This is a simple sounding but utterly counter-intuitive problem.The following explanation is the clearest I have found -so much so that even my dense head can now fully understand. Might this principle be somehow incorporated into a magician's routine? Is there anything already out there that uses an idea such as this?
??????

http://www.youtube.com/watch?v=mhlc7peGlGg&feature=related

Scott Cram posted this a while back. Now we can try it out for real to see if the principle works in action!

http://www.nytimes.com/2008/04/08/science/08monty.html?_r=1#

I have had a think myself about how to incorporate this subtle principle into a useful effect. Originally I was thinking around a 'Hummer 1-2-3' type procedure with this added on as an extra phase.However I kept coming back to the reality that this principle-when all is said and done- only provides us with a 66.66% average chance of success,in other words not solid enough to stand on it's own.

The following idea may hold promise though. It might be worth utlising the principle in conjunction with one of three playing cards being very subtly marked.

Look at the backs of any borrowed deck and you will be sure to find a card that is naturally distinguishable by virtue of a mark, piece of dirt/stain or a slight crease. Use this to your advantage.Failing that use your own cards.These should preferably be three blank backed playing cards.Very slightly blunt the lower right corner of one three blank backed playing cards with a nail clipper.This should be done so that all the card can be closely examined but you yourself can distinguish it from a few feet since you alone know what you are looking for.

Follow the procedure for swapping your choice once one of the dummies has been eliminated from the choice to leave the two cards.The fact that the cards are blanked backed should allay all suspicions that the cards might be marked.However you might want to add yet another layer of deception to the proceedings.Make the habit of sometimes turning your back whilst the cards are mixed about and choose the card with your back still turned.On other occasions you might simply choose to keep your eyes closed during both the initial mixing and the choosing.Communicate your choice verbally to the spectator.I would advise that you undertake these techniques/ploys only after a winning choice has been made previously.

When (as you will on average about 33.33% of the time)you make a wrong choice you then keep your eyes open for the very next choice or two so as to neutralise this loss by slyly noting the location of the marked target card and then winning for certainty for the next couple of rounds.Note that on these few ocassions you purposely DO NOT swap your final choice thus inherently disguising the fact that you have been doing so as a general rule.

Your overall success percentage will inevitably rise up and up to WELL ABOVE 80% or more over time.

I would personally choose to undertake this effect using blank backed colour cards.Red, blue and green, with the red faced card subtly blunted/dulled at one corner.The fact you are using colour lends itself to an interesting presentation centred about 'psychic visualisation' if you so choose.

This procedure:-
1)Strengthens the basic principle in practice moving the percentage of success to well above what was formerly possible (ie on average 66.66%),
2)The use of colour adds interest to the presentation and adds to the counter-intuitive flavour of the effect,
3)For most of the time you appear not to be looking at the cards either standing with your back turned or with your focus fixed on the ground as if looking deep within.You might even choose to be out of the room on occasion whilst the cards are being mixed up and whilst you choose by seemingly communicating your 'psychic' choices from afar.Those watching will be left puzzled as you keep succeeding against all odds.There will be no rational explanation for this.
4)There can be no suspicion of the cards being marked.
5)Subtle disguise to the fact that you always seem to be swapping your initial choice.

Anyway this is the best I can do with this mathematical puzzle! I do hope the above has not been too confusing!!

Can't remember if this new Oxford University Press book has already been mentioned in this forum:
Monty Hall

Understanding the Monty Hall Problem and its implications makes a person a significantly better bridge player.

Andy,

That's the best idea I've read so far in turning this into mental magic, with all the methods cancelling eachother out nicely. Thanks.

Regarding Hummer's "Mental Monte", one of the best uses I've seen for it is a trick near the beginning of the book Ahead of the Pack by Jack Avis and Lewis Jones called "The full monte". It's not mentioned in that trick but you can allow the spectator to randomly move the three cards around both before and after the selection procedure, which I think adds to the trick.

/Tomas

Thanks to you all.

Tomas "The Full Monte" sounds strong. I agree with you that randomly moving the three cards about seems to be most desireable in such a routine or otherwise the 'moves' seem too stilted. The Hummer procedure -whilst it might use a different methodology- does reach the same end as that of the Monty Hall routine.You could even use the same colour card idea with one of the blank backs being very subtly blunted (nail clippers) or dulled (alcohol) at one extreme corner.Indulge me for a moment.

1)A spectator mixes the three cards randomly face down on a table whilst the magician's back is turned,
2)The magician turns around (naturally to communicate with the spectator as to what he wants them to do next) and whilst pointing at the cards makes a sly peak of the location of the target card,
3)The magician turns his back again,
4)The spectator is instructed to turn any of the three cards face up (her free choice)and to remember the colour showing.The rest of the people around the table importantly can also see what the colour is.Then the spectator is instructed to place a hand each on the other two remaining cards and to turn these two face up swapping their positions at the same time.The fact the card are moved about face up means that everyone can feel a part of it.
5)The pretext for the guided moves might be so that the magician can visualise 'in their mind's eye' the colours as the spectator's hand moves them around.This gives a rational for what would otherwise be the weakness in the routine,
6)The rest of the procedure of moves flow on smoothly (without hesitation) ending up importantly with the face up cards visually seen to be in a different colour order than that at the oft. An order that looks random.
7)The spectator is then instructed to turn all the cards face down again,
8)Only then does the magician turn around again for a brief moment as he instructs the spectator to mix up all the cards once again. He makes a gesture of mixing the cards up with his hands waved towards the three cards for a moment before he turns around again. Thus he gains his second sly peak and with it the necessary information,
9)The three cards are then mixed up again randomly face down on the table by the spectator whilst the magician casually half turns to speak to the others around the table.He is not seen to even look at the cards as the actual mix takes place.
10)The spectator might then be instructed to visualise their colour as the magician tries to link into their mindstream.The magician might even fein difficulty and ask for everyone around the table for their support-to participate in the visualisation.This would make for good audience rapport.

As you can probably tell by now tell I like using colour !!

Andy.

You know, I was at Barnes & Noble the other day and I came very close to buying this book. But at the last minute I switched to another book.

Ron
:)

I don't know what the routine would be, but I know what it would be called...

Three Card Monty

===================

Seriously, I can't imagine a way to incorporate the principle as the way of effecting a routine, since it just increases the odds of something occuring, but doesn't make it a 100% sure thing.

However, I think the principle might make a good story in some type of a modified 3-Card Monty (sic) routine used to explain why the magician always is able to pick the money card, but the spectator isn't.

I have just put up a trick based on the Monty Hall Problem in the Inner Thoughts forum. Of course it isn't sure-fire but is quite deceptive nonetheless.

http://www.themagiccafe.com/forums/viewt......rum=82&0

Mike

Perhaps you should restrict your choice of partners to those who understand the problem and its implications.

;)

Heh heh...just saw that one, Bill.

I knew you'd like that.

This thread is 1 year old but I have a way to do the exact "Monty Hall" plot:
3 business cards are put on the table.
The Spectator can examine them. 2 of them have a goat drawn on the back ( or any other booby LOSE prize ) and the third one has a Ferrari on it (or any other WIN prize of high value). The cards are mixed by another Spectator from the audience .The Magician takes them back and aligns them on the table after looking at their back so that he is the only one to know where is the WIN prize .
The Spectator chooses one of the cards, WITHOUT LOOKING AT IT. He can change his mind.
The Magician turns a LOSE card face up , and invites the Spectator either to keep the card he has in hand, or to swap it against the unturned remaining one .
Whatever he does, he will always lose.

Sadly not true.

Mike

I think the poster was saying that in his trick, the Spectator always loses. He's just borrowed Monty's plot, not the actual experiment.

OK, if that's the case, I apologise.

Mike

It will work 100% (not 66,66%)

The secret is now available at a very special price on Pablo Amira's web site(and only here , with 4 other effects .)
For more details, go to:
http://amusingmysteries.blogspot.fr/

Gerry

I recently helped a friend work through the Monty Hall problem in a visual way, and thought you might enjoy seeing it here.

I explained it by building a tree, and then pruning the tree appropriately.

Building The Tree

We start with the probability of where the car itself is located. It could be behind door #1, #2, or #3, and has a 1 in 3 chance of being behind any given door, so we draw a tree diagram like this.

Next, we account for the fact that the player, unaware of where the car is located, also has a 1 in 3 chance of choosing door #1, #2, or #3. We add those probabilities to the diagram like this.

Now, the host (who knows where the car is located) is going to show you a door with a goat behind you to tempt you to switch. We know he'll never open the door you chose, since he needs to build suspense, and he'll also never show you the door with the car behind it for the same reason. So, if you chose the door with a car behind it, the host can show you either of two doors, and has a 50% chance of choosing either of those. We put that in the diagram like this.

If you choose a door with a goat behind it, then another door must have the car behind it. As mentioned, at this point the host must show the only remaining door with a goat behind it. For example, if you choose dorr #1, and the car is behind door #2, the host can ONLY build suspense by showing you door #3. We add those possibilities to the diagram like this.

Now, as with any tree diagram involving possibilities, we need to find the probabilities for every possibility, and we do that by multiplying the probabilities of each branch. For example, the odds of the car being behind door #1 (1/3) AND the person choosing door #1 (1/3) AND the host showing door #2 under those circumstances (1/2) is (1/3)*(1/3)*(1/2), or 1/18. Calculating and filling in the probabilities of every possibility, the diagram winds up looking like this.

If you add up all those probabilities at......up to 1. This is just some simple double-checking, and it looks like we've done everything right so far.

Now we've got the model prepared, it's time to use it to answer some questions.

Pruning The Tree

While playing the game, two new pieces of information are provided: Which door the player chooses, and what door the host shows. To update the tree when this new information is provided, we can prune the tree to see what effect this has, and examine the remaining percentages.

As an example, let's say the player chooses door #1, and the host reveals a goat behind door #2.

To find out what happens when we don't switch, we ask: "Given that the player chooses door #1 and given that the host reveals door #2, what is the probability that the car is behind door #1?"

OK, let's trim the tree to answer this question. First, let's remove all the possibilities that don't involve the player choosing door #1. At this stage, the pruned tree looks like this.

Next, we need to also prune the tree of any branch that doesn't involve the host showing door #2. The removes a couple of more branches from our tree.

Since the host has shown there's a goat behind door #2, we can trim away any possibility involving the car being behind door #2, leaving us with these possibilities.

Only 1/6 (1/18 + 1/9) of the original possibilities remain, and only 1/18 of the original possibilities involve the car being behind door #1. To work out the possibilities, we divide 1/18 by 1/6, which gives us 1/3.

In other words, given that the player chooses door #1 and given that the host reveals door #2, the probability of the car is behind door #1 is 1/3.

From the diagram, it's pretty easy to see that the probability of the car being behind door #2 is 0. The fact that the host showed a goat behind this door supports this.

What is the probability that the car is behind door #3, given that the player chose door #1 and the host showed door #2? In other words, what if you switch?

There's the shortcut, subtracting the probabilities of the other two possibilities from 1 to get this probability: 1 - 0 - 1/3 = 2/3.

Or, we could refer back to the trimmed diagram, and note that the same 1/6 of the original probabilities remain, but the probability of the car being behind door #3 is 1/9 of the original probabilities, so we work out 1/9 divide by 1/6, for a probability of 2/3, the same answer via a different approach!

Going back to the complete diagram, it's not hard to see that any pruning of the door the player chooses and the door the host shows (always 2 different doors, of course), there will always be a 1/9 and 1/18 probability remaining, so the math will always work out like above.