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Originally Posted by feudallord
I am trying to go step by step because i Really want to absorb this. So skalnsky starts with an example of the flip of a coin. He basically says that you and a friend will wager 1 dollar for every flip. Since the odds of heads and tails is 50 50 or "1:1" which I'm not sure why, because 50/50 = 1?

er, not exactly. "50 50" is a colloquialism meaning "both possibilities have a 50% chance of occurring". it's not actually a fraction, it's a ratio, even though it's sometimes erroneously written as a fraction.
let us begin at the beginning.
fractions are expressed in the form "numerator/denominator". for our purposes, the denominator is going to be equal to the total number of things (cards, chips or dollars) and the numerator is going to be equal to the number of relevant things (again; cards, chips or dollars). so if we wanted to express the chances of pulling a spade out of the deck, the denominator is going to be 52 because there are 52 total cards, and the numerator is going to be 13, because there are 13 spades. 13/52.
fractions can and often should be simplified. if the numerator and denominator have a common factor (a number that can be divided into both the numerator and denominator to give a whole number in both cases) this is pretty straightforward. in this case, 13 has only 2 factors; 1 and 13. 52 has far more factors, 1, 2, 4, 13, 26, 52. the highest common factor is 13, ergo 13/52 becomes (13 divided by 13)/(52 divided by 13) = 1/4 (verbalised as "one over four")
if there is no common factor, say you are working with the fraction 9/46, it may still be convenient for you to work with a fraction in the form 1/x, in which case just divide both sides of the fraction by the numerator. (9 divided by 9)/(46 divided by 9) = 1/5.111recurring.
on to ratios. whereas fractions are expressed as (subset)/(total), ratios are expressed as (subset) : (subset). usually this will be of the order (thingswedontwant) : (thingswedowant) or (moneyicanwin) : (moneyihavetoinvest).
so if we were trying to pull a spade out of a new deck, there would be 13 cards which are spades and 39 cards which are nonspades, and our ratio would be 39:13 which simplifies to 3:1 ("three to one")
now consider the relationship between fractions and ratios. the former is the subset of things over all things, the other is the ratio of two complete subsets of things which adds up to all things. which is to say, the two values either side of the ":" in a ratio added together are equal to the denominator in a fraction. we had a ratio of 39:13 and a fraction of 13/52. 39+13 = 52.
going the other way, if you have a fraction, you can construct your ratio by subtracting the numerator from the denominator to give the "things we dont want" side of the ratio. we have 13/52, so 52  13 = 39, yielding a ratio of 39:13. note that it doesn't matter if we work with the original fraction or its simplified version. 13/52 = 39:13 just as 1/4 = 3:1.
you might also want to work with a decimal or percentage sometimes. the easiest way to calculate these is start with your value in the form of a fraction, and divide the numerator by the denominator. so 13 divided by 52 (which can be written as 13/52 of course. this is what a fraction is; a division) equals 0.25, which is a decimal. to get from a decimal to a percentage, multiply by 100. 0.25 x 100 = 25%.
when are we going to use all of these? well, for the most part you are going to be working with fractions and decimals while doing the math, and ratios and percentages when making the decisions based on the math.
say someone offers you a prop bet, you have to pull two cards out of a deck, and if they're both spades, you win. if only one is a spade, or neither is a spade, he wins. he offers to lay you odds of 10:1.
the easiest way to calculate the chances of you winning is to work with the fractions. as we said earlier, the chances of pulling the first spade are 13/52. for the second spade there are only 12 spades (numerator) in a deck of 51 cards (denominator).
when working out the probability of multiple events all occurring, you first work out the probability of each event of itself, and then multiply those together to give the probability of them all happening. note that these are
dependent events which is sufficiently important to be bolded. in dependent events, the probability of successive events is dependent on the outcome of previous events. in this instance, when we draw the second card, the composition of the deck has changed because of the first card we drew. to wit, we now have a 51 card deck with 12 spades in it. in independent events, previous events do not affect subsequent events, i.e. if we drew a card, returned it and then drew another card, the composition of the deck has not changed. you are probably never going to be working with independent events in poker, of course, because cards don't go back in the deck, but if we're learning this shit we might as well learn it properly. so we have the probability of event one, and of event two, and now we put them together thusly:
13/52 x 12/51 = ?
to multiply fractions, you multiply the numerators together to give a new numerator (there is probably a name for this but i forget what it is), and then multiply the denominators together to give a new denominator:
13/52 x 12/51 = (13x12)/(52x51) = 156/2652
we now have a single fraction that describes the chances of pulling two spades out of the deck. we could construct a ratio right here (2652156):156 = 2496:156 but that's not going to make our decision any easier, so first we simplify that fraction down to the form 1/x, by dividing both sides by the numerator.
(156/156) / (2652/156) = 1/17
and now we turn this into a ratio, (171):1 = 16:1
the chances of us successfully pulling two spades out of the deck are 16:1.
pausing to note, that we could have made our lives easier by simplifying our initial fractions. 13/52 we already know simplifies to 1/4. 12/51 simplifies to 4/17 (dividing both sides by the highest common factor, which is 3). this gives us the equation:
1/4 x 4/17
= (1 x 4) / (4 x 17)
here we again have a common factors situation. since one of the numbers we are multiplying together in the numerator is a 4, and one of the numbers we are multiplying together in the denominator is a 4, then ldo whatever those values turn out to be, they're both going to have 4 as a factor...so we might as well just cancel those 4s out right now, leaving us with:
1/17
tada.
back to the problem at hand. we know the chances of doing this are 16:1. the guy is laying us odds of 10:1. is this a good bet?
well, when we have everything expressed neatly in ratios like this, you can just eyeball it and say no. if the odds he is laying are longer (bigger ratio) than the odds of it actally happening, then the times we win, we win more than we need to cover the times we lose. conversely, if the odds he lays are shorter (smaller ratio) than the deck is giving us, then he isn't going to be paying us enough to cover the times we lose.
lets have a math proof of that. the actual chances of it happening are 16:1 or 1/17, which means the chances of it not happening are 16/17. we're being paid at 10:1, which means every time we win we win 10, and every time we lose, we lose 1.
so for every 17 times we play, we lose $1 16 times and we win $10 once. that means in total we win $10(16x$1) = $1016 = $6. we played the game 17 times, so $6 divided by 17 = $0.35 per game, which is our expectation, and it sux0rs.
i will show you how and when to use decimals/percentages in a later post. lets make sure we have this down first.
first, if there are questions from anything i've written here, ask them now.
then attempt the following:
1. (easy) rolling a six sided die, you are invited to nominate a number for $1, and should that number be rolled you will be paid $6 and have your $1 stake returned. if any other number is rolled, you lose your $1 stake.
a. is this a good bet?
b. what is your expectation?
2. (hard) rolling the same die _twice_. if the second number you roll is equal to or higher than the first number you roll, you win $1. if it is lower than the first number, you lose $1.
a. is this a good bet?
b. what is your expectation?
3. (very hard). player A rolls three dice, player B rolls two dice. both A and B's dice are then ordered from highest to lowest (yes, we are playing risk), and the lowest of A's three is discarded to give two dice each. you will then consider your highest vs his highest, and lowest vs lowest. whoever has the higher number wins that pair, but if you both have the same number, B (who began with 2 dice) wins. if both players win one roll, the game is a draw and no money changes hands. if either player wins both rolls, they win $1 from the other.
a. how often will A win both rolls?
b. how often will B win both rolls?
c. how often will it be a draw, one roll each?
d. would you rather be player A or player B? and what is your expectation?
i would be obliged if others would also attempt these so I can see exactly how difficult 2. and 3. are. PM me your answers.
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He says that over the course of an hour on average you would win nothing and your expectation is zero.
Then he changes the scenario to you betting 1 dollar still but the other guy bets 2. So does that mean that, okay so is he basically saying that I am wagering lets just say on heads and the friend tails? That every time its heads I win and every time its tails he wins? And if thats the case then in the 2 dollar scenario every time its tails he will only win 1 dollar, but when its heads I win 2 dollars? Let's start there for now

yes, exactly.
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Wow...I completely overlooked the concept of pot oddds, and as I slowly get a small grasp of it I realize how fucking important it is holy shit no wonder ive lost so much money, this game is way more complicated thatn I thought to win at it

yup.
Kc