Fun Problem # 3

[LetYouDown]

Standard deck of cards. You shuffle them and draw cards until you have three of a kind. What is the most likely number of cards drawn when this happens? What if you shuffle two decks together?

[r3vbr]

hmm let me make a wild guess... about 15
no math done hehehe tell me if im close
with 2 decks its slightly less cards. id guess 13

[moomoocow]

I could not solve the problem analytically, but using a simulation - I'm getting roughly between 13 and 14 cards - so ... excellent guess!

Moo!

[r3vbr]

good to know /images/graemlins/smile.gif

[AaronBrown]

There is a 10.35% chance of getting on the 14th card, that's the maximum. You have a 55.06% chance of getting it on the 13th card or earlier, with only a 44.77% chance on the 12th card or earlier. So the median wait is between 12 and 13 cards.

I solved this by representing the state as (a,b,c) where a is the number of ranks for which you have zero cards, b is the number of ranks for which you have one card and c is the number of ranks for which you have two cards. You start at (13,0,0) and move to (12,1,0). After that you have 3/51 of going to (12,0,1) and 48/51 of going to (11,2,0). Your expected state is (11.0588,1.8824,0.0588).

This is tedious by hand (and to type) but simple in Excel. You run the expected state forward (conditional on not having a three to date) to figure out the probability of getting your first three on any given card.

[nycplayer]

tried to work out the math.
For 5 cards dealt:
(13*((4*((48!)/((2!)*(46!))))+(((49!)/((2!)*(47!)))-((48!)/((2!)*(46!))))))/((52!)/((5!)*(47!)))
I get: 0.0228091236
This seems correct (see http://www.poker1.com/mcu/tables/Table12.asp )
my number is slightly higher because it also counts quads and boats.

For 13 cards dealt:
(13*((4*((48!)/((10!)*(38!))))+(((49!)/((10!)*(39!)))-((48!)/((10!)*(38!))))))/((52!)/((13!)*(39!)))
I get
0.569939976

For 14 cards dealt:
(13*((4*((48!)/((11!)*(37!))))+(((49!)/((11!)*(38!)))-((48!)/((11!)*(37!))))))/((52!)/((14!)*(38!)))
= 0.712268908

I'd probably be really confident after 15, but I'm not as confident of my math.

[nycplayer]

14 can't be the max:
A/images/graemlins/spade.gif 2/images/graemlins/spade.gif 3/images/graemlins/spade.gif 4/images/graemlins/spade.gif 5/images/graemlins/spade.gif 6/images/graemlins/spade.gif 7/images/graemlins/spade.gif 8/images/graemlins/spade.gif 9/images/graemlins/spade.gif T/images/graemlins/spade.gif J/images/graemlins/spade.gif Q/images/graemlins/spade.gif K/images/graemlins/spade.gif A/images/graemlins/heart.gif

only 1 pair, no trips.

26 is the absolute max, at 27 you must have trips. (This is extremely low probability)

But I got numbers similar to you for probability (~50% at 13), see my math, above.

[AaronBrown]

Sorry. I meant the 14th card has the maximum probability of being the one you hit on, not that it was the largest possible number of cards to deal without a three of a kind.

[nycplayer]

Oh, I see. My numbers went over 100% chance around 15, so I guess it was the same thing - more likely to already be there than not.
I wish I understood the math better.