Probability (and intro to EV)
Poker is all about probability and gambles, so we need to review some basics. Probability is actually a very tricky subject and it can be non-intuitive. I recommend thinking about probability in terms of a branching tree. You start at the root of the tree. There is a 100% chance of starting at the root. Now, at each branch, there is an A probability of going one way and a B chance of going another. Take those chances and multiply them by the probability of being at that branch. So, if it was P coming in, it's P*A going down one branch and P*B going down the other. When you get to the leaves of the tree, all the probabilities of all the leaves should sum to 1 (100%), because probability is conserved. You can think of probability as like a fluid that's flowing from the root of the tree. At each branch, some goes left, some goes right, but the total stays the same. The chance of winding up at any leaf is just the chance of a speck of dust getting there in this fluid flow.
Note that we use % to talk about probabilities a lot. When you see a % that means "divide by 100", so 50% is the same as 0.5 ; in mathematics we use probabilities that sum to 1. That's the same as 100%, but you need to use the normalized (sum to 1) numbers when doing the math. In this book we'll convert back and forth between percents and normalized probabilities, because the %'s are easier to think about. Just remember they are the same thing and they are interchangeable,
Let's look at some simple examples. What is the chance of flipping a coin heads 3 times in a row? Each branch (heads or tails) has a 50% chance of going each way. In order to flip 3 heads, you need to go on a 50% branch three times. You multiply each time, so it's 0.5*0.5*0.5 = 1/8. The chance of any exact sequence of 3 flips is the same 1/8 ; eg. the chance of flipping tails then tails then heads. What's the chance of flipping 2 heads and 1 tails, in any order? Well, you can flip either HTH or HHT or THH , so there are 3 ways; each way has a 1/8 chance, so it's 3/8. What's the chance of flipping any mixed set (eg. not 3 heads and not 3 tails). Well, once way to compute it is to take the chance of flippin the pure cases - that's just 2/8 , since there are two pure cases. Now the chance of not flipping a pure set is just one minus that = 1 - 1/4 = 3/4. The full set of leaves is 3 heads : 1/8 , 2 heads : 3/8, 1 heads : 3/8, 0 heads : 1/8. Those probabilities add up to 1, as required.
Now we can look at some examples with real poker cards. Say you have O outs and there are 47 unknown cards in the deck. The chance of making one of those outs on the turn is O/47. What's the chance of making it on the turn *or* the river? We have to do branches here. First look at the two cases on the turn. On branch A, you make your card, with an O/47 chance. On branch B, you don't make it, with a (47-O)/47 chance. When you do make your card, it doesn't matter what happens on the river, so you're done. When you don't make your card on the turn, you can still draw on the river. Now there are 46 unknown cards, so your chance of making it on the river is O/46. This multiplies by the chance of being in branch B, so you have (47-O)/47 * O/46. The total chance of making your card on the turn or the river is
O/47 + (47-O)/47 * O/46
Another way of getting this same result is to look at the chance of *not* making your card on the turn or the river. In order to not make your card, you must miss on the turn *and* the river, so that's (1 - O/47) * (1 - O/46). Thus, your chance of making your card is :
1 - (1 - O/47)*(1 - O/46) = 1 + O/46 + O/47 - O/46*O/47
The whole tree looks like this : (@@ draw the picture)
This is a pretty simple example, but it helps you see the idea of the tree and the various ways you can sum and subtract probabilities.
A general rule of thumb is "or" means add probabilities while "and" means multiply probabilities. When there are several ways to make the result in question, you sum the chance of each way. When several things in a row must happen, you multiply the chance of each. Another good rule of thumb is that it's often easier to compute the chance of it *not* happening, and then invert that.
A classic example is the birthday paradox. If you have 20 people in a room, what's the chance that any of them has exactly the same birthday as someone else? The easy way to compute this is by looking at the chance that noone has the same birthday. The first person can have any birthday. The next person can't have the same, so they have 364/365 chance of not having the same birthday. The next person must not have either of those two, so they have a 363/365 chance of not matching birthdays. For all of them to not match, you multiply all these terms :
1 * (364/365) * (363/365) * (362/365) ... (345/365)
= 0.58856
Now the chance of someone having the same birthday is one minus that, = 0.41144, or about 41% . This is called the "birthday paradox" because it's surprising how large that number is. Now you know it's just simple probability.
A common concept we deal with is a bunch of "weights". Weights are just un-normalized probabilities. If you have a set of weights {Wn} you can turn those into probabilities by dividing by the sum :
Pn = Wn / Sum[i] { Wi }
All probabilities must always sum to one. That's the same as saying the chance that *something* (no matter what it is) will happen is always 100%. So :
Sum[n] { Pn } = Sum[n] { Wn / Sum[i] { Wi }} = Sum[n] { Wn } / Sum[i] { Wi } = 1
This is always one, so we made good probabilities. Another condition for them to be proper probabilities is that none of the W's should be negative. For example, if you have a set of weights {1,1,2} that says the first two elements are equally likely and the third is twice as likely as either of them. That is the same as the probabilities {0.25,0.25,0.5} or {25%,25%,50%}.
A lot of the time when we're doing probabilities, we can just use the weights and then divide by the sum. When we discuss a random variable, we can describe its chance of occuring either with a probability, which must be normalized (sum to one), or with a weight, in which case we'll have to divide to normalize. For example, if you were weighting by the probabilities P0 and P1 with P0+P1 = 1 :
x = P0 * a + P1 * b
You can just use the weights instead, if you divide by the sum :
x = (W0 * a + W1 * b) / ( W0 + W1 )
This is the idea of a "Weighted Sum". That is, if you have a set of weights {Wn} and a set of values {Vn} the weighted sum is :
V = Sum[n] { Wn * Vn } / Sum[i] { Wi }
V = Sum[n] { Pn * Vn }
This is a common thing that we'll use. It lets us know the "expectation value" of some random variable. Let's say each value Vn occurs randomly with a weight Wn, then this weighted sum gives us the expectation value of V - it tells us what value we will see on average if we repeat the random draw many times.
For example, if you're flipping a coin, what's the expected number of heads? Well, there are two cases, you flip heads, then the value is1, if you flip tails the value is 0. Both cases have a weight of 1, which is just a probability of 50% for each, so we can construct the expectation value :
V = Sum[n] { Pn * Vn } = 0.5 * 1 + 0.5 * 0 = 0.5
the average value of heads is 0.5 for each flip. How about when you draw a card from a deck, what's the average rank? The ranks are 2-14 (an Ace is a 14 in numeric ranking). The chance of each rank is 1/13, so we get :
V = Sum[n] { Pn * Vn } = 1/13 * 2 + 1/13 * 3 ... 1/13 * 14 = 1/13 * (2 + 3 + ... 14) = 8
So the average rank of card you draw is an 8. What about when things aren't equal? Let's say you have a bag of 5 red marbles and 7 blue ones. You draw one and then put it back. Your buddy pays you two dollars each time you draw a red and you pay him 50 cents when you draw blue. What's the average amount of money you make on each draw? The weights are {5,7} and the values are {2, -0.5}, so the expectation is :
V = (5 * 2 + 7 * (-0.5)) / (5 + 7) = 0.54167
Obviously a very profitable enterprise. This is the sort of expectation value math you have to think about all the time in poker.