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#1
08-09-2002, 06:51 PM
 Guest Posts: n/a
A Three Card Trick.

if you deal three cards off a shuffled pack they will come out in one of the following 2x2x2 = 8 possible ways:

(R = red; B = black)

RRR

RRB

RBR

RBB

BBB

BBR

BRB

BRR

if you keep dealing, they might come out, say:

RBBRB RBBBR RBBRR etc (they are only put into groups of five for ease of reading)

now here's the game:

ask your opponent to choose any one of the eight groups of three above and tell him that you'll choose one of the seven remaining groups - he can choose whichever he feels is the best group

then you'll start dealing cards off the top of the shuffled deck one at a time keeping track of the R/B order

whose ever chosen group of three is dealt out in order first wins \$1 from the other

if there has been no winner after having dealt out about half the deck, it is reshuffled - this is to try to keep the chance of an R or a B being dealt next near to 50%/50%

so, if he chose BBB and you chose RRB he would win \$1 from you, as his group came out as cards 7,8,9 before yours at 10,11,12 in the above example series

what best strategy would you use and who would win how much in this game?
#2
08-09-2002, 07:40 PM
 Guest Posts: n/a
Re: A Three Card Trick.

I would have my last two letters equal his first two. So any time he has a shot to win, i will already have had a 50/50 shot to have won and ended the game.

I can't figure out how much I would win though. I'd approximate in the 66-80 cent range per \$1.
#3
08-10-2002, 07:51 AM
 Guest Posts: n/a
Re: A Three Card Trick.

This may be a good game to experiment with the value of position.

#4
08-10-2002, 04:15 PM
 Guest Posts: n/a
Re: A Three Card Trick.

I'd do SDplayer's trick, and also make it so that my letters had both red and black in them.

For example, say SDP had "BBR" as his letters. I would choose "RBB" to cover his last two letters and have a mix...

the mix makes it so that pulling off my first card doesn't decrease my chances for getting the last two.

if it were the same game with 4 cards, you would match the other guy's first 3 and choose the other one to come closest to 50-50 red & black (any order)

--Dan
#5
08-10-2002, 08:02 PM
 Guest Posts: n/a

yes - the correct response is to make the first two of his choice the last two of yours - AND - never choose a palindromic set for your response

your wins are between 66% and 87% depending on what he chooses

if your sucker realises he is at a disadvantage you could offer to pay him \$1 when he wins but collect only 80c when you win - as long as you can get away with at least \$1 to 70c you will win in the long run
#6
08-11-2002, 05:15 PM
 Guest Posts: n/a
4 Card Trick

Haven's Solution is perfect for the 3 card game. What is the solution for the 4 card game? If your opponent chooses RBRB should you choose RRBR or BRBR? Both RRBR and BRBR obey Haven's rule:

make the first 3 of his choice the last 2 of yours - AND - never choose a palindromic set for your response.

#7
08-12-2002, 03:13 PM
 Guest Posts: n/a
Re: 4 Card Trick

Here is my question:

If player A picks BBB, then player B should pick RBB. But should he also pick RBB if he knew a third person with perfect knowledge would get in right after he does?

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