# Ask The Wizard #144

During a 4-hour layover in Vegas, what’s my best strategy to double a \$2000 bankroll? what game, large or small bets, etc.?

Tom from Culver City

First I would take a taxi to the Hard Rock, the closest major casino to the airport. I’m not sure how much odds the Hard Rock allows in craps but I would guess 3-4-5. If that is the case then bet 1/7 of your bankroll on the don’t pass bet, or \$275 to round down. If a point is established then lay the maximum on the odds, or \$1650. If you win you’ll be a lot closer to your goal, the amount will depend on the point. Win or lose bet the lesser of 1/7 of your bankroll and 1/7 of how far you are from \$4000. If you get close to either extreme just get it over with and bet everything if you’re low, or whatever you need to close the gap on \$4000 if you’re high, and forget about the odds. Four hours should be enough time. However don’t dilly dally. The lines at security can get pretty bad. If your outbound flight is in terminal C be sure to ask an agent about the secret entrance.

As I read your analysis of the Royal Match side bet in blackjack, am I correct that your odds are for the first hand of the shoe? If so, wouldn’t the real-world odds of an easy match be tilted a bit more toward the player? It seems to me that if the suits get unbalanced in any direction it would slightly lessen the house edge, and the suits will certainly fluctuate through the shoe.

Frank from Michigan

This is not true. The remaining deck needs to be exhibit more than a certain degree of skewness for the odds to swing to the player's favor. Consider a hypothetical side that pays 3 to 1 for any suited pair in a one-deck game. Of the top of the deck the probability of winning is 4*combin(13,2)/combin(52,2) = 23.53%. However if you burn two cards of different suits the probability of winning goes down to 2*(combin(13,2)+combin(12,2))/combin(50,2) = 23.51%. If you burn two cards of the same suit the probability of winning increases to (3*combin(13,2)+combin(11,2))/combin(52,2) = 23.59%. If one card of each rank were removed the probability of winning would go down to 4*combin(12,2)/combin(48,2) = 23.40%. What all this shows is that if cards are removed at a uniform distribution the odds of winning go down, however at a very skewed distribution the odds go up. As the deck is played down sometimes your odds get better, and sometimes worse, but in the long run they average out and stay at a 23.53% chance of winning.

In full pay deuces wild the probability of getting a royal flush is about 1 in 40,000. Could it be said the probability in 5-play would be five times easier, or 1 in 8000?

TS from Santa Barbara

Almost. If more than one royal per deal in 5-play counts as only one sighting then you will have sightings slightly less than 5 times as often. This is because the total number of royals will be five times as much, but sometimes they will be clumped together in the same play, usually when you get a royal on the deal, and thus 5 on the draw.

The following table shows the probability of making a royal in 1-play according to the number of cards to the royal held, assuming full pay optimal strategy.

### Royal Flush Probability in 1-Play Video Poker

 Card Held Probability on deal Probability on draw Total probability 0 0.19066396 0.0000014 0.00000027 1 0 0.00000561 0 2 0.01969711 0.00006167 0.00000121 3 0.01299751 0.00092507 0.00001202 4 0.0003309 0.0212766 0.00000704 5 0.00000154 1 0.00000154 Total 0.22369101 0 0.00002208

What this table shows is that 22.37% of the time you will have a possible royal draw. The rest of the time a royal will be impossible, for such reasons as you held a wild card or a pair. The lower right cell shows the overall royal probability is 0.00002208, or 1 in 45282.

The next table shows the same thing but for 5-play, and the probability of at least one royal.

### Royal Flush Probability in 5-Play Video Poker

 Card Held Probability on deal Probability on draw Total probability 0 0.19066396 0.00000698 0.00000133 1 0 0.00002803 0 2 0.01969711 0.00030832 0.00000607 3 0.01299751 0.0046168 0.00006001 4 0.0003309 0.10195134 0.00003374 5 0.00000154 1 0.00000154 Total 0.22369101 0 0.00010268

Note the probability of at least one royal is 0.00010268. This is 4.65 as high as the probability for one-play. The reason is the probability of making at least one royal is always less than five times that of 1-play. For example the probaiblity of hitting a royal holding for to the royal is 1/47 in 1-play. However in 5-play the probability of making at least one royal is 1-(1-(1/47))5 = 0.101951341, which is about 4.79 times as high.

During my last two sessions at my local casino of choice, I’ve been on the positive end of dealer error for two different poker games: 4-card poker and 3-card poker. Both times, I remained silent and in the case of 4-card poker, certainly didn’t want to ruin it for the other players. Not to rationalize, but sometimes I also feel that bringing these mistakes to light can make the dealer look worse than letting it slide -- assuming of course, the eye in the sky doesn’t catch the error -- with a subsequent reprimand behind the scenes. I have a few questions related on this topic. Has there been any study or estimation relating to how dealer error affects the house edge? Certainly all but the most incoherent or novice player will catch errors that help the house, but it’s been my experience that most errors that help the player are not brought to light, by players, anyway. Looking back, dealer error has given back a significant amount of money to me over the past year. Thanks for any insights!

Tim from Cleveland

I do not know of any formal study. As you would expect dealer errors tend to decrease as the quality of the casino goes up. I have personally played thousands of hours behind the tables since I turned 21 years old 19 years ago (it seems like just yesterday). Based on all that play I strongly feel that most errors favor the house, probably about 80%. For example many dealers do not know that you still pay the ante bonus in Three Card Poker even if the player lost or the dealer didn’t qualify. (Bluejay got shortchanged this way because he wasn’t sure of the rule himself.) I’ve had a few dealers disagree with me on this rule, who were later correctly overruled by the pit boss. I wonder how many players, who don’t know the rules as well as I do, were shortchanged by these same dealers before I played. Of course the error is more likely to be corrected if the error favors the dealer. I tend to think the cost of errors to the casino is not very high because of the higher percentage in favor of the dealer. In fact I wouldn’t be surprised if the casinos actually made money from errors overall. If anyone in casino management has another point of view then I’m all ears to hear it.

They say the probability of winning the Powerball lottery is 1 in 146,107,962. In the recent drawing for a \$340 million jackpot the local media said the number of tickets sold was 105,000,000. My questions are if you win what is the probability you will have to share the jackpot, and how much does this reduce the expected value?

Mitch F. from Hopkins, MN

First let’s confirm that probability. The player must match 5 regular numbers from a pool of 55, and one Power Ball from a pool of 42. The probability of winning would be 1 in combin(55,5)*42 = 1 in 146,107,962. So I agree with your probability. I like to use the Poisson distribution for questions such as yours. The mean number of winnings would be 105,000,000/146,107,962 = 0.71865. The general formula for the probability of n winners, with a mean of m, is e-m*mn/n!. In this case the mean is 0.71865 so the probability of zero is e-0.71865*0.718650/0! = 0.48741. So the probability of at least one winner is 1-0.48741 = 0.51259. So 0.71865 winners will have to 0.51259 of a jackpot to split up. That is 0.51259/0.71865 = 0.71327 jackpots per winner. So jackpot sharing reduces your expected win to 71.327% of the jackpot amount, or a reduction of 28.673%.

Wizard, I’ve got a question regarding my "true" winning percentage for NFL picks. I tell my friends that I have not had a losing season for the past 5 years (which is true), betting from Weeks 3-16 with an average of 2-3 picks per week. My thing is that I have a great feel for games and I bet substantially more in games where I believe the odds are in my favor. As an example, I have made 9 bets this year, 4 of which were \$55, hitting 3 of those picks. I have made 3 \$110 bets, winning 1 of those. But I loved two games in which I bet \$330 on one, and \$600 (even line), and won both. Absolutely, I’m 6 of 9 for 66% clip, but being that I hit on the games I was more sure about, it really signifies around an 80% clip (money wise). Does this make sense to you? I’ve roughly done this, it turned out, when I calculated it for the last 5 years (60-70% in picks, but ~80% when factoring how much money I wagered). Can you tell me an easy way to factor in the amount bet, and If i’m correct in my assumptions? Thank you very much.

Luke from Chicago

I have been thinking a lot about this lately. In my opinion a winning percentage should have an equal weighting per game. You should also have a separate statistic on your overall return of investment, however any statistic should be backed up with a list of the side, date, line source, point spread, and the odds (usually -110). Another issue you don’t bring up is what to do if you have to lay -120 off of a 3 or 7 point spread. It would be easier to attain a good win percentage if you loaded your picks with such bets. So I believe a return on investment figure should be kept even if flat betting. Another thing that bothers me about some other handicappers is they quote lines that are nowhere to be seen. I think it is okay to shop around a bit but quoted lines should be not difficult to find. I admit I don’t do some of these things myself this season, because I didn’t think of these things when I started. Next year, if I do this again, I will document my results as an investment as well.