# Ask The Wizard #5

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.

In the first 20 blackjacks the probability that you get exactly x of them is (1/2)^{20} * combin(20,x). Let me save you the trouble. Here are the odds of exactly 0 to 3 blackjacks out of the first 20:

Pr(0) = 0.0000010

Pr(1) = 0.0000191

Pr(2) = 0.0001812

Pr(3) = 0.0010872

Pr(3 or less) = 0.0012884

So, the odds of 3 or less are 1 in 776. That is not enough to warrant an accusation of cheating. It could easily be simple bad luck.

Personally, I like to see four standard deviations south of expectations (probability of 1 in 31,574) before I get suspicious. It would take five standard deviations (probability of 1 in 3.5 million) before I made a formal accusation.

According to your blackjack software tables, both the Unified Gaming 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?

How does the house have the advantage in Blackjack?

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

Single deck blackjack rules are usually tight. Cards are dealt face down. Doubling is usually restricted to 9 to 11, or 10 to 11. The dealer will hit a soft 17 and double after a split will probably not be allowed. Be sure to NOT play if blackjack pays less then 3 to 2, which is usually the case with single-deck games.

You can always see the relatively current blackjack rules in Vegas at my Wizard of Vegas site.

If you are rolling 6 six-sided standard dice what are the odds of rolling six of a kind?

The answer is 6*(1/6)^{6} = 6/46,656 = 1/7,776 =~ 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.

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.

Normally, 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 - 0.00199521^{75} = 1 - .9980048 ^{75} = 1 -.8608886 = .1391114.

However, in bingo we can't use the method above because all cards go against the same draw of balls. It is hard to explain, but because the cards are arranged in five columns of 15 possible numbers each, the expected number of balls is correlated. It would take a random simulation to properly answer your question. Without doing that, 13.9% is a good rough guess.

When I calculate the combos of player and dealer hands for Caribbean 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.

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) = 19,933,230,517,200.

I had the good fortune of hitting four 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 $5,000 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.

Table game tournament strategy is very complicated. However, briefly, I would bide by time in the early hands of each round. Sometimes your opponents will all burn themselves out and you can advance without any effort. When it gets down to about five hands to go you'll need to make your move on any players way ahead of you. This is the time when you want to pull into first or go bust trying. It is also good to wait to save your big bets for when you act AFTER your biggest competitors.

What are the odds in drawing three cards to a pair and getting a full house at five-card draw poker?

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, let's 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)=16,215 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 165/16,215 = 0.0101758, or about 1 in 98.

For more information on the combin() function, please see my section on probabilities in poker page.

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.

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, then the dealer will pay off your bet and the tip separately. 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 and gave the dealer his money.

Given that a lottery has 10 million potential combinations, what are the odds that someone will win with 90% confidence given that 10 million 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 methodology used to solve it.

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%, which is very closely approximated as 1-(1/e).