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Reason #5 why the Wizard likes Bovada:

Intelligent Bonuses

Many online casinos offer huge signup bonuses, but there’s a catch. Buried in the fine print is that play on the most popular games doesn’t count towards earning the bonus. It’s common for play on blackjack, baccarat, roulette, craps, and video poker to be excluded. In many cases, only slots count.

And that’s if you can even find the terms and conditions. Many casinos put their 100% bonus in big flaming letters but make you hunt all over their site to find the rules.

Bovada allows play on all games to count towards the wagering requirement. It’s that simple. Just no opposite betting. All casinos ought to be as easy as Bovada about this. The bonus offer itself is simple too: on your first deposit, they’ll give you an extra 10%. If you deposit $100, you’ll wind up with $110 in chips or tokens.

Finally, in the unlikely event that Bovada feels you’ve been abusing their bonuses they won’t seize your winnings like most other casinos will. In the worst case scenario they will politely tell you that they will not be offering you any future bonuses, but you are welcome to keep playing and keep everything you have made already.

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Ask the Wizard #5

Edition Date: Mar 4, 2000

Do you really trust the online gambling sites for such games as blackjack. I played one a few weeks back. The dealer got 14 blackjacks to my 3 during that session. I’m just suspicious and interested in your comments. Thanks.
— Mitch

What you should be worried about from online casinos is not that they are cheating with the cards. The casinos don’t need to cheat to make money, the built in house edge is sufficient to ensure financial security. I speculate that if they did cheat they would actually make less money because their players would not return. However I do not claim that nobody is cheating, anything is possible. What you should be worried about is bad customer service. Some online casinos have a lot to learn about answering e-mail timely and understanding their own rules. Regarding your question about getting only 3 blackjacks out of 17 the probability of getting 3 or less is:

(1/2)17*(combin(17,0)+combin(17,1)+combin(17,2)+combin(17,3)) =

(1/131072) * (1 + 17 + 136 + 680) = 834/131072 = .0063629.

In other words this is going to happen to 1 person in 157 on average. Still this is nowhere near the kind of evidence needed to fairly acuse foul play. With any kind of gambling your money will go up and down like a roller coaster. To truly see if the long run looks fishy you need to log thousands of trials and their outcome. I may be willing to analyze the results of such data if it were summarized properly.

 

According to your bj software tables, both the perplexa and boss media systems give the player a slight edge. If this is so, how can the casinos that use these software systems turn a profit? Is it because relatively few players use basic strategy and/or good money management techniques?
— trekon3

Online casinos are not the only ones to offer positive expectation games. Using basic strategy some Las Vegas casinos offer games with a negative house edge. There are video poker games all over the country with a positive expectation assuming optimal strategy. The reason the casinos can afford to do this is that the vast majority of players make errors in strategy. As someone who has played hundreds of hours of blackjack in casinos all over the US I seldom see other players playing proper basic strategy. Money management has nothing to do with it.

 

How does the house have the advantage in Blackjack?
— Rob

The dealer has the advantage because the player has to go first. If both of you bust it isn’t a tie, you lose.

 

With single-deck blackjack, are the cards dealt face-down? And what are the general rules for single-deck? I know you mentioned double downs only on 9-11 or 10-11, but do you know what the general rules are for double-after-split, dealer-hitting-soft-17, etc. for single-deck? I’m a basic strategy player, and wanted to find out just what the rules tend to be for single-deck (I’ve only ever done 6 or 8 deck) so I can prepare the perfect spreadsheet from blackjackinfo.com. Thanks so much for your help!
— Alex from Berkeley, CA

Single deck blackjack rules are usually tight. Cards are dealt face down. In southern Nevada you can usually double on any two cards and in northern Nevada it is often 9-11 or 10-11 only. The dealer will hit a soft 17 almost always and doubling after a split is usually not allowed.

 

If you are rolling 6-six sided standard dice what are the odds of rolling six of a kind?
— Jeff Blum from Miami, Florida

The answer is 6*(1/6)6 = 6/46656 = 1/7776 =~ 0.0001286 .

 

Consider a bingo game with 75 random cards. Draw 12 random numbers, according to standard bingo rules. Is the probability of a bingo 75 * 0.00199521? (I got the 0.00199521 from your table of bingo probabilities for a standard occurring within 12 numbers called) If not, what is the probability that a bingo will occur? You have a great page.
— Charlie

You’re right, according to my table of my probabilities in bingo the probability of any one person getting a bingo within 12 numbers drawn is 0.00199521 . If the probability of an event happening is p the probability that it will happen at least once in n times is 1-(1-p)n. In this case the probability that at least one person will get a bingo is 1 - 0.0019952175 = 1 - .9980048 75 = 1 -.8608886 = .1391114 .

 

When I calculate the combos of player and dealer hands for Carib Stud Poker, I get only 3,986,646,103,440 vs your 19, etc. I’m off by exactly a factor of 5. I used combin(52,5)*combin(47,5). Where did I go wrong? Thanks and I think your site is just great.
— Bob from Lake Charles, Louisiana

Thanks for the compliment. You are off by a factor of five because the dealer can have any one of 5 cards face up. In other words order does matter with the dealer’s hand since the first card is dealt face up. The correct derivation of the total combinations is combin(52,5)*47*combin(46,4) = 19933230517200.

 

I had the good fortune of hitting 4 of a kind at a local casino, and was subsequently invited to play in a Let It Ride tournament, where approximately 300 players will compete for quite substantial prize money. My question is, what do you think would be the optimal strategy? Each player is to be given $5000 in play chips, and the minimum bet will be $25 a hand. There will be "heats", with the first round having all but 100 eliminated, second all but 25, third will have 6 left, and then the final round.
— Donald from Rochester, New York

Good question. The key to winning each heat will depend much more on how you bet then how you play your cards. I would suggest not deviating from the strategy presented in my let it ride section. A good betting strategy to increase your odds for a big win is to bet the minimum most of the time to keep the house edge from grinding you down. However you should have moments of very aggressive betting. I would bet as much as you can enough times so that if you win you’ll advance to the next heat. Let me know what happens.

 

What are the odds in drawing 3 cards to a pair and getting a full house at 5 card draw poker?
— Nick

There are two ways to get a full house in this situation: (1) draw a three of a kind or (2) draw one more to the pair and another pair. I’m going to assume you discard three singletons. First lets work out the number of combinations under (1). There are 3 ranks with only 3 suits left (remember you discarded 3 singletons) and 9 ranks with 4 suits left. The number of combinations is thus 3*combin(3,3)+9*combin(4,3) = 3*1 + 9*4 = 39. Next lets work out the number of combinations under (2). There are 2 suits left to add to the existing pair. There are combin(3,2) ways to form a pair from the 3 ranks with 3 cards left and combin(4,2) ways to form a pair to from the ranks with 4 cards left. So the total combinations under 2 is 2*(3*combin(3,2)+9*combin(4,2)) = 2*(3*3 + 9*6) = 126. The total number of ways to arrange a full house is the sum under (1) and (2), or 39+126=165. There are combin(47,3)=16215 ways to arrange the 3 cards on the second draw. The probability of drawing a full house is the number of ways to draw a full house divided by the total combinations, or 165/16215 = 0.0101758, or about 1 in 98. For more information on the combin() function see my section on probabilities in poker.

 

When is the best time to tip and where should you place that tip. Should you tell the dealer that you are tipping them? I am often concerned about the eye in the sky and where I place a tip, how much to tip, and what to say about the tip.
— Josiah from South Haven, Michigan

In general almost everything at the table should be communicated through hand signals and chip placement, including tipping. The vast majority of the time players make a bet for the dealer. To do this place the tip on the edge of the betting circle, along with your own bet in the middle. The tip is not subject to the table minimum since it is treated as part of your own bet, just earmarked for the dealer. If you double or split your own bet you should do the same for the dealer’s bet. If you win the dealer will pay off your bet and the tip sepparately. Don’t touch the tip or the winnings on the tip, let the dealer collect them. Once I forgot that I had made a bet for the dealer and started to put the tip and the winnings in my stack when the dealer said, "I thought that was for me!" Needless to say I was very embarrassed at this fauxpas and gave the dealer back their money.

 

Given that a lottery has 10mm potential combinations, what are the odds that someone will win with 90% confidence given that 10mm tickets are sold. Clearly it would not be 100% since some tickets would be duplicates. I am less interested in the answer than in the methodolgy used to solve it.
— Scott from New York, New York

Lets try to rephrase the question. Assuming the lottery has 10 million combinations and all players choose their numbers randomly (allowing for duplicates) how many tickets would the lottery need to sell so that the probability of at least one person winning is 90%? Lets let p be the probability of winning and n be the number of tickets sold. The probability of 1 person losing is 1-p. The probability of all n people losing is (1-p)n. The probability of at least one winner is 1 - (1-p)n. So we need to set this equal to .9 and solve for n.

.9 = 1 - (1-p)n
.1 = (1-p)n
ln(.1) = ln((1-p)n)
ln(.1) = n*ln(1-p)
n = ln(.1)/ln(1-p)
n = ln(.1)/ln(.9999999)
n = 23,025,850.

So the lottery would need to sell 23,025,850 tickets for the probability of at least one winner to be 90%. In case you were wondering if the lottery sold exactly ten million tickets the probability of at least one winner would be 63.2%.

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