Probability Theory Poker
- Probability Theory Practice
- Probability Theory Pierre De Fermat
- Probability Theory Poker Rules
- Probability Theory Predicts That There Is
- Probability Theory Poker Strategy
Intro
So welcome to 15.S50, Poker Theory and Analytics. So this is going to be Monday, Wednesday, Friday from 3:30 to 5:00. I just got a room for a review session on Tuesday, Thursday for anyone who needs to catch up a little bit. A Tutorial on Probability Theory 4. Conditional Probability The probabilities considered so far are unconditional probabilities. In some situations, however, we may be interested in the probability of an event given the occurrence of some other event. For instance, the probability of R: “Tomorrow, January 16th, it will rain in Amherst. Indeed, when the economist Ingo Fiedler analyzed hundreds of thousands of hands played on several online poker sites over a six-month period, he found that the actual best hand won, on average.
This is a problem concerning basic probability calculations from the text: “A First Course in Probability Theory” by Sheldon Ross (8th addition).
Sample Problem
Chapter 2 #16. Poker dice is played by simultaneously rolling 5 dice. Show that:
a) P[no two alike] = .0926, b) P[one pair] = .4630, c) P[two pair] = .2315, d) P[three alike] = .1543, e) P[full house] = .0386, f) P[four alike] = .0193, g) P[5 alike]=.0008.
Solution
Notation, I will let ^ designate the power function. For example, 6^5 is 6 to the fifth power. 6! is 6 factorial, 6! = 6 * 5 * 4 * 3 * 2 * 1.
To calculate the Probabilities here, we will divide the number of occurrences for a particular event by the total possibilities in rolling 5 dice.
N[total] = total possibilities in rolling five dice = 6^5 = 7776
Note: N[total] is the number of ordered rolls. For example, if we rolled the dice one by one and rolled in order 3,4,5,6,2, it would be considered as different from if we rolled 2,3,4,5,6 in order.
a) P[no two alike]
There are 6 choices of numbers for the first dice. The 2nd dice must be one of the remaining 5 unchosen numbers, and the 3rd dice one of the 4 remaining unchosen numbers, and so on… This gives an ordered count, so:
N[no two alike] = 6 * 5 * 4 * 3 * 2 = 720
P[no two alike] = N[no two alike]/N[total] = 720/7776 = 0.09259259
b) P[one pair]
First we count the possible sets (unordered) of numbers. Here we have one number that is the pair, 6 choices. Then we must choose 3 different numbers from the remaining 5 values, as these three are equivalent, we have choose(5,3) = 5!/(3! * 2!) possibilities. So 6 * choose(5,3) is the total combinations of numbers. Since we are dealing with ordered counts, we must consider the orderings for each set of numbers. We have 5 die with 2 the same so the number of orderings is 5!/2!. Multiplying these together gives us the number of ordered samples:
N[one pair] = 6 * (5!/(3! * 2!)) * (5!/2) = 3600
P[one pair] = N[one pair]/N[total] = 0.462963
c) P[two pair]
Here we have 2 pairs which are equivalent, so we must choose 2 values from the 6 possible values, choose(6,2)= 6!/(2! * 4!). Then we must choose 1 value from the remaining 4 for the single value, 4 ways. Now we must consider the orderings, 5 die with 2 sets of 2 the same so the number of orderings is 5!/(2! * 2!).
N[two pairs] = (6!/(2! * 4!)) * 4 * (5!/(2! * 2!)) = 1800
P[two pairs] = N[two pairs]/N[total] = 0.2314815
d) P[three alike] (the remaining two cards are different)
There are 6 choices for the three of a kind. Then we must choose 2 different values from the remaining 5 choices, choose(5,2) = 5!/(2! * 3!). The number of orderings is 5 items with 3 being identical, which is 5!/3!.
N[three alike] = 6 * (5!/(3! * 2!)) * (5!/3!) = 1200
P[three alike] = N[three alike]/N[total] = 0.154321
e) P[full house] 3 alike with a pair.
We need one value for the three that are alike, 6 ways, and then we must choose from the remaining 5 values for the pair, 5 ways. The orderings are given by 5!/(3! * 2!).
N[full house] = 6 * 5 * (5!/(3! * 2!)) = 300
P[full house] = N[full house]/N[total] = 0.03858025
f) P[four alike]
We need one value for the four of a kind, and then one value from the remaining 5 for the last die. The number of orderings is given by 5!/4!.
N[four alike] = 6 * 5 * (5!/4!) = 150
P[four alike] = N[four alike]/N[total] = 0.01929012
g) P[five alike]
Here we need 1 value for the five alike, 6 ways. There is just 1 possible ordering as all five die are the same.
N[five alike] = 6
P[five alike] = N[five alike]/N[total] = 0.0007716049
Check: the numbers of each type must sum to N[total] = 7776:
720 + 3600 + 1800 + 1200 + 300 + 150 + 6 = 7776
Extra:
Straight
We can have two possible straights: one composed of (6,5,4,3,2) and one composed of (5,4,3,2,1). Each of these straights can be permuted 5! ways.
N[straight] = 2 * 5! = 240
P[straight] = N[straight]/N[total] = 0.0308642
Now let’s run a simulation in R:
Results of simulation:
notwoalike: 0.09194
onepair: 0.464
twopair: 0.23187
threealike: 0.1554
fullhouse: 0.03727
fouralike: 0.01889
fivealike: 0.00063
straight: 0.03072
The market for uncertified physicians was not very big in 16th-century Italy. So when a young Girolamo Cardano found himself without a College membership or cash, he did what any reasonable person would do: he became a professional gambler.
Over his career, Cardano meticulously documented observations about the games he played. Specifically, he identified patterns that corresponded to wins and losses. Cardano’s observations – outlined in his work “Book on Games of Chance” – form the basis for what we now call mathematical probability.
Knowledge of probability allowed Cardano to earn a steady income from gambling, and professional gamblers today rely on probability to make winning bets. I doubt there’s one WSOP winner alive who lacks at least a basic understanding of probability.
Probability is the tool that allows us to win money at poker. So let’s take a look at how poker probability works, and learn how to apply it to our game. We also welcome you to check out our remaining 3 parts to this 4-part series on probability, which can be found below:
What Is Probability?
As a field of mathematics, probability is the study of uncertain events. What’s the chance that it will rain tomorrow? How likely am I to get an A on this test? If you’re dealt AA preflop, how often will you win the hand? The field of probability helps us to answer these questions.
When we obtain an answer to a question of chance, it’s called a probability. For example the probability of seeing rain tomorrow might be 50%. Half the time it will rain, and half the time it won’t.
In poker, probabilities help us to estimate the likelihood of certain events happening. For example the probability of getting aces in the hole is 1/221, or 0.45%. Knowing this particular probability helps us to guide our hand selection at the tables. More importantly, it helps us to guess our opponent’s hand distributions with stunning accuracy.
Representing Poker Probability
There are three main ways to express a probability:
• As a decimal (e.g. 0.03)
• As a percentage (e.g. 3%)
• As a fraction (e.g. 3/100)
All of the above expressions mean the exact same thing, they’re just written in different forms. How you choose to express probabilities boils down to personal preference.
On forum discussions and in poker strategy articles, you’ll most often see poker probabilities in percentage or fraction form. For example, the chance of As-2d winning against a random hand can be expressed as:
• 54.93% in percentage form
• 54.93/100 in fraction form
In mathematical calculations, we’ll most often use decimal form. It’s just easier to write that way, since calculators most often return decimal values. For example the expected value of going all-in against one opponent for $1 with As-2d is:
= $win* Pwin – $cost
= $2(0.5393) – $1
= $0.0786
Probabilities in Plain English
All these numbers can seem confusing at first. But if you start thinking about probabilities in plain English, you’ll eventually develop an intuition for what they are. That’s especially helpful with poker probabilities, where you often have to calculate them on the fly.
For example, say we have an event A with the probability P(A) = 1/52. This just means that for every 52 events, we can expect event A to happen exactly once. Said differently, event A will happen once per 52 cycles.
Numerator: # of times the event occurs
Denominator: # of iterations
Numerator / Denominator = Probability
Every probability has this same basic format, which means you can convert them easily to English descriptions. Try converting these example probabilities into plain-English descriptions yourself:
1. P(Dealt Aces) = 1 / 221
2. P(Aces Win Against Random Opponent) = 85.20 / 100
3. P(Kings Make Set on Flop) = 1 / 8.5
Check your answers at the bottom of the article.
Probability Theory Practice
How Does Probability Help With Poker Strategy?
Probability is useful in almost every aspect of poker strategy. To name just a few examples, probability can:
Probability Theory Pierre De Fermat
1. Tell us the likelihood of an opponent holding certain hands, given our reads and her actions.
2. Tell us how often we’ll be dealt strong vs. weak hands.
3. Tell us how often our strong hands or draws will win by showdown.
In the first case, we can transform our reads into mathematical evaluations of poker situations. This general process is called building a hand range, which is a probability distribution of an opponent’s possible holdings.
Once we calculate an opponent’s range, we can analyze the probability of our own holding’s winning against said range. And with that information, we can calculate the average earnings we’ll expect to make by betting, checking, calling, or folding. This allows us to play profitably 100% of the time, as long as we’re mentally sharp. Obviously this is a huge advantage when we’re trying to win at poker.
In the second case, we can figure out how long we’ll have to wait for good hands preflop. This information tells us how often we should expect to play with a card advantage against opponents. From this we can deduce how often we should be raising, folding, and calling preflop. Even better, we can deduce how often an opponent should be raising, calling, and folding; and then we can exploit our opponent’s strategy, depending on how weak or strong is action range actually is.
In the third case, we can calculate the profitability of playing an unmade hand. For example we might have one card to a flush on the turn facing an opponent’s bet. Probability can tell us how likely we are to make our flush. Using this probability, we can calculate our expected value: the exact dollar amount we’d expect to win by making the call. Once we’ve got this information, we’re absolutely certain whether or not a call is profitable in the long run!
Answers to Exercises
1. We’ll get aces once per 221 deals.
2. Our aces will win 82.20 times per 100 hands.
3. Our kings will flop a set once every 8.5 hands.
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