# Ask the Wizard #5

Mitch

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.

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 U.S., I seldom see other players playing proper basic strategy. Money management has nothing to do with it.

Rob

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.

Alex from Berkeley, CA

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.

Jeff B. from Miami, Florida

The answer is 6*(1/6)^{6} = 6/46,656 = 1/7,776 =~ 0.0001286 .

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.

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.

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

Donald from Rochester, New York

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.

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, 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.

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, 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.

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