Ask The Wizard #306
The Suncoast is running a poker promotion where the player wins $50 to $100 if he gets a specific high pocket pair and loses in Texas Hold 'Em. The pocket pair depends on the day, but can be jacks, queens, kings, or aces. If the pair loses the win is $100 if both hole cards are black, $75 if both are red, and $50 if there are one of each color. What is this promotion worth on an hourly basis?
It depends on the number of players at the table. The more the better, because you'll have a greater chance of losing with more players. The following table shows the probability of each of the four pairs losing by the total players at the table, including yourself. This assumes that nobody folds. This is obviously an unrealistic assumption, so I would take these probabilities as an upper bound.
Probability of Losing in Texas Hold 'Em
| Players | Jacks | Queens | Kings | Aces |
|---|---|---|---|---|
| 10 | 80.16% | 77.34% | 73.57% | 68.64% |
| 8 | 74.87% | 71.29% | 66.74% | 60.95% |
| 6 | 65.95% | 61.70% | 56.68% | 50.49% |
| 4 | 50.37% | 46.09% | 41.41% | 35.82% |
| 3 | 38.43% | 34.71% | 30.79% | 21.22% |
| 2 | 22.85% | 20.37% | 17.88% | 15.07% |
The average win is easily calculated as $100 × (1/6) + $75 × (1/6) + $50 × (1/2) = $62.50. That said, the next table shows the expected value of each of the four pocket pairs whenever they occur, assuming no other players fold.
Expected Win per Occasion
| Players | Jacks | Queens | Kings | Aces |
|---|---|---|---|---|
| 10 | $50.10 | $48.34 | $45.98 | $42.90 |
| 8 | $46.79 | $44.56 | $41.71 | $38.09 |
| 6 | $41.22 | $38.56 | $35.43 | $31.56 |
| 4 | $31.48 | $28.81 | $25.88 | $22.39 |
| 3 | $24.02 | $21.69 | $19.24 | $13.26 |
| 2 | $14.28 | $12.73 | $11.18 | $9.42 |
The next table shows the value of this promotion per hand played. It is simply the product of the table above and the probability of getting the required hold cards, which is 6/1326 = 0.90%.
Expected Win per Hand Played
| Players | Jacks | Queens | Kings | Aces |
|---|---|---|---|---|
| 10 | $0.23 | $0.22 | $0.21 | $0.19 |
| 8 | $0.21 | $0.20 | $0.19 | $0.17 |
| 6 | $0.19 | $0.17 | $0.16 | $0.14 |
| 4 | $0.14 | $0.13 | $0.12 | $0.10 |
| 3 | $0.11 | $0.10 | $0.09 | $0.06 |
| 2 | $0.06 | $0.06 | $0.05 | $0.04 |
The next table shows the value of this promotion per hour played, assuming a rate of 30 hands per hour. Again, this assume nobody ever folds, so I would take this as an upper bound on the value per hour.
Expected Win per Hour Played
| Players | Jacks | Queens | Kings | Aces |
|---|---|---|---|---|
| 10 | $6.80 | $6.56 | $6.24 | $5.82 |
| 8 | $6.35 | $6.05 | $5.66 | $5.17 |
| 6 | $5.60 | $5.23 | $4.81 | $4.28 |
| 4 | $4.27 | $3.91 | $3.51 | $3.04 |
| 3 | $3.26 | $2.94 | $2.61 | $1.80 |
| 2 | $1.94 | $1.73 | $1.52 | $1.28 |
Consider a game with the following rules:
- A random number generator provides random numbers between 0 and 1 uniformly distributed.
- Two players each get a separate number. Each player can see his own number only.
- Player 1 may keep his initial number or swap for a new random number.
- Player 2, knowing player 1's action, has the same option to keep his original number or swap for a new one.
- Player with the higher number wins.
I have four questions about the game:
- Answer the following questions about the game:
- At what number is player 1 indifferent to standing and switching?
- Assuming player 1 switches, at what number should player 2 be indifferent to standing and switching?
- Assuming player 1 stands, at what number should player 2 be indifferent to standing and switching?
- Assuming optimal strategy by both players, what is the probability player 1 will win?
The answer and solution can be found in my page of Math Problems, problem 225.
Is the "Jackpot Only" option in the Mega Millions lottery a good value?
If we ignore the effects of taxes, the annuity on the jackpot, and jackpot sharing, then if the jackpot is greater than $224,191,728 you should invoke the "Jackpot Only" option. If we do consider those factors, then you should never invoke but the Megaplier instead.
For more information, please see my page on the Mega Millions lottery.
At the Sky City casino in Auckland, New Zealand, both the player and dealer must make use of both his hole cards in Ultimate Texas Hold 'Em. How does this effect the odds compared to the usual rules where any five cards can be used?
That rule increases the house edge from 2.19% to 7.97% and the Element of Risk from 0.53% to 1.90%. This is because the dealer won't qualify more often and it will be harder to win on the Blind bet, which requires a straight or better.
For more details of my analysis, please see my new page on the Auckland variant of Ultimate Texas Hold 'Em.
For discussion about this question, please see the thread ULTIMATE IN NEW ZEALAND in my forum at Wizard of Vegas.