Fun Problem # 2

[LetYouDown]

Start with 13 cards, all the same suit. Choose three cards at random, and discard the highest and the lowest, while returning the middle card to the original set. Repeat with the smaller set until only a single card remains. What is the probability for each of cards you started with being last one in the set?

[TomCollins]

1-0%
2-.344%
3-1.26%
4-3.56%
5-8.95%
6-19.7%
7-32.5%
8-13, same as above, but flipped.

[LetYouDown]

How did you approach it? Computer program or number crunching? If number crunching, what were your exact numbers in fractional terms?

[TomCollins]

This was a simulation. 1 million tests. Number crunching on this seems incredibly long for each one.

[DougOzzzz]

well for 1 and 13 its obviously 0, 2 and 12 are easily to calculate. For 2 for instance, the final 3 numbers must be 1, 2 and any other number (card). So you're really choosing 3 cards from 11 to start, then 3 cards from 9, then 3 cards from 7, then 3 cards from 5, then 3 cards from 3, and again 3 cards from 3. And you divide that by the total number of combinations. what you end up getting is a bunch of cancellations and 1/13C3 or about 0.35 percent.

It gets much more complicated for 3 through 11.

[LetYouDown]

[ QUOTE ]
It gets much more complicated for 3 through 11.

[/ QUOTE ]
That's the fun part! I'm hoping BruceZ or Siegmund will chime in on this. I'd like to see how they'd approach it.

[TomCollins]

It is just tedious work. You enumerate all the ways to draw.

[LetYouDown]

I've seen a shortcut posted that led to the exact answer...which is the only reason I mentioned it.

[AaronBrown]

I get, out of 2,554:

0 for 2 and A
11 for 3 and K
38 for 4 and Q
80 for 5 and J
110 for 6 and 10
608 for 7 and 9
860 for 8

I'm not sure it's right, I did it pretty sloppily.

[pzhon]

[ QUOTE ]
How did you approach it? Computer program or number crunching? If number crunching, what were your exact numbers in fractional terms?

[/ QUOTE ]
I used recursion. I don't see a shortcut.

0
1/286
20/1573
337/9438
1264/14157
505/2574
4604/14157
505/2574
1264/14157
337/9438
20/1573
1/286
0

lud[row_, offset_] :=
lud[row, offset] =
If[row == 0, If[offset == 0, 1, 0],
If[Abs[offset] > row, 0,
Binomial[row + offset, 3]/Binomial[2 row + 1, 3]lud[row - 1,
offset - 1] +
Binomial[row - offset, 3]/Binomial[2 row + 1, 3] lud[row - 1,
offset + 1] + (row + offset)(row - offset)row/
Binomial[2 row + 1, 3] lud[row - 1, offset]]]