Easy Probability Problem

[jpps1]

Probability that of the first four cards dealt out of a deck, there will be three of the same number dealt, in any order (i.e. 7,7,7,x or 7,x,7,7, etc...).

[tarheeljks]

i have a feeling this is wrong b/c my combinatorics are rusty but here goes.

52 C 4 four card combos = 270,725

4 C 3 = 4 combos of three cards w/the same value (e.g. 777)

52 total cards, so 52*(4 C 3)= 52*4=208 combos of 3 cards w/ the same value

48 remaining cards

52*4*48 = 9984 combos of 4 cards with 3 being the same #.

9984/270,725= ~.0367

approx 4%


this seems a little high and like i said my combinatorics are rusty.

[Gullanian]

I would have to say thats wrong by the look of it? I dont think you see trips on the turn that often.

[CT11]

If you mean exactly 3 of the same cars (ie 7777 does not count) then probably the most clear way is to calculate
yyyx yyxy yxyy and xyyy and sum them.
yyyx: 1/1*3/51*2/50*(1-1/49)
yyxy: 1/1*3/51*(1-2/50)*2/49
yxyy: 1/1*(1-3/51)*3/50*2/49
xyyy: 1/1*(1-3/51)*3/50*2/49

Its intuitively obvious that these four probabilities will come out to be the same. Indeed they do.
the total is 0.0092.

[CT11]

There is a counting argument that may be more clear.
There are 13C2 ways of choosing two different face valued cards if order matters then there are 2*(13C2) ways.

Choose two different cards, each with different face values and make sure they are ordered. The first card you choose find the number of ways you can have three of them (4C3 = 4) the second find the number of ways you can have one of them (4C1 = 4). Therefor the number of 4 card hands that include three of a kind (but not four of a kind) is 2*(13C2)*4*4/(52C4) = 0.0092.

at least thats what google calculator gives me

[tarheeljks]

[ QUOTE ]
I would have to say thats wrong by the look of it? I dont think you see trips on the turn that often.

[/ QUOTE ]

yeah i figured it was and i have figured out what i did incorrectly. just needed to get the wheels turning. . .

[DPP LDR]

is the ? about the cards on the board or the first 4 dealt down up or otherwise after a standard wash/shuffle/cut?

[rJ_]

A slightly different method but same answer.

13* 4C3 * 48C1 / 48C4 = .0092

13 different trips x 4 of any value in a deck and we want 3 to make trips x 48 possible other cards choosing 1 of them / all possible 4 card combinations.

google calculator

rJ_

[tarheeljks]

[ QUOTE ]
A slightly different method but same answer.

13* 4C3 * 48C1 / 52C4 = .0092

13 different trips x 4 of any value in a deck and we want 3 to make trips x 48 possible other cards choosing 1 of them / all possible 4 card combinations.

google calculator

rJ_

[/ QUOTE ]

this is what i meant to do. i multiplied by 52 instead of 13, oops.