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rockoon
10-25-2003, 06:51 AM
Consider a standard 52-card poker deck with the suit information removed. There are 4 aces, 4 kings, 4 queens, and so forth down to 4 dueces.

How many ways (combinations.. order does not matter) can you draw 7 cards from this deck?

well
10-25-2003, 08:55 AM

Let a=(a1,a2,a3,a4) denote a combination of 7 cards consisting of a1 singles, ..., a4 quadruples.
Here's the list of all those formats.
The right column shows the number of combinations for each format. Summation on this column produces the answer.

7 0 0 0 | 1716
5 1 0 0 | 10296
3 2 0 0 | 12870
1 3 0 0 | 2860
4 0 1 0 | 6435
2 1 1 0 | 8580
0 2 1 0 | 858
1 0 2 0 | 858
3 0 0 1 | 2860
1 1 0 1 | 1716
0 0 1 1 | 156

Sorry for the simplicity of my answer, but this took less time than coming up with an answer in the form of a more general formula. Besides that, I am trying to get quicker in using matlab...

Next Time.

Nottom
10-25-2003, 03:41 PM
This doesn't seem right, if I'm reading this correctly you are nearly 3 times as likely to draw quads as you are to draw no-pair.

No-Pair:
7 0 0 0 | 1716

3 0 0 0 | 2860
1 1 0 1 | 1716
0 0 1 1 | 156

Am I interpreting this correctly?

well
10-25-2003, 04:26 PM
[ QUOTE ]
This doesn't seem right, if I'm reading this correctly you are nearly 3 times as likely to draw quads as you are to draw no-pair.

No-Pair:
7 0 0 0 | 1716

3 0 0 1 | 2860
1 1 0 1 | 1716
0 0 1 1 | 156

Am I interpreting this correctly?

[/ QUOTE ]

No, not completely.

The figures you quote tell you that there are nearly 3 times as many combinations for quads as there are for 7 distinct ranks (there are also straights in these combinations, so make it more like 2.4 (guess)).

Although I think this is what you wanted to know, I'll show you how you do get probabilities.
If you want to know how likely you are to draw quads, you must know how likely you draw each of the distinct combinations!
For (7 0 0 0) multiply by 4^7, for each single card can be drawn in 4 ways.
For (1 1 0 1) multiply by 4*6, for each double can be drawn in 6 ways.
In general, for each non-zero element ai of (a1 a2 a3 a4) you should multiply the number of combinations by ai times comb(4,i).

Next Time.

Nottom
10-26-2003, 12:39 AM
[ QUOTE ]
The figures you quote tell you that there are nearly 3 times as many combinations for quads as there are for 7 distinct ranks (there are also straights in these combinations, so make it more like 2.4 (guess)).

[/ QUOTE ]

So are you confirming the fact that there are more combinations containing 4 of a kind than containing no-pair (straights and flushes included)?

well
10-26-2003, 08:05 AM
[ QUOTE ]
So are you confirming the fact that there are more combinations containing 4 of a kind than containing no-pair (straights and flushes included)?

[/ QUOTE ]

I do not understand what you mean by flushes, since the suitproperty of the card has been discarded.
But I hereby confirm that indeed there are more distinct combinations containing a four of a kind,
than there are distinct combinations containing no-pair (straights in- or excluded).

I see that you have difficulty understanding this.
Let's look at two combinations, these are no poker combinations, but 7-card combinations:

(a) Four of a kind fives with three tens on the side.
(b) A 'straight' from five to ten with an ace.

Both combinations have been counted exactly once in the previous posts.

For (a), there are 4 ways to draw this for there are 4 choices for the tens to pick.
Or: one of the four tens is left in the deck.

For (b), there are 4*4*4*4*4*4*4 ways to draw this for there are 4 choices for each rank.

I hope you see now that you should not confuse these numbers of distinct combinations with the probabilities to draw them.

Next Time.