Hello everyone ,

I have spent quite along time coming up with a premise for an effect that relies on a self working method which can be found in Aldo Columbini's impromptu card magic , card colledge and even Scarne on card tricks.

If we assume that all the picure cards, the the procedure is as follows:

The performer holds the deck in hand and in his mind as the mental count of 10 . He deals the top card of the deck face up onto the table . If the faceup card has a 10 value , the dealing is stopped and a second pile started . If the faceup card is not the same value , the mental count is dropped to 9 and the next card turned face up and placed on the previous card . Again if the value is a 9 it matches the count the dealing for this pile is stopped and the procedure started again with a second pile starting from a mental count of 10. If the value and mental count do not match dealing continues on the first pile until 10 cards have been dealt face up and the count has reached 1 . A facedown card is now placed ontop of this pile .

The procedure is started again with a second pile and then a third pile .

At the end the outcomes can be 3 piles with a face up card on each , 2 face up cards and a facedown card , 1 face up card and 2 facedown or all 3 facedown .

My question is, what is the probability of each of the outcomes occuring

Any help appreciated

Andy

Sorry the title should have said .

Help sort for a maths problem

I think this is a tough problem, a lot of things to take into account. Unless someone has a nice shortcut, I think the best way to tackle this is to do many computer simulations and then see the experimental results. TomasB on this board is particularly good at running these simulations. You might want to send him a PM if he doesn't see this thread first.

Andy,
Your second sentence reads:

"If we assume that all the picure cards, the the procedure is as follows:"

Could you please clarify?
Leonard

Hello Leonard

If we assume picture cards have the value of 10

Andy

Thanks for the suggestion landmark

Starting with a "wrong" assumption that after each card is dealt it is returned to the deck and the deck shuffled, I calculate the probability of each pile ending face up to be just less than 2/3. The probability of a pile ending up face down is then just over 1/3. I will refine these calculations as time allows to account for cards already dealt. I think there are 1,000 (=10^3) case to consider.

Assuming that P(face up) = 2/3 and P(face down) = 1/3, I find the following:

P(3 face up & 0 face down) = 1 * (2/3)^3 = 8/27 = 0.296
P(2 face up & 1 face down) = 3 * (2/3)^2 * (1/3)^1 = 12/27 = 0.444
P(1 face up & 2 face down) = 3 * (2/3)^1 * (1/3)^2 = 6/27 = 0.222
P(0 face up & 3 face down) = 1 * (1/3)^3 = 1/27 = 0.037

I hope this helps.
Leonard

Leonard, even with that "wrong" assumption, I'm not clear how you got those probabilities. Could you explain further, how you got the probability of 2/3 for a face up pile?

Landmark,
Here is what I did. Start with:

P(x) = probabilty of thinking of "x" and dealing a card of value "x" at that time.
There are 16 cards of value 10, and 4 cards of each other value (1 through 9), so that:

P(10) = 16/52 = 30.8%
P(9) = (1 - P(10)) * 4/52 = 5.3% [no match followed by a match]
P(8) = (1 - P(10) - P(9)) * 4/52 = 4.9% [no match, no match, match]
.
.
.
P(1) = (1 - P(10) - ... - P(2)) * 4/52 = 2.8%

The sum of P(1) through P(10) is the probability of the stack remaining face up (one match), and is equal to 66.3%.
I think the next step would be to update the individual probabilities as cards are flipped, and either match or don't match.
Leonard

Hi there.

This is an interesting problem, and the probability involved is not particularly easy. I haven't had a chance to think about it properly yet, but in the time that I have been thinking I haven't managed to come up with a 'clever' way to make the calculation simpler.

In the meantime I thought I'd run the simulations for you. In the case you are interested in (treating court cards as 10s) the probabilities work out at approximately

0.0395 0.2294 0.4443 0.2867

(reading left to right is 0 'successes' to 3 successes). So the case that no matches are made should happen only around 4% of the time.

When I perform 'Numerology' (the effect from Card College which uses this procedure), I still count court cards as 11,12,13 etc...and leave them in the deck... in this situation the probabilities become approximately

0.0906 0.3347 0.4125 0.1622

And finally in the case that you remove the court cards from the deck (so you have a pack of 40 cards, Ace-10 of each suit) you have

0.0445 0.2403 0.4428 0.2724

One point of note for anyone who is wanting to do the calculations explicitly, this last form will be the easiest to prove... and the original form of the question is the hardest... although a nice proof should work for decks of arbitrary many cards etc.

Owen

PS. Interestingly these match quite closely to the approximations Leonard made.

To all thhose who haev heped sort thus out my sincerest appreciation

When I have fully written up the scripted version of my effect , I shall send each of you a copy

kind regards

Andy