Ask The Wizard #390

In the last Ask the Wizard column, you said the Trips bet is supposed to always have action in Ultimate Texas Hold 'Em, including when the player folds. I know of many casinos that require a Play bet for the Trips to have action, otherwise they take it down. Sometimes they justify it by comparing it to Three Card Poker, where the Pairplus bet is taken down if the player folds. My question is how does this rule violation affect the house edge?


Good question! I did only some cursory math on this one, because it takes my computer days to cycle through every combination of cards in Ultimate Texas Hold 'Em, not to mention my own re-coding time.

If the player plays by correct strategy for maximizing the value of the base game, I find the increase in house edge of the Trips bet is 0.27%. However, the player might also make a bad small raise to save the Trips bet. I show the increase in house edge of the base game to be 0.11% if the player never folds with a three of a kind on the board. The player should consider just how bad the raise is and the ratio of his Trips to Ante bet in deciding how to play a three of a kind on the board with two low kickers as his own cards. Of course, such a player probably wouldn't make the Trips bet to begin with.

I would like to repeat my sources, from the last column, on the rule that the Trips bet always has action.

If this situation happens to you in Nevada or Washington, I would protest it at the table and file a dispute with the gaming authorities if it doesn't go your way.

This question is asked and discussed in my forum at Wizard of Vegas.

What is the probability of being up betting even money bets in double-zero roulette after 100 to 1000 spins, in groups of 100?


The following table shows the probability of a net win, loss and being exactly even after 100 to 1000 spins, in groups of 100. For example, the probability of being up after 500 spins is 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

Such calculations are easy in Excel with the BINOMDIST function. Here is the format of how to use it:

BINOMDIST(number of occurrences, number of trials, probability of success, cumulative?).

For the last term, put in 0 for exactly that many occurrences and a 1 for that many or less.

Here is an example how to use it for the 500 spin case:

Probability of net loss = Probability of 49 or less wins = BINOMDIST(249,500,18/38,1) = 0.871510.
Probability of being even = Probability of exactly 250 wins = BINOMDIST(250,500,18/38,0) = 0.017826.
Probability of net win = Probability of 49 or less losses = BINOMDIST(249,500,20/38,1) = 0.110664.

This question is asked and discussed in my forum at Wizard of Vegas.

You are allowed a total of one square foot of metal to make a can, including the top and bottom. What is the radius of the can that maximizes the volume?


In interpret your question to mean you can shape the one square foot of metal any way you like, including two circles and a rectangle for the side of the can.

The radius should be 1/sqrt(6π) =~ 0.230329433 feet.


The height is approximately 0.690988299 feet and the volume approximately 0.115164716 cubic feet.


Recall the volume of the can is πr2h, here r is the radius and h is the height.

Recall also the surface area, with top and bottom, is 2πr2+2πrh

Set the surface area equal to 1: 1 = 2πr2+2πrh

Solving for h: h = (1-2πr2)/2πr.

Plus that into the equation for the volume: V=πr2 * (1/(2πr) - r)

= r/2 - πr3

DV/dr = 1/2 - 3πr2

Set the derivative equal to 0 and solve for r:

3πr2 = 1/2

r = 1/sqrt(6π)


This question is asked and discussed in my forum at Wizard of Vegas.

What is the formula for 1+2+3+...+n?




Click here for my solution (PDF).