# Ask The Wizard #14

How do you go about working out the separate probabilities for getting: (1) queen high, (2) king high, and (3) ace high in three card poker?

I'll just work out the king high first and briefly show the formula for the other two. The probability will be the number of king high hands divided by the total number of hands. The number of ranks less than king is 11. The king high hand must have two different of these ranks. The number of ways to arrange 2 out of 11 is combin(11,2) = 55. However, one of these combinations is king-queen-jack, resulting in a straight, so subtracting that combination there are 54 left that do not form a straight. Next, there are four suits for each rank, or 4^{3}=64 possible combinations of suits. However, four of these 64 result in a flush so there are 64-4=60 combinations of suits left. So, the total number of king high combinations is 54*60=3240. There are a total of combin(52,3)=22,100 to arrange 3 cards out of 52. So, the probability of forming a king high is 3,240/22,100 = 0.1466063 . The probability for ace high is: (combin(12,2)-2)*(4^{3}-4)/combin(52,3)=0.1737557. Note the -2 instead of -1 because of both the a-2-3 and q-k-a straights.

The probability for queen high is: (combin(10,2)-1)*(4^{3}-4)/combin(52,3)=0.119457.

Hi there -- love your site! If I have blackjack, and the dealer is showing an ace, I am given the option of taking even money or continuing the hand. Should I take the even money or play on?

This is another way of looking at the decision to take insurance. The expected return of taking even money is obviously 1.0 units bet. Lets assume an infinite deck of cards for the sake of simplicity. The probability the dealer will have a blackjack is 4/13 and the probability otherwise is 9/13. If the dealer gets a blackjack, then you push. If not, then you win 1.5 units. The expected value of not taking insurance is (4/13)*0 + (9/13)*(1.5) = 13.5/13 = 1.0384615, which is greater than 1.0. Thus declining insurance, or even money, and playing the hand is the better bet. In an actual game with a finite number of decks the odds are even better since one 10 is already out of the deck (in your hand) which lowers the dealer's probability of having a blackjack.

"Thou shalt not hedge thy bets." -- Ten Commandments of Gambling (#7)

Hi, I'm doing a project on Bingo and I would like to know how to find the probability of Bingo. The probability of getting a line, horizontally, diagonally and vertically, a coverall, and the four corners. I have already seen your probability table and i wanted to know the formula you used.

The probability of getting a bingo (5 in a row) is complicated to explain, mainly due to the free square. I used a computer to do it. Four corners is much easier. The probability of having 4 corners, given x marks on the card, is combin(20,x-4)/combin(24,x). In other words, it is number of ways to put 4 of the marks in the corners and the rest anywhere else divided by the number of ways to put all x anywhere on the card. The probability of getting four corners within y calls is the sum for i=4 to y of the product of the probability that given y calls there will be x marks on the card and the probability that these x marks will form four corners (above). The probability of getting x marks in y calls is combin(24,x)*combin(51,y-x)/combin(75,y). Following this logic, you should be able to see the math for a coverall.

I have heard it said that with one come bet a player should call off the odds on a come out roll, but with two or more the player should leave them on. The rational is that with two or more on the odds of rolling one of the come points is greater than rolling a 7, but with one bet the player is more likely to win than lose.

The player should always leave the odds on regardless of how many come bets are active. When considering the options, looking at the probability of winning alone is not enough. Yes, with one come bet the odds of losing the come bet are greater than winning, however the potential win is greater than potential loss. The reason the player should always leave the odds on are because it is a bet with zero house edge. By turning the odds off the player is making the overall game more heavily weighted towards the bets with a house edge, thus increasing the overall ratio of the expected loss to the total amount bet.

I do not like to gamble. Any look at the math will show that I am bound to lose, the exception being blackjack, if I could do the complex count. However, I do enjoy the casino atmosphere. So, the question is how do I stretch my money to lose as slowly as possible?

I seem to recall an article that stated that by placing two bets on the same roll in craps that I could cut down the house odds to the minimum. I'm not going to win big, but I won't lose big either. I know it doesn't sound very exciting, but I'm a boring guy. I figure my wife and I can side up the table, separately, and in effect cancel out each other, one will win big, the other will lose big. If we each bring a big enough roll, maybe we can stretch it out into a couple of hours of play. I think the bets were Pass & Don't Pass.

Fred, betting on opposite sides of the same bet is no fun. If you and your wife talk to each other, then you will look like fools by betting against each other. If you pretend to not know each other it will take away from the fun.

Your priority to minimize potential losses, yet still play, is not unusual. Personally, I would find a place with low minimums where you can feel comfortable with and play a low volatility game with a low house edge. Two slow games with lots of pushes are pai gow poker and pai gow (tiles). You may not have the patience to learn tiles, so my perscription is to take up pai gow poker.

I was just wondering, if you know everything, then what was the date that my boyfriend and I first started dating and broke up (the very last one)? Will we ever be back together?

First starting dating on August 17.

Last break up on February 4.

You will break up four more times.