Ask the Wizard #31

Ask and ye shall receive. Please see my page on Catch a Wave.

How should I know? However don't make any accusations unless you have sufficient evidence to back them up.

If, indeed, they won it was because of luck and not because it was a winning system. As I have said a thousand times before, any system based on a negative expectation game in the long run not only can't overcome the house edge, it can't even dent it.

Please see my hole carding strategy for Three Card Poker. Following the strategy, you will enjoy a 3.48% advantage!

Thanks for the compliment. The formula for the house edge in buy and lay bets is the commission divided by the bet plus commission. In this case, the best bet is to bet $39 for the $1 commission. On the buy bet the house edge would be 1/40 = 2.5%. Assuming you can lay $78 to win $39 on the 4 and 10, and still only pay $1, the house edge would be 1/79=1.27%. I'll leave the other situations as an exercise for the reader (I hated it when my math books would say that).

I have never seen solitaire played for money in Vegas. I understand in the early days of Vegas people wagered on the standard Klondike variation of solitaire but I don't anything else about it.

No.

Not counting a three of a kind and two pairs, the following are the ways to get a three pair and number of combinations.

No wild card: combin(13,3)*10*6³*4 =2471040
Wild card used to compete pair of aces: combin(12,2)*10*6²*4² = 380,160
Wild card used as singleton ace: combin(12,3)*6³ = 47,520

The total number of combinations is 2,898,720. This is less than half of the 747,0676 combinations for a three of a kind.

Each frame in these video slots is weighted equally. Any given line is equally likely to produce any given combination. Thus, the return is the same regardless of the number of coins played.

The casinos switch dealers when it is time for someone to go on a break or go home. Switching dealers does not change the player's odds unless the player is a card counter and the game is single- or double-deck, where a new dealer necessitates a fresh shuffle.

The machine picks one number for each reel.

Thanks for the compliment. Any introductory probability and statistics book should give good treatment to the binomial distribution. Briefly, the binomial distribution is the probability that any given number of events will happen given a specific probability for each event and a specific number of trials. Specifically if the probability of each success is p, the number of success is s, and the number of trials is n then the probability of s successes is p^s * (1-p)^n-s * combin(n,s). The combin function is explained in my glossary. For example, suppose you want to know the probability that in 100 spins of a roulette wheel the number of reds will be exactly 60. According to the binomial distribution, the probability is (18/38)⁶⁰ * (20/38)⁴⁰ * combin(100,60) = 0.003291.

Excel also has a function for the binomial distribution. It is =BINOMDIST(x,n,p,0), where:

x=number of positive trials. n=total number of trials. p=probability of success in any given trial.

Use a 0 in the fourth position of the function for the exactly probability of x wins. For the probability of x or less wins, use a 1.

In the roulette example above, the function would be =BINOMDIST(60,100,18/38,0)

The house edge is the same regardless of how many come bets you make assuming you always take the maximum allowable odds and leave the odds turned on during a come out roll. How many come bets you make should be a matter of personal preference.