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.
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