone else wants to PM me, I'll show you the math--it's not that hard. Interestingly, if you cut and make four piles, the probability of getting an honors card--Ace, King, Queen, Jack or 10 is over 98%.

Jack Shalom
Nir Dahan
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Jack,

I was talking about making 3 piles and doing repeating rounds of the game. In each round the top 3 piles are exhausted.
You continue that until one of the piles is gone.

I am sorry it wasn’t clear.

Then the odds are hard to calculate. Of course in the single FIRST round it is very easy.

Nir
sashain
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Similar to Nir's case, but much more interesting I think, is the Monty Hall Paradox. With three cards it would go like this:

Three cards face down, one is a queen. Spectator selects a card (puts finger on it) but does not look at it. Dealer turns up one of the other cards and shows it is not the queen. The spectator is given a chance to switch his/her choice to the other face down card. Should the spectator stick with the initial choice or would his/her odds of winning increase by switching?

The intuitive answer is that the odds on sticking started at 1/3 for the choice made from the 3 cards, and became 1/2 (even) for either of the two cards remaining cards. Therefore it doesn't matter if he/she switches.

The correct answer is that the odds improve greatly by switching. The queen is most often under the card that was not selected and was not shown by the dealer (2 out of 3 in the player's favor).

When published (as the Monty Hall Problem, which door has the prize behind it) in the New York Times some years back, many university math professors wrote in to defend the incorrect solution.

Steve
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MichelAsselin
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Ben, I remember something like that in "Scarne On Cards", which I do not have on hand anymore. But I am sure of it.
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rgranville
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Quote:
On 2003-08-23 17:42, The Original Countelmsley wrote:
Is that really so counterintuitive? OF the 8 outcomes, 2 have all the colors the same. So, 6/8 times the other color will not match the like pair. That gives you the probability of 3/4 as you stated. I think it is interesting but not something that would be the basis of a baffling magic effect.


Counterintuitive to whom? Many people (incorrectly) reason that the face down card is either red or black, and therefore the probability of being opposite of the two face up cards is .5. And indeed, if one were to deal out three cards face down, turn up the first, and then turn up the second, the probability fo the third card being a specified color is .5. I admit I don't see a basis for magic here, either, but that could just be my limitations.
Smile
Nir Dahan
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The Monty Hall problem is always a great one to start a debate with.
Zeiros
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How about staging the whole thing as a test of their psychic abilities, much like those 'trying to guess which symbol is on a card' but with 'same or different'. Go through the whole deck, dealing out three cards, turning over two and asking the question: 'Same or different?' This obviously needs a lot of work to figure out a way of getting it to flow smoothly but with the unexpected probabilities it has the potential to be quite confusing.
Greg Owen
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I think the three card/red/black setup can be VERY deceptive. Many are taught the gambler's fallacy - that there are NOT "runs" and that previous trials do not affect the outcome of future trials. That is, a run of blacks on the roulette wheel does NOT mean that "it is now time" for a red.

While this is not what is happening when two cards the same color are turned up, many will reason that the color of the third card should be independent of the color of the other two (they are RANDOMLY chosen) and so it is 50/50 for red/black.

How about this: Ask everyone to concentrate on the thought, "DIFFERENT". I need everyone to THINK DIFFERENT (can you see a corporate use for this?).

Now have someone shuffle the deck as everyone is thinking DIFFERENT to influence the final position of the cards.

To check on how effective the group was, deal out the deck in 3's and proceed as in the above post.

The group is very likely to succeed in influencing the cards to beat 50/50 by a good margin. And a round of applause for the audience!

- Greg Owen
Author of The Alpha Stack ebook - the balanced memorized stack
gobeatty@yahoo.com
Roy J Hopwood
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Pages 216 - 220 in "Scarne on Cards" describe "Ace - Duece - Jack"

Scarne sums it up thus:

"When a man bets that the Ace, Duece, or Jack will appear in the three card combination comprised of the three bottom cards of the cut piles in this game, the chances are 12,220 to 9,880 in his favor.

That is, the percentage in his favor is 10 1/123."

Hope this helps.
Take Care

Roy J Hopwood
Magical Entertainer



"He who stops being better stops being good."

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LobowolfXXX
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Quote:
On 2003-08-23 17:42, The Original Countelmsley wrote:
Is that really so counterintuitive? Of the 8 outcomes, 2 have all the colors the same. So, 6/8 times the other color will not match the like pair. That gives you the probability of 3/4 as you stated.

That's not intuition; it's math. Talk to someone who doesn't understand probability theory. It's counter-intuitive. To someone who doesn't know the math, it's 50-50.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley.

"...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us."
balducci
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Umm ... speaking as someone with advanced training in the field of mathematical statistics I would like to state that the probability of 3/4 given for the original problem is not quite correct if you are using a deck of 52 cards, with 26 blue and 26 red.

Yes, there ARE 8 possible outcomes and 6 give you the desired result BUT if you think about it carefully you will observe that the eight outcomes are NOT equally likely. Getting 3 blue cards is less likely than getting 2 blue and 1 red, for instance.

The correct probability works out to be 16900 divided by (5200+16900) which is about 76.47059 percent Smile

Again, this assumes a regular deck of 52 cards as stated in the inital post.

P.s. If you tell the spectator that you are using 3 cards and each is equally likely to be red or blue, then the originally stated result of 3/4 would be correct.
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