odds no one can beat board. |
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Posted Tue Jul 12, 2005 2:33 am GMT by rebuyman
What are the odds of dealing a board where no one can beat it.
Example, four of a kind with Ace kicker on the board or four aces with a King on the board. Just asking because within one hour i saw boards of A3333 and AAKAA.
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Posted Tue Jul 12, 2005 5:00 am GMT by Muck
The following hands can’t be improved upon.
1 hand = 4 of a kind A K
12 hands = 4 of a kind K-2 A
10 hands = Spade straight flushs
10 hands = Club straight flushs
10 hands = Heart straight flushs
10 hands = Diamond straight flushs
4 hands = Royal straights
Total 57 hands
The odds depends on the number of active players. If we assume heads up the total number of possible boards hands = 205,476,480.
So the answer is about 1 in 3604850 (correct me if wrong)
Posted Tue Jul 12, 2005 5:27 am GMT by improv
| Muck wrote: | The following hands can’t be improved upon.
10 hands = Spade straight flushs
10 hands = Club straight flushs
10 hands = Heart straight flushs
10 hands = Diamond straight flushs
|
A striaght flush can always be improved upon unless it is a Royal.
e.g
BOARD: 2  3 4 5 6
a player could hold a 7
which would improve the board.
Posted Tue Jul 12, 2005 6:43 am GMT by Muck
| improv wrote: | | A striaght flush can always be improved upon unless it is a Royal. |
DOH!  Thanks improv.
So that's 4 instead of 40.
So 21 in 205,476,480
or 1 in 9,784,594 (correct if wrong)
Posted Tue Jul 12, 2005 7:15 am GMT by improv
Hmmm...
1 hand = 4 of a kind A K
12 hands = 4 of a kind K-2 A
4 hands = Royal straight flush
I think these are the only hands that cannot be improved upon... 17 in total...
1 in 9,784,594
Posted Tue Jul 12, 2005 7:41 am GMT by Muck
What about the non-flushable royal straights?
Wait a minute now that I look at it the maths for that is wrong too. Since you could use any rainbow combination of AKQJT :
That’s a lot of combinations.
Posted Tue Jul 12, 2005 8:08 am GMT by improv
| Muck wrote: | What about the non-flushable royal straights?
Wait a minute now that I look at it the maths for that is wrong too. Since you could use any rainbow combination of AKQJT :
That’s a lot of combinations. |
alot of royal straights could be improved upon.
any straight with 3 of any suit in could become a flush if someone held suited cards....
Too many combinations for me to even contemplate... done my A-Level in maths but didnt take the statistics module, doh!
Posted Tue Jul 12, 2005 8:34 am GMT by Muck
| improv wrote: | | Muck wrote: | What about the non-flushable royal straights?
Wait a minute now that I look at it the maths for that is wrong too. Since you could use any rainbow combination of AKQJT :
That’s a lot of combinations. |
alot of royal straights could be improved upon.
any straight with 3 of any suit in could become a flush if someone held suited cards....
Too many combinations for me to even contemplate... done my A-Level in maths but didnt take the statistics module, doh! |
How could you improve on a royal straight (NB: I did say non-flushable and rainbow)?
I had a go at the math but I’m not sure how to allow for repeated combinations. If I was going to estimate I’d say there are approx 186 different hands to make a non-flushable straight.
96 definite ways + a guess at how many combinations (48 + 24 + 12 + 6)
Posted Tue Jul 12, 2005 8:51 am GMT by Loonbat
| Muck wrote: |
1 hand = 4 of a kind A K
12 hands = 4 of a kind K-2 A
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I don't like this math!
4 of kind As with a K = 4 hands (K clubs, spades, hearts, diamonds)
4 of kind (K-2) with an ace = 48 hands (same logic as above)
Posted Tue Jul 12, 2005 9:01 am GMT by Muck
| Loonbat wrote: | | Muck wrote: |
1 hand = 4 of a kind A K
12 hands = 4 of a kind K-2 A
|
I don't like this math!
4 of kind As with a K = 4 hands (K clubs, spades, hearts, diamonds)
4 of kind (K-2) with an ace = 48 hands (same logic as above) |
LOL! So apart from everything my first post was bang on :D
Posted Tue Jul 12, 2005 11:03 am GMT by Loonbat
| Muck wrote: | | LOL! So apart from everything my first post was bang on :D |
Quite right ... good job 
Posted Tue Jul 12, 2005 2:30 pm GMT by cono
it has to be less than 1 in 8 million even without including royal flushes/straights.
Suppose the first card is an ace. 4/52
Next card is not an ace. 48/51
Next card is same as second card. 3/50, 2/49, 1/48
(4/52)x(48/51)x(3/50)x(2/49)x(1/48)=1 in 270,725, and that is just the chance of having the board show AXXXX where X is a constant.
Posted Tue Jul 12, 2005 7:54 pm GMT by racquet000
Do we have to call it a Roal Stright Flush??? 
Posted Wed Jul 13, 2005 6:06 am GMT by Tadzio
Anyone's odds (a1)
KAAAA (4/52, 4/51, 3/50, 2/49, 1/48 ) = 96/311,875,200 = 1/3,248,700
AXXXX (4/52, 48/51, 3/50, 2/49, 1/48 ) = 1152/311,875,200 = 1/270,725
AKQJTs (20/52, 4/51, 3/50, 2/49, 1/48 ) = 480/311,875,200 = 1/649,740
Odds of any of these occurring:
1728/311,875,200 = ~1/180,483
Note: these are the odds of any pot will be won by one of these hands. For the board-- and only the board-- to make these hands you have to make a few assumptions to make the problem managable. This first assumption to make is the number of players. I'm assuming 10. The second assumption to make is that none of the players has a card the board needs to make its hand. Which is a pretty big assumption in its own right. Assuming these, it follows:
Board's limited conditions odds (bL)
KAAAA (4/32, 4/31, 3/30, 2/29, 1/28 ) = 96/24,165,120 = 1/251,720
AXXXX (4/32, 48/31, 3/30, 2/29, 1/28 ) = 1152/24,165,120 = ~1/20,977
AKQJTs (20/32, 4/31, 3/30, 2/29, 1/28 ) = 480/24,165,120 = 1/50,344
Odds of any of these occurring, keeping in mind our assumptions:
1728/24,165,120 = ~1/13,984
Odds that none of the players at a 10 hand table will take one of the cards the board needs to make its hand during the pre-flop dealing stage:
Player's odds of failing to ruin the board's limited conditions (pR)
KAAAA (44/52, 43/51, 42/50, 41/49, 40/48, 39/47, 38/46, 37/45, 36/44, 35/43, 34/42, 33/41, 32/40, 31/39, 30/38, 29/37, 28/36, 27/35, 26/34, 25/33) = (130,320,960/311,875,200; 69,090,840/184,072,680; 33,390,720/102,080,160; 4,071,600/52,307,640) = 1,224,124,620,360,967,980,387,532,800,000/3,065,325,832,220,095,439,942,000,064,000,000 = ~1/2,504
AXXXX (44/52, 43/51, 42/50, 41/49, 40/48, 39/47, 38/46, 37/45, 36/44, 35/43, 34/42, 33/41, 32/40, 31/39, 30/38, 29/37, 28/36, 27/35, 26/34, 25/33) = (130,320,960/311,875,200; 69,090,840/184,072,680; 33,390,720/102,080,160; 4,071,600/52,307,640) = 1,224,124,620,360,967,980,387,532,800,000/3,065,325,832,220,095,439,942,000,064,000,000 = ~1/2,504
AKQJTs (32/52, 31/51, 30/50, 29/49, 28/48, 27/47, 26/46, 25/45, 24/44, 23/43, 22/42, 21/41, 20/40, 19/39, 18/38, 17/37, 16/36, 15/35, 14/34, 13/33) = (24,165,120/311,875,200; 9,687,600/184,072,680; 3,160,080/102,080,160; 742,560/52,307,640) = 549,331,853,867,906,767,257,600,000/3,065,325,832,220,095,439,942,000,064,000,000 = ~1/5,583,471
Using these numbers you can figure out the odds of an occurance happening where you'll have the opportunity (at a 10 hand table) to apply the ~1/13,984 chance that the board will catch an unbeatable hand. In general, the fewer hands at a table, (bL) will reflect worse odds.
(bL) is the board's best odds to get these hands, but once (bL) is adjusted for (pR), the board's odds quickly begin to approach (a1).
Posted Wed Jul 20, 2005 11:33 am GMT by gol4pro
These numbers are so far off it's mind-boggling to me. If you don't know what you're doing, then don't try.
Ok. First off, it depends on your hole cards. If you're holding an Ace, the probability that nobody will beat the final board goes down obviously. If you're not holding an Ace, it goes up.
We're just going to assume that you're not holding any hole cards.
Total number of boards = 52c5 = 2,598,960
a) Royal flush = 4c1*5c5/52c5 = 4/2,598,960 = 1/649740
b) Broad way straight with no 3-flush. I'm not COMPLETELY sure I did this right, but I'm pretty sure. 1c1*4c1*4c1*3c1*3c1*2c1 = chance of XcXcXdXdXs (where board 2-flushes twice) + the probability that it only 2-flushes once = 1c1*4c1*4c1*3c1*3c1*2c1 + (1c1*4c1*3c1*3c1*2c1*1c1)= 360/2,598,960
c) 4 of a kind, with an Ace on the board. (1c1)12c1*4c4*4c1 = 48 combinations
d) 4 of a kind ACES, with a KING on the board. (1c1)(1c1)4c4*4c1 = 4 combinations
all in all we have, (4+360+48+4/2,598,960) = 424/2,598,960 or about the odds of making a straight flush on the flop.
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