Sorry, you're not allowed to be embarrassed. This is Beginners. We're all learners here. (Besides, if you're an artist type, safe to assume your math skills will still be a lot better than my drawing skills.)

People learn stuff in different ways so it's a case of finding a website that teaches this stuff in a way you find helpful.

You could do worse than check out these links to start with:

Fractions

Decimals, Fractions and Percentages

Don't hesitate to come back here and ask for whatever else you need.

Good Luck.

Ok, I'll start there, and thanks a lot

Late to the party but just want to thank you for fantastic article. Really helped me a lot!

Great work, helps alot. Thanks!

Thank you!

wait so what happened to the other sections that were supposed to be posted???

There's a link to the index in the first line of the article.

DD, I've made a PDF of all the (main) posts in this series so that I can read it on my phone. Would you mind if I uploaded it somewhere and provided a link for everyone?

I wouldn't mind at all. That'd be cool.

Brilliant, thanks. Here is the link for everyone:

http://www.freefilehosting.net/mathenoobics

Solid post! Nothing new, but a great refresher, especially of basic probability terminology.

Should mention that in your OP, p(Heart AND Ace) is a great poker example of two independent events! No need to turn to sports, dice or coins!

Because the suit you choose has no limitation on which ranks you can choose afterwards for the same card, we can say that p(Heart) and p(Ace) are mutually independent of each other!

That is,
p(Heart AND Ace) = p(Heart) x p(Ace) = 1/4 x 1/13 = 1/52

Thus, p(Heart OR Ace) = p(Heart) + p(Ace) - p(Heart AND Ace) = p(Heart) + p(Ace) - p(Heart) x p(Ace)

Replacing Heart and Ace with general variables A, B:
p(A OR B) = p(A) + p(B) - p(A AND B) <- in general where A, B are any two events
p(A OR B) = p(A) + p(B) <- A, B are mutually exclusive: p(A AND B) = 0
p(A OR B) = p(A) + p(B) - p(A) x p(B) <- A, B are mutually independent: p(A AND B) = p(A) x p(B)

Note that two non-trivial events, p(A), p(B) not= 0, cannot be mutually exclusive and mutually independent simultaneously.
Thus, we can say for non-trivial events, exclusiveness implies dependence and inexclusiveness implies independence.

That is to say, the concepts of mutual exclusiveness and independence are, in fact, not independent.

Now, on to post #2!

^^^ Well, I quite like using non-poker examples (coins, sports) because there's a good chance (one might say a high probability ... ) that people are familiar with those already. Building on what they already know, or are familiar with. Also useful to let people know that this isn't just poker mumbo-jumbo, that these concepts are at work all around us.

I wouldn't disagree with anything else you said, although I suspect you may be making things more complex than they need to be at this stage.

Thanks for the feedback.

Hey,

I understand the formula for the first two questions when just adding and subtracting, but can't get my head around the last 2.

Can someone explain why you use the times method? I know it's to do with two circumstances that don't affect each other but can't grasp that in poker terms.

Secondly, on the third example you have to calculate 52 x 52. I can't do that in my head haha.

Thanks

Here is a little trick that might help for squaring numbers:

52*52 = (50 + 2) * (50 + 2) = 50*50 + 2*(50*2) + 2 *2 = 2500 + 200 + 4= 2704

Another example:

18 * 18 = (10 + 8) * (10 +8) = 10*10 + 2*(10*8) + 8*8 = 100 + 160 + 64 = 324

Think about what happens when we perform a simple 'experiment'.

We're going to spin a coin, followed by rolling a die.

Let's do the experiment 120 times.

How often do we get Heads on the coin spin?
We know p(Heads) = 0.5
and we're spinning the coin 120 times
so we'd expect to see Heads, er, 0.5 x 120 = 60 times

Now, each time we get Heads, we follow up with a roll of the die.
How often do we roll a 5 after getting Heads?
We got Heads 60 times
and we know p(rolling a 5) = 1/6
so we'd expect to get a 5, after getting Heads on the coin, 10 times.

So in our 120 runs of this experiment we get Heads AND a 5, 10 times

so p(Heads AND 5) = 10/120
=1/12
=1/2 x 1/6
= p(heads) x p(roll a 5)

cliffs: half the time we get Heads on the coin, and then one sixth of that half the time, we follow up with getting a 5.

Hope that helps.
Shout if it doesn't.

I can grasp that more because in my head rolling a coin and rolling a die are completely different.

I can't in Poker because in my head all the cards are linked somehow.

Haha sorry but Maths doesn't come natural to me.

My problem is knowing when to use the plus formula and when to use the times formula.

Can you explain using poker examples so I can understand which for e.g. The first formula (The addition/takeaway one) wouldn't work when you need to use the other formula.

Thanks Joe

P.S. These are great.

Yeah, I wouldn't sweat it too much.

All I tried to do here was introduce some basic probability concepts because probability is so key to understanding equities (see the thread about equity in this series).

Right now, if you can do the four questions (a) - (d) in my OP, you got what I was hoping you'd get out of this thread.

You're never going to be sitting at a table thinking ah, right, my decision now is based on the addition rule, not the multiplication one (or words to that effect).

What you ARE going to be thinking about is stuff like a shove here with my draw is profitable because villain doesn't continue often enough blah blah blah, and all of that is heavily dependent on the probability of him folding and your equity when called. Hence, probability as the first topic in the series. It's super important later on.

(And even when you're facing those kind of decisions, you're not working them out at the table, but you'll have done a ton of them away from the table and developed a feel for correct play. Not something I claim any particular expertise at, but like a lot of us, it's something I work on.)

I never really gave my feedback on this series of articles even though I've probably read them three times, but I will do so with some very strong words:

One of the ways you can tell the difference between someone who's serious about getting better at this game and someone who just SAYS they're serious about getting better at this game is by whether or not they've read these articles.

Hi DiamondDog, hi everyone

I'm going to be having couple of questions if that's ok

#1 what does those two examples illustrate/represent?

Anything. Absolutely anything. A coin flipping heads, me being hit by a bus, Sklansky solving the Riemann hypothesis, some moron noob posting a question here about GTO, doesn't matter. Something happens or it doesn't.

Don’t know the context but one of the important consequences of the second equation is that if an event happens or does not, for the event occurrence probability it is often easier to calculate the probability of the event not happening and then subtract that from 1.0.

Example The probability you will be dealt one or two aces in hold ‘em is

Direct.............P = 2*(4/52)*(48/51) + (4/52)*(3/51) = 14.9%

Or

Complement. P = 1-(48/52)*(47/51) = 14.9%

Sorry, this confuses me so I asked the wrong question
I understand that "A" is like an event and it can either happen or not happen.

#2 But why probability of A + probability of no A = 1? What does that "1" mean? Does it mean that after adding it up, A will always happen? (0=never, 1=always)

ps:
"p" stands for probability, right?

1=100%

Ok cool I get it, thanks, so it's like:

(if 25% is A and 75% is nonA)

25% + 75% = 100%
1/4 + 3/4 = 4/4 = 1



Following same logic, how would this example look like, expressed as odds? Can it be expressed as odds and summed up at all?
1:3 + 3:1 = 4:4 = 1:1? (that can't be right)

To convert fractions into odds, I've found this formula:
(denominator - numerator) : numerator, so

1/4 into odds is:
(4-1):1 = 3:1

3/4 into odds is:
(4-3):3 = 1:3

But how to add them up together?