Once the hierarchy poker hands is clear, then it is time to know how to calculate the chances that your hand will improve as the game progresses. This mental activity requires a combination of logic and math, or, in the alternative, so much playing experience that the results of the logic and math can be used out without having to calculate them all over again. Calculating the chances for hand improvement is just one example of how much technical thinking must go on in poker:

Imagine a game of five card draw with five other players. The player in last position is dealt a hand with four miscellaneous low spades and a fifth low card in an off suit. To decide whether to play the hand, the player needs to calculate the odds of receiving a spade on a one-card draw.

Many might think the odds are 1 in 4. This could be true if the correct idea were to pull a spade from a shuffled deck of 52 cards. The probability that the first card will be a spade is 1 in 4 or 25% (assuming the bottom card is not visible).

This is not the answer for the poker example. It will be important to consider as well the 4 spades in the hand, and the fact that there are 47 unseen cards in the hands of other players or in the deck. It is appropriate to consider the unseen cards of the other players in conjunction with the unseen cards in the deck, as the probabilities are the same.

Thus, the probability of receiving a spade as the next card is (13-4) divided by 47. This number is 19.1%. This is a little worse than 5:1 odds against. ("Worse" here means more remote a possibility.)

Whether it makes sense to bet on a possible 19.1% outcome depends on how much is in the pot and the cost to stay in the play (the "pot odds"). Five to one pot odds can not easily occur on the initial betting round with only six players in the game, and the chance of success in this exact venture is slightly worse that 5:1 odds anyway. For this reason, these cards should not be played.

Pot odds of 5:1 are unlikely on an initial betting round with six players because the number of bets in the pot (the numerator) can only be five or more if everybody else is in the hand already or if another player or two have raised. In the latter event, the denominator, which is the number of bets required to play, would go to 2 or more. So the only scenario for 5:1 pot odds would be if the player in question was last to bet, and everybody else was in the game, with no raises.

Experienced players, having already confronted this situation before, probably do not need to work through the math again. They just know. The methodology for determining the possibilities for all deals or draws is essentially the same as that just described, except that the specifics can become complex in calculation. For example, in a game of seven-card stud, with "Fourth Street" revealed (so that each player has two down cards and two up cards), what is the chance of getting at least one ten in the next three cards? First, of course, the number of known tens needs to be figured out. Suppose you, as the player, have two and there are no other visible tens on the table. Suppose further, that there were four other players in the game, but one folded before the fourth card was issued. That means that 19 cards have been dealt. Eight of them are not visible to you (the hole cards of the other players). That leaves 11 "known cards" and hence 41 unseen cards.

Of the 41 unseen cards, two of them are tens. The chance that the next card will be a 10 is 2/41. Later you will learn how to calculate multiple-card odds. For current purposes just accept the statement that there are a total of 10,660 different ways that three cards can be combined out of a deck of 41 cards. Of these, 1482 combinations have one ten in them, and 78 have two tens in them. So the chance that at least one of the next three cards dealt to the player will be a ten is about 14.6% or approximately 6:1 odds against.

If the pot odds are more favorable than 14.6%, then it would be worth trying to stay in the game for a third ten. Embedded in this probability is the 0.73% chance of having four of a kind (picking up both tens).

Now an experienced player might be able to take the number of cards left (41), and divide it into 2 for the odds of a ten in one card. Forty-one is close to 40. Two divided by 40 can be done in the head. That's 5%. With three cards, the chances are about 15%. That is a really quick approximation that may be a little optimistic, but certainly comes close. Without using a calculator the player will know that the pot odds have to be 6:1 or better to justify hanging in there.