Practical Probability - Part II

In this column, I will cover some of the most basic concepts of probabilities and odds. First, I'll give you a quiz with a few easy questions, and then one that should be easy but has proven to create conceptual problems for a lot of people, including some very experienced mathematicians.

1. If you are dealt one card from a normal deck, what is the probability that it is an ace?
2. What are the odds against it being an ace?
3. Two normal coins are flipped. There are three possible outcomes – two heads, two tails, and one of each. What is the probability of each outcome?
4. What are the odds against each outcome?
5. This is the tough one. It is known as "The Monte Hall Problem," after the old TV show Let's Make a Deal. You're on a game show, and you're given the choice of three doors. Behind one door is a car; behind the others, goats. You pick a door – say, No. 1 – and the host, who knows what's behind the doors, opens another door – say, No. 3 – which has a goat. He then asks you, "Do you want to switch to door No. 2?" Is it to your advantage to switch your choice?
6. What is the probability of winning the car on your first guess? Is the probability of winning the car the same if you switch? If not, what is it now?

A simple, not too rigorous, way to approach probability is to think of something that has a number of possible outcomes. Each outcome is equally likely. Then, specify a specific outcome or group of outcomes. The probability of the specified group is its number of outcomes divided by the total number of outcomes. For example, a deck has 52 cards. There are 13 spades. The probability of drawing a spade is 13 divided by 52. This can be written as a fraction, 1/4; a decimal, .25; or a percentage, 25%.



Odds are often the easiest way for a gambler to apply probability to making betting decisions. The odds compare the outcomes that aren't in the specified group to those that are in it. Thus, there are 39 non-spades. If we divide this by 13 spades, we come up with 3-to-1. (As an item of trivia for rock-music buffs, The Doors miscalculate the odds in one song. "The odds are five to one, yes one in five, that no one here gets out alive." If the odds are 5-to-1, there is one chance in six.)



Answers:

1. 1/13 or .077 (4 aces, 52 cards; therefore, 4 divided by 52)
2. 12-to-1 (48 non-aces, 4 aces; therefore, 12-to-1)
3. This question has a small trap in it. There are really four outcomes: head, head; tail, tail; head, tail; and tail, head. Therefore, the probabilities are .25 for head, head; .25 for tail, tail; and .50 for one of each.
4. 3-to-1 against two heads, 3-to-1 against two tails, and 1-to-1 against the split (also known as even money or pick 'em)
5. It is right to switch. More discussion of this follows the answers.
6. The probability of your first guess being right is 1/3. The probability of the second being right is 2/3.

The Monte Hall Problem: This problem assumes that you prefer a car to a goat. (Judging from the hands my opponents have been playing lately, I wouldn't be surprised to learn that a lot of players find goats pretty attractive.) Figuring out the probability of your initial guess being correct is fairly straightforward. You will guess correctly 1/3 of the time. The confusing part relates to deciding whether to switch. When first presented with this problem, a lot of people realize that it is correct to switch. It appears, however, that the elimination of one door has changed the probability from 1/3 to 1/2 (a choice from three doors down to a choice from two). But remember that one of the losing doors has been eliminated. The best way to arrive at the answer is to compare switching to your initial choice. If your initial choice was right (1/3), your switch costs you the car. If your initial choice was wrong (2/3), the switch guarantees that you will win, since the only other wrong choice has been eliminated. So, switching will work two-thirds of the time. This problem has generated a great deal of interest in academic circles and among casual puzzle-solvers. It crops up in the excellent novel The Curious Incident of the Dog in the Night-Time by Mark Haddon.



Using Probabilities and Odds in Simple Poker Situations: New players, and even some more experienced ones, often get confused by discussions of outs and odds. Basically, outs are the cards that will win for you. If there is more than one card to come, outs are sequences that will win. For example, you have the A K and your opponent has the A A. He moves all in on fourth street with a board of Q 8 4 2 and one card to come, so you have nine outs or winners (any spade). You have him covered. Should you call? We have specified that there are eight known cards, and therefore 44 unknown ones. So, you have nine winners and 35 losers. The odds against you are 3.9-to-1. Now you have to calculate the total amount in the pot, including his last bet, and divide by the amount needed to call. This gives you pot odds. For example, if there is 800 in the pot and he bets 200, making the total pot 1,000, you are getting 5-to-1 (1,000 divided by 200). Since your hand is less than a 4-to-1 underdog, you should call.



Outs may also have to be downgraded to account for the fact that a card that improves your hand may improve your opponent's hand. His improved hand may beat or tie yours, so what may have appeared to be an out really isn't. In real life, you seldom know his exact hand, so some of your outs will often be tainted. In the above example, suppose that your hand and the board are the same, but your opponent might have any of these four hands: (a) A A, (b) A Q, © Q J, or (d) 8 8. How many outs are there for each case?



We have already discussed (a), and there are nine outs (spades.) In (b), there are 12 outs (nine spades and three kings). In ©, there are 14 outs (eight spades, three aces, and three kings). In (d), there are only seven outs (there are nine spades that appear to be outs, but the 4 and the 2 not only make a flush for you, but a full house for your opponent, and must be eliminated, leaving only seven. I will continue the discussion of outs and odds in my next column.



Steve "Zee" Zolotow, aka The Bald Eagle, is a successful games player. He currently devotes most of his time to poker. He can be found at many major tournaments and playing on Full Tilt, as one of its pros. When escaping from poker, he hangs out in his bars on Avenue A – Nice Guy Eddie's on Houston and Doc Holliday's on 9th Street – in New York City.