• Chance of any one opponent getting dealt 2 aces pre-flop is the same, 221-1, but we need to take into account that there are 2 kings ‘accounted for’ here. That is to say their cards came from 50 unknown cards and not the full 52. So, the chances are a little bit bigger – 0.48% in.
• The probability you will get pocket aces in any one hand is 6/1326, or once every 221 hands. According to my 10-player Texas Hold ’em section (/games/texas-hold-em/10players.html) the probability of winning with pocket aces is 31.36%, assuming all players stay in until the end. However that is a big if.

In 10mil sets of 3 hands you should get AA 3 times in a row at some point, and for that matter 22 3 times in a row, and it is the same odds to get AA followed by KK followed by 22. Your way is correct in that after the first hand you ask what are the odds that my next two hands will be AA also. 2) You are dealt pocket Aces, less than half of a percent of the time (1 out of every 221 hands) 3) You’ll flop a set or better with a pocket pair about 12% of the time (a little worse than 1 time out of 8) 4) If you get all the way to the river, you’ll hit a set or better about 1 in every 5 times www.advancedpokertraining.com.

Assuming you draw five cards, and count all hands with exactly two jacks, then the probability would be combin(4,2)*combin(48,3)/combin(52,5) = 6*17296/2598960 = 3.99%.

For probability questions, I like to take the number of combinations the event you're interested can happen divided by the total number of combinations. First review the combin function in my probabilities in poker section. The number of ways to get a four of a kind is simply the number of singletons in the deck, or 48. The number of ways to get a three of a kind (not including a full house) is the product of the number of ways to get the third card, 2, and the number of ways to get two other singletons, 2*combin(12,2)*4² = 2,112. The total number of ways the cards can come up in the flop are combin(50,3)=19,600. So, the probability of a four of a kind is 48/19600=0.0024, and the probability of a three of a kind is 2,112/19,600=0.1078.

There are 52*51/2 = 1326 ways to arrange 2 cards out of 52. There are 4*3/2=6 ways to arrange 2 aces out of 4. So the answer is 6/1326 = 1/221. The probability of this happening twice in a row is (1/221)² = 1 in 48,841.

Odds Of Getting Pocket Aces Back To Back

First, there are 10*9/2=45 ways you can choose 2 players out of 10. The probability of two specific players getting four aces is 1/combin(52,4)=1/270725. So the probability of any two players getting a pair of aces is 45/270725=0.0001662.

For those unfamiliar with the terminology, each player gets two cards to himself and the three flop cards are shared among all players. So this is the same as asking if you dealt three community cards, all of different ranks, and ten 2-card hands, what is the probability three of the 2-card hands would be pairs that match one of the three community cards.

The probability player 1 has a set is 3*combin(3,2)/combin(49,2). Then the probability player 2 has a set is 2*combin(3,2)/combin(47,2). Then the probability player 3 has a set is combin(3,2)/combin(45,2). However, any three players can the three sets, not necessarily the first three. There are combin(10,3) ways to choose the 3 players out of 10 that have sets. So the answer is combin(10,3)*(3*combin(3,2)/combin(49,2))*(2*combin(3,2)/combin(47,2))*(combin(3,2)/combin(45,2)) = 0.00000154464 = 1 in 64,740.

For any given person the probability of having AA is combin(4,2)/combin(50,2) = 6/1,225 = 0.0049 because there are 6 ways to pick 2 aces out of 4, and 1225 ways to pick any 2 cards out of the 50 left in the deck. The probability is the same for a pair of kings. For A-K the probability is 4*4/1,225=0.0131, because there are 4 ways to get an ace and 4 ways to get a king. For A-Q the probability is 4*2/1225=0.0065, because there are 4 aces but only 2 queens left in the deck. So the probability any given player will have one of these hands is (6+6+16+8)/1225 = 0.0294. Now the next step is clearly not perfect because if one player doesn't have one of these hands the odds the next player does is a little bit higher. Forgetting this for the sake of simplicity the probability no player has one of these hands is (1-0.0294)⁸ = 78.77%. So the probability at least one player has one of these hands is 21.23%.

Let’s assume you have a pair of aces. Before considering that the other player has another pair the probability of flopping a three of a kind is the [nc(one ace)*nc(two ranks out of 12)*nc(one suit out of 4)² + nc(any other three of a kind)]/nc(any three cards), where nc(x) = number of combinations of x. This equals [2*combin(12,2)*4²+12*combin(4,3)]/combin(50,3) = (2112+48)/19600 = 11.020%. Now lets assume the other player has any other pair, but not the same as yours. Then probability becomes [2*(combin(11,2)*4² + 11*2*4 + 11*combin(4,3)]/combin(48,3) = 11.4477%.

The probability for any given hand is (combin(4,2)/combin(52,2))*(1/combin(50,2)) = 1/270725. So, the probability of this happening twice in a row is 1 in 270,725² = 1 in 73,292,025,625.

13*12*combin(4,2)*4/combin(52,3) = 3744/22100 = 16.941%.

The following table shows estimated probabilities that a pair will be beaten by at least one higher pair according to the number of players (including yourself). These probabilities are not exact because the hands are not independent. However to find the exact probabilities would get complicated and I think these are pretty close. My formula is 1-(1-r*combin(4,2)/combin(50,2))^(n-1), where r=number of higher ranks than your pair, and n = total number of players. The table shows the probability of another player having a pair of aces, when you have a pair of kings, in a 10-player game, to be 4.323%.

Probability Pair Beaten by Higher Pair

Pair	2 Pl.	3 Pl.	4 Pl.	5 Pl.	6 Pl.	7 Pl.	8 Pl.	9 Pl.	10 Pl.
KK	0.49%	0.977%	1.462%	1.945%	2.425%	2.903%	3.379%	3.852%	4.323%
QQ	0.98%	1.95%	2.91%	3.861%	4.803%	5.735%	6.659%	7.573%	8.479%
JJ	1.469%	2.917%	4.344%	5.749%	7.134%	8.499%	9.843%	11.168%	12.473%
TT	1.959%	3.88%	5.763%	7.609%	9.42%	11.194%	12.934%	14.64%	16.312%
99	2.449%	4.838%	7.168%	9.442%	11.66%	13.823%	15.934%	17.992%	20.001%
88	2.939%	5.791%	8.56%	11.247%	13.855%	16.387%	18.844%	21.229%	23.544%
77	3.429%	6.74%	9.937%	13.025%	16.007%	18.887%	21.668%	24.353%	26.947%
66	3.918%	7.683%	11.301%	14.776%	18.115%	21.324%	24.407%	27.369%	30.215%
55	4.408%	8.622%	12.65%	16.501%	20.181%	23.7%	27.063%	30.279%	33.352%
44	4.898%	9.556%	13.986%	18.199%	22.205%	26.016%	29.64%	33.086%	36.363%
33	5.388%	10.485%	15.308%	19.871%	24.188%	28.273%	32.137%	35.794%	39.253%
22	5.878%	11.41%	16.617%	21.517%	26.13%	30.472%	34.559%	38.405%	42.025%

Let's call the players A, B, and C. The probability A has a pair of aces is combin(4,2)/combin(52,2) = 6/1326. The probability B has a pair of kings is combin(4,2)/combin(50,2) = 6/1225. The probability C has a pair of queens is combin(4,2)/combin(48,2) = 6/1128. However there are 3! = 1*2*3 = 6 ways you can arrange three pairs between three players. So, the answer is 6*(6/1326)*(6/1225)*(6/1128) = 0.000000707321.

Thanks for the kind words. The probability you will get pocket aces in any one hand is 6/1326, or once every 221 hands. According to my 10-player Texas Hold ’em section (/games/texas-hold-em/10players.html) the probability of winning with pocket aces is 31.36%, assuming all players stay in until the end. However that is a big if. If forced to make a guess I’d estimate the probability of winning with aces in a real 10-player game is about 70%. So the probability of getting pocket aces and then losing is 0.3*(1/221) = 0.1357%. So, at $100 per incident that is worth 13.57 cents per hand. Over ten people that costs the poker room $1.36 per hand on average, which cuts into the rake quite a bit. I tend to agree with your strategy of calling, which will keep more players in the hand, and increase your chance of losing.

The odds of a specific player having aces is combin(4,2)/combin(52,2) = 6/1326. The odds of the next player having a pair of kings is combin(4,2)/combin(50,2) = 6/1225. However, in a ten-player game there are 10 possible players who could get the aces, and 9 possible players for the kings. So a strong approximation would be 10*9*(6/1326)*(6/1225) = 0.001995, or 1 in 501. This answer is slightly too high, because it double counts the situation where two players have aces, or two have kings, or both.

The math of this gets very messy due to the possibility of more than one playing having a higher pair, including the same type of pair. For example if you have pocket kings two players could have pocket aces. However it is easy to show the expected number of players who will beat you. This would be n*r*(6/1225), where n is the number of opponents, and r is the number of higher ranks. The following table shows the average number of players who will have a higher pocket pair according to your pocket pair (left column) by the number of opponents (top row).

Expected Number of Higher Pocket Pairs by Number of Opponents

Pair	1 Opp.	2 Opp.	3 Opp.	4 Opp.	5 Opp.	6 Opp.	7 Opp.	8 Opp.	9 Opp.
2,2	0.0588	0.1176	0.1763	0.2351	0.2939	0.3527	0.4114	0.4702	0.529
3,3	0.0539	0.1078	0.1616	0.2155	0.2694	0.3233	0.3771	0.431	0.4849
4,4	0.049	0.098	0.1469	0.1959	0.2449	0.2939	0.3429	0.3918	0.4408
5,5	0.0441	0.0882	0.1322	0.1763	0.2204	0.2645	0.3086	0.3527	0.3967
6,6	0.0392	0.0784	0.1176	0.1567	0.1959	0.2351	0.2743	0.3135	0.3527
7,7	0.0343	0.0686	0.1029	0.1371	0.1714	0.2057	0.24	0.2743	0.3086
8,8	0.0294	0.0588	0.0882	0.1176	0.1469	0.1763	0.2057	0.2351	0.2645
9,9	0.0245	0.049	0.0735	0.098	0.1224	0.1469	0.1714	0.1959	0.2204
T,T	0.0196	0.0392	0.0588	0.0784	0.098	0.1176	0.1371	0.1567	0.1763
J,J	0.0147	0.0294	0.0441	0.0588	0.0735	0.0882	0.1029	0.1176	0.1322
Q,Q	0.0098	0.0196	0.0294	0.0392	0.049	0.0588	0.0686	0.0784	0.0882
K,K	0.0049	0.0098	0.0147	0.0196	0.0245	0.0294	0.0343	0.0392	0.0441

To get the probability that at least one player will beat you I will make the not entirely correct assumption that the number of players with a higher pocket pair is a Poisson random variable with a mean in the above table. Given that assumption the probability that at least one player will beat you is 1-e^-µ, where µ is the mean. For examle if you have pocket queens and there are 9 other players the expected number of players with a higher pocket pair is 0.0882, so the probability of at least one player having a higher pocket pair is 1-e^-0.0882 = 8.44%. The table below shows those probabilities.

Probability of Higher Pocket Pair by Number of Opponents — Wizard's Approximation

Pair	1 Opp.	2 Opp.	3 Opp.	4 Opp.	5 Opp.	6 Opp.	7 Opp.	8 Opp.	9 Opp.
2,2	5.71%	11.09%	16.17%	20.95%	25.46%	29.72%	33.73%	37.51%	41.08%
3,3	5.25%	10.22%	14.92%	19.39%	23.62%	27.62%	31.42%	35.02%	38.42%
4,4	4.78%	9.33%	13.67%	17.79%	21.72%	25.46%	29.03%	32.42%	35.65%
5,5	4.31%	8.44%	12.39%	16.17%	19.78%	23.24%	26.55%	29.72%	32.75%
6,6	3.84%	7.54%	11.09%	14.51%	17.79%	20.95%	23.99%	26.91%	29.72%
7,7	3.37%	6.63%	9.77%	12.82%	15.75%	18.59%	21.34%	23.99%	26.55%
8,8	2.9%	5.71%	8.44%	11.09%	13.67%	16.17%	18.59%	20.95%	23.24%
9,9	2.42%	4.78%	7.08%	9.33%	11.52%	13.67%	15.75%	17.79%	19.78%
10,10	1.94%	3.84%	5.71%	7.54%	9.33%	11.09%	12.82%	14.51%	16.17%
J,J	1.46%	2.9%	4.31%	5.71%	7.08%	8.44%	9.77%	11.09%	12.39%
Q,Q	0.97%	1.94%	2.9%	3.84%	4.78%	5.71%	6.63%	7.54%	8.44%
K,K	0.49%	0.97%	1.46%	1.94%	2.42%	2.9%	3.37%	3.84%	4.31%

So my approximation of the probability of at least one higher pocket pair is 1-e^{-n*r*(6/1225)}.

P.S. After this column appeared one of my fans, Larry B., wrote a brute force combinatorial program to solve this problems. Here are his results.

Probability of Higher Pocket Pair by Number of Opponents — Larry B.'s Exact Probabilities

Pair	1 Opp.	2 Opp.	3 Opp.	4 Opp.	5 Opp.	6 Opp.	7 Opp.	8 Opp.	9 Opp.
2,2	5.88%	11.41%	16.61%	21.5%	26.1%	30.43%	34.5%	38.33%	41.94%
3,3	5.39%	10.48%	15.3%	19.87%	24.18%	28.26%	32.12%	35.77%	39.22%
4,4	4.9%	9.56%	13.99%	18.2%	22.21%	26.03%	29.66%	33.12%	36.4%
5,5	4.41%	8.62%	12.66%	16.52%	20.21%	23.73%	27.11%	30.35%	33.45%
6,6	3.92%	7.69%	11.31%	14.8%	18.15%	21.38%	24.48%	27.47%	30.34%
7,7	3.43%	6.74%	9.95%	13.05%	16.05%	18.95%	21.76%	24.47%	27.09%
8,8	2.94%	5.8%	8.58%	11.28%	13.91%	16.46%	18.95%	21.36%	23.71%
9,9	2.45%	4.84%	7.19%	9.47%	11.71%	13.9%	16.04%	18.13%	20.17%
T,T	1.96%	3.89%	5.78%	7.64%	9.47%	11.27%	13.04%	14.77%	16.48%
J,J	1.47%	2.92%	4.36%	5.78%	7.18%	8.57%	9.93%	11.29%	12.63%
Q,Q	0.98%	1.95%	2.92%	3.88%	4.84%	5.79%	6.73%	7.67%	8.6%
K,K	0.49%	0.98%	1.47%	1.96%	2.44%	2.93%	3.42%	3.91%	4.39%

Later Stephen Z. suggested a simple approximation. Take the number of higher pairs, multiply by the number of other players, and divide by 2. That is the percentage probability that there will be at least one higher pair. For example, with a pair of jacks in a 10 player game the probability of a higher pocket pair is 3*9/2 = 13.5%. Using that formula you get the following for all situations.

Probability of Higher Pocket Pair by Number of Opponents — Stephen Z. Approximation

Pair	1 Opp.	2 Opp.	3 Opp.	4 Opp.	5 Opp.	6 Opp.	7 Opp.	8 Opp.	9 Opp.
2,2	6%	12%	18%	24%	30%	36%	42%	48%	54%
3,3	5.5%	11%	16.5%	22%	27.5%	33%	38.5%	44%	49.5%
4,4	5%	10%	15%	20%	25%	30%	35%	40%	45%
5,5	4.5%	9%	13.5%	18%	22.5%	27%	31.5%	36%	40.5%
6,6	4%	8%	12%	16%	20%	24%	28%	32%	36%
7,7	3.5%	7%	10.5%	14%	17.5%	21%	24.5%	28%	31.5%
8,8	3%	6%	9%	12%	15%	18%	21%	24%	27%
9,9	2.5%	5%	7.5%	10%	12.5%	15%	17.5%	20%	22.5%
T,T	2%	4%	6%	8%	10%	12%	14%	16%	18%
J,J	1.5%	3%	4.5%	6%	7.5%	9%	10.5%	12%	13.5%
Q,Q	1%	2%	3%	4%	5%	6%	7%	8%	9%
K,K	0.5%	1%	1.5%	2%	2.5%	3%	3.5%	4%	4.5%

For those not familiar with the hold 'em terminology, you are asking for the probability of at least a pair in six cards, given that the first two are (the hole cards) of different ranks. I hope you'll forgive me if I just do the probability of getting exactly a pair, including hands that also form a straight or flush.

The number of ways to pair one of your hole cards is six (2 hole cards * 3 suits remaining). The other three cards must all be of different ranks from the 11 left. There are combin(11,3)=165 ways to choose 3 ranks out of 11. For each of these there are four suits to choose from. So the number of ways to pair one of your hole cards is 6*165*4³=63,360.

Now let's look at the number of ways to get a pair outside of the two hole cards. There are 11 ranks to choose from for the pair. Once the pair is chosen there are combin(4,2)=6 ways to choose 2 suits out of 4. For the other two cards there are combin(10,2)=45 ways to choose 2 ranks out of the 10 fully intact ranks left. For both of those ranks there are 4 possible suits. So the total combinations for a pair, not including the hole cards, is 11*6*45*4²=47,520.

The total number of ways to choose 4 cards out of the 50 left in the deck is combin(50,4)=230,300. So the probability of getting exactly a pair in six cards is (63,360+47,520)/230,300 = 48.15%.

The probability of three different ranks in the flop is combin(13,3)×4³/combin(52,3) = 0.828235. There are combin(10,3)=120 ways you can choose three players out of ten. Of the three, the probability the first will have a set is 3×combin(3,2)/combin(49,2) = 0.007653061. The probability the second will have a set is 2×combin(3,2)/combin(47,2) = 0.005550416. The probability the third will have a set is combin(3,2)/combin(45,2) = 0.003030303. Take the product of all this and the probability is 0.828235 × 120 × 0.007653061 × 0.005550416 × 0.003030303 = 0.00001279, or 1 in 78,166.

For readers who may not know, a 'set' is a three of a kind after the flop, including a pocket pair. The probability of not making a set is (48+combin(48,3))/combin(50,3) = 17,344/19600 = 88.49%. So the probability of making a set is 11.51%. In 2,787 pairs you should have made a set 320.8 times. So you are 47.8 sets under expectations. The variance is n × p × (1-p), where n = number of hands, and p = probability of making the set. In this case the variance is 2,787 × .1176 × .8824 = 283.86. The standard deviation is the square root of that, or 16.85. So you are 47.8/16.85 = 2.84 standard deviations south of expectations. The probability of luck this bad or worse can be found in any Standard Normal table, or in Excel as norsdist(-2.84) = 0.002256, or 1 in 443.

The probability of being on the losing end of KK vs. AA is (combin(4,2)/combin(52,2)) × (combin(4,2)/combin(50,2)) = 0.000022162, for each opponent at the table. That is once every 45,121 hands, so your math was right. The expected number of times that would happen in 400 hands is 400 × 0.000022162 = 0.008865084, per opponent. The following table shows the probability of 3 or more instances of having KK against AA in 400 hands, by the number of opponents.

3+ KK vs AA probability in 400 hands

Opponents	Probability	Inverse
1	0.0000001145	1 in 8,734,376
2	0.0000009133	1 in 1,094,949
3	0.0000030658	1 in 326,182
4	0.0000072234	1 in 138,438
5	0.0000140202	1 in 71,325
6	0.0000240728	1 in 41,541
7	0.000037981	1 in 26,329
8	0.0000563277	1 in 17,753
9	0.0000796798	1 in 12,550

So, yes, I would say this looks fishy. The fewer the players, the more fishy it looks. I would be interested to know where this game was.

Thanks. There are 50 cards left in the deck, and 42 of them are not aces or kings. The probability of not seeing any aces or kings in five community cards is combin(42,5)/combin(50,5) = 850,668/2,118,760=40.15%. So, the probability of seeing at least one ace or king is 100% - 40.15% = 59.85%.

The probability of a specific other player having pocket aces, given that you do, is (2/50)×(1/49) = 1 in 1,225. Given 9 other players, the probability is 9 times that, or 1 in 136. This might seem like an abuse of taking the sum of probabilities. However, it is okay if only one player can get the two aces. To answer your question, the probability that another player had pockets aces three out of the three times you had pockets aces is (9×(2/50)×(1/49))³ = 1 in 2,521,626.

Did your pocket aces get cracked again? Are you wanting to know what are the odds of aces getting cracked?
This is something that many of us wonder about when our aces seem to get cracked again and again and again.
So let's just jump right into it!

Is Pocket Aces a Good Hand?

The first thing you should know is that pocket aces is a very good hand, the best hand in the game in fact.
And just because you lost with it, doesn't mean you should limp preflop with it or even fold it next time.
In fact, pocket aces is over an 80% favorite versus all other pocket pairs like KK, QQ, JJ and so on.
And it will also be your biggest long term winning hand in terms of total profit.
You can actually just go check this for yourself by filtering for the profitability of all hands inside a program like PokerTracker.

What Are the Odds of Pocket Aces Getting Cracked?

These are the odds your pocket aces getting cracked:

14.82%

When you are dealt pocket aces you have a 14.82% chance of losing versus a completely random hand.
So in other words, not very often but it will happen from time to time!

Is the Poker Site You Are Playing On Rigged Because Your Pocket Aces Got Cracked Again?

Not necessarily.
A lot of people jump to this quick conclusion without really thinking about the math first.
Like we just mentioned, you actually have almost a 15% chance of losing with pocket aces versus a completely random hand.
And since most people will have something a little bit better than a 'completely random hand' versus your pocket aces, they will often have an even higher chance to win against you than this.
This is why when people say that online poker is rigged or some major site like PokerStars is rigged, they often haven't actually thought it through enough.
They are instead just reacting in the moment and not using the rational part of their brain which clearly tells us that pocket aces is not some 'invincible' hand.
Your pocket aces will actually lose frequently in poker, this is simply a fact. It does not necessarily mean you are playing on a rigged poker site.

Basic Pocket Aces Odds

Here are some other basic stats you need to know when you have pocket aces:

• You will be dealt pocket aces 0.0045% of the time (1 in 221 hands)
• You will lose with pocket aces versus pocket kings approximately 18% of the time
• You will lose with pocket aces versus a broadway hand like KJ approximately 14% of the time
• You will lose with pocket aces versus a random suited hand approximately 18% of the time
• You will lose with pocket aces to ace king approximately 8% of the time
• You will lose with pocket aces to a suited connector like 98 of hearts approximately 24% of the time

So these are some pretty good odds with pocket aces versus other common hands. This is why most professional poker players actually list pocket aces as their favorite hand!
Daniel Negreanu for example has mentioned this several times before including in his advanced poker training course.

More Important Pocket Aces Odds You Need to Know

Here are some more important odds for pocket aces that I recommend you study and memorize:

Odds Of Getting Pocket Aces In Texas Holdem

• You will beat a flopped straight draw with pocket aces approximately 65% of the time
• You will beat a flopped flush draw with pocket aces approximately 61% of the time
• You will beat a flopped straight flush draw with pocket aces approximately 43% of the time
• You will beat a flopped two pair hand with pocket aces approximately 24% of the time
• You will beat a flopped trips hand (three of a kind) with pocket aces approximately 8% of the time

One of the good parts about getting dealt pocket aces is that you almost always have a chance to win.
The odds aren't always great, but there is almost no scenario where you will be drawing completely dead.
Now please don't get me wrong here.
There definitely is a time when you need to fold your pocket aces, like I discuss in this video I made recently:

But for the most part, pocket aces are a great hand to have in almost all situations in Texas Hold'em because even if somebody flops a huge hand against you, you almost always have outs still (chance to win).

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That is why I wrote this free little 50 page poker cheat sheet to give you the exact strategies to start consistently making $1000 (or more) per month in low stakes poker games right now.
These are the exact poker strategies that I have used by the way as a 10+ year poker pro. And I lay them all out for you step by step in this free guide.
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What Are the Odds of Getting Pocket Aces Twice in a Row?

So we already know that the odds of getting dealt pocket aces are only 1 in 221.
In other words, not good.
But how about the odds of getting dealt pocket aces twice in a row?
The odds of getting dealt pocket aces twice in a row is:
1 in 48,481
And then to take it one step further...
The odds of getting dealt pocket aces three times in a row is:
1 in 10,793,861
So as you can see, the odds of getting dealt pocket rockets twice in a row is exceedingly rare and the odds of getting dealt aces three times in a row is like winning the lottery.However, as we all know, anything can happen at the poker tables. This is why it is a good idea that you are using a good poker HUD so that you can quickly get reads on your opponents.

What is the Best Strategy to Win Big With Pocket Aces?

The best pocket aces strategy is to play them hard and fast. This means raising or re-raising with them preflop and then betting big on the flop, turn and river.
From time to time you will need to fold them as I already mentioned above. This is specifically versus very tight and passive players who raise you on the turn or river.
This is something I have discussed in detail before on this blog as well in my guide to when to fold an over pair in poker.
But for the most part, the best strategy with pocket aces is very simple. Play them fast, bet often and bet big with them!

Should You Ever Slow-Play Your Pocket Aces?

What about slow playing your pocket aces? Yay or nay?
In general this is a bad idea especially at the lower stakes. And the reason why is because you miss out on a ton of value.
Most players at these limits like to call way too much and never fold any pair or draw.
This is something that was very clear to see in my recent video series where you can watch me absolutely crush the 1c/2c online cash games.
I have some of the best results of all-time in these games and I can tell you that I almost never slow play my pocket aces.
And this is because the problem with slow playing your pocket aces then is that you will often end up winning a very small pot, when you could have won a much bigger pot by simply betting more.
And over the long run, this really, really starts to add up.
It is really important to remember that when you see world class poker pros like Phil Ivey slow-playing their pocket aces, it is usually because they are playing against another world class pro.
They have to employ some trickery from time to time in order to constantly keep their opponents guessing.
But if you play in the typical small stakes online or live poker games, the best course of action is going to almost always be to avoid slow playing your pocket aces.
Just play them straight-forward every time and you will win much more.

Final Thoughts

So what are the odds of your pocket aces getting cracked?
Well, against a random hand it is almost 15% and can be as high as 25% versus suited connectors for example. This might not sound like much, but it will come through from time to time.
Because the truth is that pocket aces are really just a one pair hand that can be beat.
And this is why it is important not to get too emotionally invested when you get dealt this beautiful hand.
With all that said though, pocket aces are still the best hand in the game and you should play them hard and fast in almost all scenarios.
Lastly, if you want to know the complete poker strategy that I have used as a 10+ year poker pro, make sure you grab a copy of my free poker cheat sheet.