Note: All odds tables at Poker1.com were personally calculated by Mike Caro, unless otherwise specified. Many were done in the early 1970s and included in the original Doyle Brunson Super/System — A Course in Power Poker, first published in 1978.
At the time of publication, there were few, if any, reliable sources for poker odds. The 50 often-cited tables stood the test of time and remain unchallenged to this day.
Index of all poker odds tables
TABLE 3 – Five-card draw:
Basic one-card high drawing chances
(53-card deck – including Joker)
|
You’re drawing one card to… |
Probability of making a complete hand expressed in percent (%) is… |
Odds against making a complete hand are… |
|
8♠ 7♠ 6♠ Joker (22-way hand) |
45.83 |
1.18 to 1 |
|
9♥ 7♥ 6♥ Joker (19-way hand) |
39.58 |
1.53 to 1 |
|
6♣ 5♠ 4♦– Joker or |
33.33 |
2 to 1 |
|
K♣ J♣ 10♣ 9♣ (13-way hand) |
27.08 |
2.69 to 1 |
|
6♦ 5♣ 3♠ Joker (12-way Straight) |
25.00 |
3 to 1 |
|
K♥ 8♥ 7♥ 4♥ (Flush) |
20.83 |
3.8 to 1 |
|
9♦ 8♣ 7♠ 6♦ (Open-end Straight) |
18.75 |
4.33 to 1 |
|
J♠ 10♠ 8♦ 7♦ (Inside Straight) |
10.42 |
8.6 to 1 |
|
A♦ A♥ 7♣ 7♠ (Aces-up) |
10.42 |
8.6 to 1 |
|
7♦ 7♣ 4♠ 4♦(Small Two-Pair)* |
8.33 |
11 to 1 |
|
A♦ A♥ A♠ 4♦ (Three Aces with kicker)* |
10.42 |
8.6 to 1 |
|
8♣ 8♥ 8♠ A♦ (Trips with Ace kicker)* |
10.42 |
8.6 to 1 |
|
5♦ 5♥ 5♣ K♣ (Trips with non-Ace kicker)* |
8.33 |
11 to 1 |
* Four-of-a-kind counts as a complete hand
(Note: Mike Caro added category for trips with non-ace kicker to this table 2010-03-28. They are the same as improving when drawing to two small pair. Addition is only at new Poker1.com.)
Note: All tables in this series follow the same logical construction; however tables 1 through 4 are formatted slightly differently. This is an experimental application of the new Poker1 typography and may eventually be applied to the other tables.
Everything is everywhere
Any Poker1 page takes you anyplace you want to go!
↓ Search Poker1 ↓
♠ Poker1 Megadex tools ♠
— main navigation departments —
Collections
Related groups of Poker1 content
↓ Major collections ↓
↓ Tip collections ↓
↓ Contributor collections ↓
↓ More collections ↓
* Any collection followed by an asterisk ( * ) has no entries yet.
Poker1 everything
Browse alphabetically
[a-z-listing display=”posts” post-type=”post”]
Poker1 library
Content in categories
- ALL (newest first)
- Exclude ads
- Private
- SPECIAL INFO
- SPOTLIGHT
≡ Content above: Poker1 Phase2a specification ≡
Mike.
Quick question re: 5 card stud vs 5 card draw. I know all the odds for the hands in a stud game and I know how to calculate them.
How do you figure them out for draw when there are so many actions you can take after the initial deal? I imagine you have to assume optimal play (as in video poker).
I don’t even know where to begin. I have briefly looked online so far. I have found pay tables. But I have yet to find any math on it.
Any help would be greatly appreciated.
d
Please see reply to Jerry above. — Mike Caro
Can you please help me calculate a few poker odds?
If I’m playing 5 card draw, specifically a video poker machine, and I have been dealt 4 cards to a royal flush, I assume the odds are 1 in 47 of drawing 1 card and completing the royal flush. Correct?
But what if I am dealt 3 cards to a royal and draw 2 cards. What are the odds of drawing 2 cards and completing the royal?
Also, what if I am dealt 2 cards to a royal and draw 3 cards. What are the odds of drawing 3 cards and completing the royal?
Finally, what if I am dealt 1 card to a royal and draw 4 cards. What are the odds of drawing 4 cards and completing the royal?
What formula do you use to calculate these odds?
Thanks!
Jerry
Hi, Jerry —
Sorry I overlooked your comment earlier. Thanks for joining our Poker1 family.
Drawing one to four parts of royal flush. (46-to-1 against, one chance in 47. You are correct. Formula is self-evident — you’ve seen five cards, 47 remain, and only one completes the royal flush.)
Drawing two cards to three parts of a royal flush. (1,080-to-1 against. Formula: 47 cards for the first times 46 remaining cards for the second, and that answer divided by two possible orders of arrival. So, (47 x 46) / 2 = 1,081 combinations. Of these, only one will contain the exact two cards needed to complete the royal flush.)
Drawing three cards to two parts of a royal flush. (16,214-to-1 against. Formula: (47 x 46 x 45) / (3 x 2 x 1 — the orders of arrival) = 16,215 combinations of which only one contains the exact three cards needed to complete the royal flush.)
Drawing four cards to one part of a royal flush. (178,364-to-1 against. Formula: (47 x 46 x 45 x 44) / (4 x 3 x 2 x 1 — see previous formula descriptions) = 178,365 combinations of which only one contains the exact four cards need to complete the royal flush.
Hope this helps.
Straight Flushes,
Mike Caro