Poker Combination Calculation Exercise

I think it’s 1745 .

The lowest possible nuts-on-the-board is the broadway straight.

There are 13 quad combinations and 4 straight flush combinations.

There are no full house combos because quads are always possible if the board is paired, and no other flushes besides the four straight flush combos.

That’s the easy part.

For the broadway straights, there are two basic suit patterns where no flush is possible: xxyyz, and xxyzw.

For xxyyz, there are 16 ways for the suits to distribute across it and 27 ways for the ranks to distribute across of each of those. That makes 432 combinations.

For xxyzw, there are 24 ways for suits to distribute across it and 54 ways for the ranks to distribute across each of those. That makes 1296 combinations.

So summing it all together:

4 straight flushes
13 quads
432 xxyyz broadway straights
1296 xxyzw brodway straights
—————————–
1745 nuts-on-the-board

Did I get it right?

There are 4 different royal flushes. There are 13 ways to put four of a kind with the top kicker on the board.

I think that covers it. So 17 / (52 choose 5).

I don’t know remember how to find 52 choose 5. I should have paid more attention in 6.042.

I believe there are 4 ways to make each of the 13 different 4 of a kinds. 4 aces has 4 different king kickers, and all of the rest have 4 different ace kickers, no?

I’ll take the 1st shot at this:

There are 52×51x50×49x48/5! = 2,598,960 combinations of 5 cards if you ignore order.

The only way to get the nuts on the board are:

royal flush = 4 combinations (1 in each suit)

quads with an ace kicker = 48 combinations (12 quads x 4 suits for the ace)

quad aces with a king kicker = 4 combinations (one for each king)

Now for the hard ones:
AKQJT straight with all 4 suits represented. Two of the cards have to be of the same suit, so there are 10 combinations of which two cards are suited (AK,AQ,AJ,AT,KQ,KJ,KT,QJ,QT,JT). Then there are 24 ways to divvy up the suits between the 4 card sets. 10×24=240 combinations.

AKQJT straight with 2-2-1 suit distribution. There are 5 choices for the singleton card. Once the singleton has been established, there are only 3 ways to pair up the other 4 cards. For example, if the singleton is the T, you can pair the ace with the K, Q, or J. That leaves 5×3=15 ways to combine the cards without assigning specific suits. Then, we again have 24 ways to assign specific suits. 24×15=360.

Note that these are the only way to have the nuts on the board. This is because:
1) Straight Flushes are never the nuts, they can always become better straight flushes.
2) Full houses are never the nuts as they can become quads
3) Flushes are never the nuts unless they are a royal flush
4) straights are never the nuts unless they are AKQJT.
5) anything less can become trips or quads.

So the total combination of nuts = 4 + 48 + 4 + 240 + 360 = 656

656/2598960 = 0.025% of total boards are the nuts, or 1 in every 3962 boards.

On the face of it, it sure seems like I see those AKQJT straights on the board more often than that, but I think the math is correct. Feel free to correct any mistakes.

I guess my reply didn’t get logged, which is good because I noticed an error or two.

The number I believe is 1784.

13 quads with 4 different suited kickers each and 4 royal flushes.

Easy peasy.

The hard part is counting the broadway straights.

There are 2 basic suit patterns for a straight where no flush is possible: xxyyz and xxyzw.

For xxyyz straights there are 16 ways to distribute the suits across them and 27 ways to distribute the ranks across each of those. That makes 432 combinations.

For xxyzw straights there are 24 ways to distribute the suits across them and 54 ways to distribute the ranks across each of those. That makes 1296 combinations.

So adding it up:

4 royal flushes
52 quads
432 xxyyz straights
1296 xxyzw straights
———————-
1784 nuts-on-the-board

Did I get it?

If I recall this correctly, we can find the number of possible hands by using the COMBIN(x,y) function in Excel, where X is the total number of cards, and Y is the number drawn. This formula is derived from the following equation:

X!/(Y!*(X-Y)!)

In this case,

52!/(5!*(52-5)!)

= 2,598,960 possible combinations

Now comes the task of finding how many board combinations make the stone cold nuts. Here’s the easy ones:

Royal Flush in all four suits = 4

Quad Deuces through Kings with each of the four Aces = 48

Quad Aces with each of the 4 Kings = 4

After these, the only nut hands that remain are broadway straights. A straight can be made with a 5 through Ace as the high card. Here we’re concerned only with the Ace version. The number of ways to arrange any specific 5 cards using 4 different suits is 4 to the 5th power, or 1024. We also need to account for the 4 possibilities of a royal flush, which we already counted. So we subtract those and get 1020.

Number of possible broadway straights = 1020

That’s pretty much it for nut hands on the board, according to the following logic:

On Board Possibly beat by holding

No Pair Match board card to pair
Pair Two pair
Two Pair Higher pair or boat
Trips Full House, Quads, etc.

You get the idea. So if we add up all the nut hands we get:

4 + 48 + 4 + 1020 = 1076 possible nut hands

1076 possible nut hands/2,598,960 total hands

= 0.000414011759

So I guess my answer is that around four 10,000ths of all possible boards result in the nuts. All this said, I reserve the right to have my math proven horribly wrong by those much smarter than me.

–TFGoose

Toonces,

That’s the same reasoning I came up with (before reading your answer), so I think that part is right.

I also confirmed your broadway counting using a different method.

I did basically the opposite of what you did. I took all possible broadway combos, and backed out the 3 flush and 4 flush boards (so my answer includes the royals you counted separately).

I count the total number of broadway straights as 4^5 == 1024.

For the 3 flush boards, I did:
number of ways to arrange the single suited cards = 5C3 = 10. Now, that can be any suit (multiply by 4), and the last 2 cards can be any of the remaining suits (multiply by 3 twice, as we don’t care if they are the same suit or not). So, 10*4*3*3 = 360.

For the 4 flush boards, using the same logic, we have:
5C4 = 5, and we have 5 * 4 * 3 = 60.

So, we have 1024 – (360+60) = 604. Including the 4 royal flushes. And 52 sets of quads + nut kicker = 656.

So the spammer’s replies get through but mine don’t? What gives?

Ack, yep, I did pretty much the same thing as you guys, but forgot to take out the 3 and 4 flush boards from my broadway straight calculation.

–TFGoose

I think the way to calculate it is much easier. You only have to calculate the broadways straights which is 4^5. This gives us all the A-T combinations including royal flushes): 1024 possibilities. The rest are quads with the nut kicker: kings to deuces with an ace: 12*4. plus Aces with a king: 4 more hands. summa summarum 1076 hands.

What boards apply and how many:
* Royal flushes: there are 4, one in each suit
* Rainbow top straights: AKQJT, no three of each suit. There’s 480 of those (see below).
* xxxxA quads (LOL Kristy Gazes) = 4 x 12 = 48
* AAAAK = 4

Why 480 straights? AKQJT can come up in 4^5=1024 ways. But not counting are:
(*) 4 royals
(*) 60 4-flushes
(*) 480 3-flushes

So there are 4+480+48+4 = 536 nut boards.

The total number of boards is 52×51x50×49x48/5×4x3×2x1 = 2,598,960

So the percentage of nut boards is:
536 / 2,598,960 = 0.0206236 %
That’s about one in every 2 in every 10,000 boards. Which means I should have seen about 50 in my lifetime now, but it seems much more ..

Fadaham,

I keep coming up with 360 3-flushes. The missing 2 card combinations to broadways =10.

There are 9 ways this can occur without making a 4 flush. The scenario for each of the 4 suits in the 3 flush, and I get 360 ?

I just read all the comments. I took the same approcah as Weasel and came up with same answer.

However, since I usually play on weekdays and stayed home from casino today for new showerdoor installation guy, I am supposed to have been cleaning house thus far.

Thanks Ed. =)

I used weasel97’s method for calculating the number of straights without flush possibilities (600). Adding in the royal flushes and quads, I also get 656 nut boards, just like weasel97 and Toonces, the driving cat. That means only 0.0252% of boards fulfill this criterion.

I guess the lesson is that we shouldn’t be overly eager to call when we see someone make a big bet on a board with a big made hand (such as a straight) — it’s very possible that it’s a value bet.

Also, let me add that I’m impressed by all of your replies. I hope NOT to see you playing against me in my crappy pokerstars 3.40 SnGs and 25NL games.

@jamleeco:
360 3-flushes is correct. I had “grunched” my answer (mot looking at any answer prior).

Hairsplitter’s finest hour: Exposed burn cards. If a dealer exposes and burns the AH we have six variations of AAA KK full house board chops. And Dude, let’s not forget, let’s not forget how many (20) cards in the deck, as the exposed burn, can’t complete royal flushes that otherwise hafta chop.(That ain’t legal either.)

i have a question. in hold-em, one player has royal flush, the other full house. how many combos are there to get this reasults ??

http://www.838dz.com