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Ask the Wizard #158
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 1e^{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 1e^{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 1e^{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% 
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 singlezero roulette if you are a roulette player. So I see no inconsistency.
Chad from Fargo
According to various sources it is required to turn your cards over in a tournament game, but optional in a cash game.
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.
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.
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.
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%.
A reader wrote in to express his comments about the assumptions I made in my answer. Here is a link to his commentary.
Hector from Miami, Florida
There is no particular time to expect the small cards. The last hand in a cut card has just about the same odds as the overall shoe. However if the dealer deals out much more than the average number of hands in a cut card game the last hands tend to be very bad for the player. This is because in the early hands the players and dealer didn’t hit much, which in turn is because lots of large cards came out, leaving more small cards for later in the shoe. So if you notice that you have already passed the average number of hands and the cut card is still a ways off then the deck is probably small card rich and it would be a good time to drop your bet or take a walk. However with other players jumping in and out of the game, and inconsistent cut card placement the practicality of this strategy is very small.
Pam from Glastonbury
Based on your version this guy seems like a real bum. Family is a lot more than DNA. If he isn’t willing to accept his responsibilities then he is nothing more than a sperm donor. If he doesn’t care the rest of his family probably won’t either. However he is obligated to help support his child, and I would hold him to that. If you press for the courts for paternity tests his family will learn about this anyway. For now I would lower the temperature and don’t give out any unsolicited information. Brining others into this may only escalate tensions.