## Wizard Recommends

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Hello, thank you for a very interesting and informative site. I have a question of my own that I hope you can answer for me. As a Texas hold 'em player, I pay special attention to pocket pairs, and have particular interest towards 10-10 or J-J or similar, as on the surface they appear strong but can be beaten easily. My question however, is how do you work out the probability of there being at least one person on your table holding a higher pocket pair to yours?

Andrew from St. Albans

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, 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%

Congratulations on a great site. I totally understand your anger over the spread of 6 to 5 Blackjack payouts but am very curious as to why Americans seem to accept 00 Roulette without any argument. This Roulette is almost criminal and should be ranked along side Keno & Slots.

Andrew from Sydney

Thanks. You make a valid point. The house edge in 6 to 5 blackjack is 1.44% under the usual rules, while double zero roulette is 5.26%. That is 3.7 times as bad. However I have learned through the years that it is almost hopeless getting players to leave a game they like, regardless of how bad the house edge is. So the best I can do is advise them how to play their game of choice. For blackjack players there is still no shortage of 3 to 2 games out there. Playing 6 to 5 is giving the casino an extra 0.8% advantage for no reason at all. I also stress the importance of looking for single-zero roulette if you are a roulette player. So I see no inconsistency.

Is it required to flip over your cards if you call a player going all in? I’ve seen this done on television numerous times, but can I just wait until then end of the hand?

According to various sources it is required to turn your cards over in a tournament game, but optional in a cash game.

I’m coming out to Vegas next month during March Madness. I’ll probably be bringing a couple thousand dollars with me, some of my own, and some for some friends I’ll be placing bets for on the NCAA games. I’m not sure I’m comfortable carrying that amount of cash around. Would I be able to get a casino marker for a few thousand dollars or are they typically only issued for ’whales’, and can you draw on it for non-table gaming such as the sportsbook? Are there any costs or pitfalls associated with getting a marker (other than having your personal banking information in another database). Thanks for the great site and your time.

Mitch

Gamblers in your range certainly do use markers. You should try to establish credit with the casino before you go. Alternatively you can wire the casino money, that way you won’t have to go through a credit check. Either way, do so at least a week in advance. The use of markers and wire is very commonplace in the casino and from what I hear the process usually goes very smoothly.

I’ve heard that casinos are looking at using RFID in chips to speed up counting, reduce errors, and defeat fakes. Is the use of this new technology expected to eliminate the ability to use those inadvertently pocketed chips from one casino at another one? Thanks for the site and your time.

Mitch from Hopkins

For those who may not know, RFID stands for Radio Frequency Identification. I’m not an expert on this topic but it is my understanding they will be used to track player betting patterns, which will help for both comping and catching card counters. However counterfeit chips seems to be a growing problem and that may be another benefit. Currently casinos like it when you leave with chips and never cash them. That is why they create so many chips for special occasions, hoping chip collectors will hoard them. Again, I’m not an expert, but I don’t think it would be cost effective to create these chips if the expense were more than the face value. So I think you’ll be safe pocketing chips.

I have been with my boyfriend for 2years and 8mths we are doing ok, we have talked about our futures and how we want to spend them together, but latley he has been acting different and today i asked him a question about a friend of mine she said that he talked to her well we always tell each other when we talk to the opposite sex but for some reason i guess he didnt and when i asked him about it he just said he didnt talk to her and he got really mad and defensive whats going on with him?

Emily from Bedford

He is probably sick of this policy of reporting to you whenever he talks to someone of the opposite sex. If you can’t trust each other to have an innocent conversation with someone of the opposite sex then the relationship is doomed. So I don’t blame him for getting mad and defensive. I think you are making a small problem into a big one over this. My advice is drop it, and drop this ridiculous confession rule while you’re at it.

In the NBA, there are three divisions per conference, and 8 teams per conference make the playoffs. The top 3 seeds in each conference are the respective division winners, and the #4 through #8 seeds are the non-division winners with the best records. This year, 2 teams from the same division in the Western Conference, the Spurs and Mavericks, have the top record in the West. If this keeps up, it means that the 2nd best team in the Western Conference will end up with the #4 seed and will have to face the best team in the 2nd round if they both win. Many people are pointing to this as a problem with the system, while the NBA considers it an anomaly. In an effort to explain that its not an anomaly because it could happen frequently, an analyst from ESPN recently made the following statement in a blog: "There are 15 teams in each conference, and five teams in each of the three divisions. That means that there's a 4-in-14 chance that the team with the second-best record will be from the same division as the team with the best record." Is he correct that there is a 4/14 chance of it happening in a particular conference? How would you figure this out? If he is correct, than it would happen in at least 1 of the conferences 57% of the time, right?

AJ from Huntington Woods, MI

Yes, he is right. There are combin(15,2)=105 ways to choose the two best teams out of 15. There are 3*combin(5,3)=30 ways to choose them from the save division. So the probability the two best teams are from the same division is 30/105 = 4/14. The probability of this happening in at least one conference is 1-(10/14)2 = 48.98%.