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Ask The Wizard #390
In the last Ask the Wizard column You mentioned that in Ultimate Texas Hold 'Em, the Trips bet is intended to remain active, even if the player decides to fold. I'm aware that several casinos enforce a requirement for a Play bet for the Trips to remain in play, otherwise it doesn’t count. They often draw parallels to Three Card Poker, where the Pairplus bet is forfeited if the player folds. My question pertains to how this rule breach impacts the casino’s advantage.
That's a good inquiry! I've only performed a brief calculation on this matter, as it takes my computer quite a while to evaluate all potential card combinations in Ultimate Texas Hold 'Em, not to mention the time I need to reprogram.
If a player adheres to the optimal strategy aimed at maximizing the core game’s value, I discover that the Trips bet increases the house edge by 0.27%. However, there's also the chance a player might make a suboptimal small raise to protect their Trips bet. I’ve calculated that the house edge for the base game rises by 0.11% in cases where the player never folds despite having a three of a kind showing on the board. The player must carefully assess the implications of that raise and the ratio of their Trips to Ante bet when faced with a three of a kind on the board with two low kickers as their hole cards. Naturally, a player in this position may not even opt to place the Trips bet in the first place.
I want to reiterate the sources I mentioned in the previous column regarding the rule that ensures the Trips bet always remains active.
- Washington State Gaming Commission (internal link). If you wish to access the document directly, click here. here .
- Nevada Gaming Control Board (internal link). To see the document directly, click here. here
Should you encounter this issue in Nevada or Washington, I recommend raising a protest at the table and lodging a formal complaint with the gaming authorities if the outcome isn’t in your favor.
This topic is being examined and discussed on my forum at Wizard of Vegas .
What are the odds of having a positive outcome when placing even-money bets in double-zero roulette after a series of 100 to 1000 spins, segmented into groups of 100?
The table below illustrates the probabilities of achieving a net win, a loss, or breaking even after completing 100 to 1000 spins in sets of 100. For instance, the likelihood of being ahead after 500 spins stands at 11.0664%.
Net Outcome in Roulette
| Spins | Net Win | Even | Net Loss |
|---|---|---|---|
| 100 | 0.265023 | 0.069282 | 0.665695 |
| 200 | 0.207117 | 0.042698 | 0.750185 |
| 300 | 0.165841 | 0.030361 | 0.803798 |
| 400 | 0.134792 | 0.022893 | 0.842315 |
| 500 | 0.110664 | 0.017826 | 0.871510 |
| 600 | 0.091518 | 0.014167 | 0.894315 |
| 700 | 0.076106 | 0.011418 | 0.912476 |
| 800 | 0.063567 | 0.009298 | 0.927135 |
| 900 | 0.053283 | 0.007631 | 0.939086 |
| 1000 | 0.044796 | 0.006302 | 0.948902 |
These types of calculations can be easily executed in Excel using the BINOMDIST function. Here’s how to format it:
BINOMDIST(number of successes, number of trials, probability of success, cumulative?).
For the last parameter, you should enter 0 for a precise number of occurrences and 1 for that number or fewer.
Here’s a demonstration of applying it to the case of 500 spins:
The probability of a net loss = Probability of 49 or fewer wins = BINOMDIST(249,500,18/38,1) = 0.871510.
The probability of breaking even = Probability of exactly 250 wins = BINOMDIST(250,500,18/38,0) = 0.017826.
The probability of a net gain = Probability of 49 or fewer losses = BINOMDIST(249,500,20/38,1) = 0.110664.
This topic is being examined and discussed on my forum at Wizard of Vegas .
You have a maximum allowance of one square foot of metal to construct a can, which includes the top and bottom parts. What radius would optimize the can's volume?
I interpret your question as suggesting that you can design the one square foot of metal in any configuration you choose, including using two circles and a rectangle for the side of the can.
The radius should be 1/sqrt(6π) = 0.230329433 feet.
The height needed is roughly 0.690988299 feet, while the volume is about 0.115164716 cubic feet.
Recall the volume of the can is πr2In this context, r represents the radius, while h denotes the height.
Also remember that the total surface area, including the top and bottom, can be represented as 2πr.2+2πrh
Set the surface area equal to 1: 1 = 2πr2+2πrh
Solving for h: h = (1-2πr2)/2πr.
Substitute that into the volume equation: V=πr.2* (1/(2πr) - r)
= r/2 - πr3
DV/dr = 1/2 - 3πr2
Set the derivative to zero and solve for r:
3πr2= 1/2
r = 1/sqrt(6π)
This topic is being examined and discussed on my forum at Wizard of Vegas .
What is the formula for 1+2+3+..+n?
Click here for my solution (PDF).