I'm working to understand the differences between Odds, Probability and Chance. I've come up with a hypothetical situation to show where I'm having a bit of an issue.

Chad shuffles a standard deck of playing cards (52 cards, 4 suits (spades, hearts, clubs, diamonds), 13 cards per suit (2 through ace)), with the deck sufficiently shuffled to produce a random outcome. Chad then looks through the cards in the deck and memorizes their order.

Joe is asked what are his odds of drawing an Ace from the top of the deck, to which he replies 1 to 12. He is then asked what is his probability and his chance of drawing an Ace from the top of the deck, and he replies about 7.69% probability and 7.69% chance. Chad is asked what are the odds of Joe drawing an Ace from the top of the deck, to which he replies 1 to 12. Chad is then asked what is Joe's probability and chance of drawing an Ace from the top of the deck, and he replies 7.69% probability and 100% chance.

Joe draws the first card, and gets an Ace of Hearts.

As I understand it, the odds and probabilities don't change based on whether the outcome is known prior to the resolution, and probabilities are calculated as (Number of Positive Resolutions) / (Total Number of Resolutions). Chances seems to be similar to probabilities, but still different enough though, based on whether the resolution is known or not.

Since, after Chad shuffles the deck, the order of the deck is static, was Joe's chance of drawing an Ace from the top of the deck ever really 7.69%? Was Joe incorrect? Is there a fallacy in there somewhere? Or is the correctness of an answer for chance truly dependent on foreknowledge of the resolution?

Is there another mathematical concept that defines or attempts to quantify what the actual resolution will be, without the participant (e.g. Joe) knowing, but others may know, the resolution beforehand?

Edit: Daniel makes a point that the semantics may be an issue, so to facilitate an answer, and avoid the slippery slope of this becoming a semantics discussion, these are the best standard definitions I could find that state how I currently understand the terms:

Odds: "Betting odds are calculated in the form r:*s*('r to s') and correspond to the probability of winning P=s/(r+s). Therefore, given a probability P, the odds of winning are (1/P) - 1:1." http://m.wolframalpha.com/input/?i=odds+mathematical+definition&x=0&y=0

Probability: "Probability is a measure or estimation of how likely it is that something will happen or that a statement is true. Probabilities are given a value between 0 (0% chance or will not happen) and 1 (100% chance or will happen). The higher the degree of probability, the more likely the event is to happen, or, in a longer series of samples, the greater the number of times such event is expected to happen." http://en.wikipedia.org/wiki/Probability

Chance: I can't find a rigorous definition, just English definitions (http://m.wolframalpha.com/input/?i=chance+definition&x=0&y=0), but I believe my problem description most closely follows Baysian probability for chance, because according to the Wikipedia page (http://en.wikipedia.org/wiki/Probability) it "includes expert knowledge as well as experimental data to produce probabilities", which allows for Chad's foreknowledge of the order of the deck.


  • $\begingroup$ Joe's odds of drawing an ace from the top of the deck are 1 in 12, not 1 in 13. Odds are the ratio of the probability of an event occurring to the probability of it not occurring. $\endgroup$ – antirealist Feb 2 '13 at 11:43
  • $\begingroup$ Odds have been fixed. Thank you for the concise explanation. $\endgroup$ – Peter Feb 2 '13 at 11:50
  • $\begingroup$ I think all those words mean the same thing. Probability theory is only a tool for working with large number of inputs and outcomes. As appealing as it is, mathematics should not be expected to give you a "real" answer. It's only the best estimate from the given assumptions. In this case, Chad and Joe have different assumptions and thus different mathematical problems. It's only natural that they should get different answers. $\endgroup$ – Karolis Juodelė Feb 2 '13 at 11:54
  • $\begingroup$ @Karolis I'm a bit baffled by the statement, "mathematics should not be expected to give you a 'real' answer." Mathematics is used to produce answers in engineering applications (e.g. calcs for stresses or loads), physics experimentation (e.g. modeling particle collisions), and software packages (e.g. spreadsheet applications). Why are these applications of mathematics not real, or does that only apply to Probability Theory? Is Probability Theory the only branch of mathematics that deals with odds, probabilities and chance? Is Statistics part of Probability Theory? $\endgroup$ – Peter Feb 2 '13 at 12:15
  • $\begingroup$ @Peter, when constructing a mathematical model of a real problem, many factors cannot be taken into account (usually because there is no way to measure them). Thus a sufficiently precise approximation is the best you can expect. This is particularly evident in probability where the central assumption of randomness is hardly true. This doesn't make probability theory a less useful tool, as it can still give good enough approximations. $\endgroup$ – Karolis Juodelė Feb 2 '13 at 12:47

This seems to be partly a semantic question. The terms "odds," "chance," and "probability" in the context of games of chance are often used interchangeably. The following convey the same idea.

The odds of drawing an ace are one in thirteen. This usage is somewhat informal. It is more common to define define "odds" as the ratio of "possible successes" to "possible failures." In this case there is a distinction between "odds" and "probability." We would express this as saying the odds of drawing an ace are 4:48 or 1:12 (one to twelve).

The probability of drawing an ace is one in thirteen.

The chances of drawing an ace are one in thirteen.

The other part of the question concerns the state of knowledge of a player. The odds of an event calculated by a person with one state of awareness (A) are not necessarily the same as the odds calculated by someone with a different state of awareness (B).

It may be worth saying that someone who is omniscient might be said to know the true probability of an event. But one who knows is not using a probabilistic calculation.

Probabilistic calculations take into account available information. As with the stock market, people have different sets of data and so their probabilistic calculations differ. The calculations may all be correct, given the respective states of knowledge. However, some sets of data are better than others.

Addition in light of comment: Three players A, B, and C, contemplate the probability of drawing a 10 of clubs from the top of a deck that has been well-shuffled. They use the law of conditional probability (Bayes' theorem) as follows.

Player A knows nothing about the top card. Given no special insight beyond the makeup of a standard deck (his data) he calculates the probabilty of the event (club AND 10) as 1/52.

Player B has some insight. He recognizes a nick on the top card. There are only four nicked cards in the deck and one of them is a 10 of clubs. Using Bayes' rule, with B the probability of a nicked card and A the probability of a 10 of clubs, he calculates: P(10 of clubs|nicked card) as $$\frac{P(B|A)\cdot P(A)}{P(B)} = \frac{1\cdot(1/52)}{4/52} = \frac{1}{4}.$$

Player C happened to see the card as it was shuffled to the top. No need to use the laws of probability, which evolved in part to improve the quality of guesses in games of chance. He knows that the probability of drawing the 10 of clubs is 1. His knowledge of the relevant facts means that his estimate of the likelihood best reflects reality.

Each player makes the calculation that available data permit. For most conceivable events the probabilities for an omniscient player are either 1 or 0. The event will (or will not) happen. This does not affect the less privileged player's calculation in the least. There may be a dozen Bayesian calculations over the same hand of cards, each using a particular set of 'givens.' All the calculations may be correctly done but those assuming the most accurate information will better reflect the actual odds.

  • $\begingroup$ "For someone who has memorized the order of the cards, the probability of drawing an ace, given that the ace is known to be on top, is 1." -- The probability is also 1 for someone who has not memorized the order (repeat this experiment $n$ times, always putting the ace on top) $\endgroup$ – alancalvitti Feb 2 '13 at 15:50
  • $\begingroup$ See my comment above. Odds and probabilities are not the same thing. In general if the probability of an event is $p$, the odds are $p/(1-p)$. $\endgroup$ – antirealist Feb 2 '13 at 16:04
  • $\begingroup$ @antirealist The odds of drawing an ace are 1 in 13 = 4 in 52, as in 4 of the 52 possible cards are aces. Odds would typically be stated as 1 to 12 (= 1:12) though, as in 1 ace to 12 not-aces. $\endgroup$ – dave Feb 2 '13 at 16:37
  • $\begingroup$ @daniel, I wrote that, provided the ace is on top, the probability is also 1 for someone who has not memorized the deck. $\endgroup$ – alancalvitti Feb 2 '13 at 16:56
  • $\begingroup$ @daniel, let's say you're the player and you haven't memorized the deck, but the ace is on top. We repeat the experiment many times, so each time you find an ace on top - always. Even though you haven't memorized it. Both in frequency and in the Bayesian sense, you will converge on 1 as the probability of picking an ace, not 4/52. $\endgroup$ – alancalvitti Feb 2 '13 at 17:13

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