Applied probability theory

I don`t stop repeating that if you are to some extent deeply planning to play, if you hope to begin win in casino you have to know the base of probability theory at least. So I`ll try to “apply” this probability theory to gambling questions. Prob theory and games are closely related indeed. For example the French word “hazard” – “case”, “risk” came probably from Arab “azar” – “dice”. Actually the science arose from gambling questions exactly and dice game exactly!

It is considered that order holder d`Mere wrote a letter to Bles Paskal asking him how many times have to drop dices to have a probability of two sixes occurring more than ½ (ho thought that 24 drops is enough). Paskal answered this question, began to raise analogous questions in his correspondence with Ferma and these two scientists` names are usually calls as founders of probability theory (it`s about the middle of 17 century). Third “father-founder” is Guigens who also answered the players` question if during three dices dropping 11 or 12 points will occur oftener. Of course certain questions of probability theory were also raise earlier but the base of harmonious theory was laid by these famous scientists.

So, let`s give the definition of probability theory, that is the mathematics section learning random magnitudes` regularity: random events, random magnitudes, their properties and operations with them. Lets` continue… Random event is event which can happen or not in present conditions and which have a defined probability p (from 0 to 1) of its occurring. Random magnitude is magnitude which assumes one of many values as a result of experience and appearance of one or other value of this magnitude is random event at that.

Since history of science development began with dices we`ll appeal to this game, it is too opportune for these. We have dice (cube) with six sides and numbers from 1 to 6 on each. Random event will be for example each number occurring for example 1. If there is normal not “charged” cube each of sides will occur with the same frequency that is one time of six. That 1/6 is probability of our random event occurring (occurring of 1). Probability designates with letter p or P usually. And the random magnitude with our experience (dice dropping) will be occurring of the one or other number exactly. Set of probable magnitudes are 1, 2, 3, 4, 5 and 6 with it. As it can`t be other numbers occurring (there are no 7 or 8 on a cube), these numbers from 1 to six will be total event system. Event probabilities sum in the total system is 1 (that`s why if events are equal probable it is easy to calculate each event probability as 1/event quantity). As it is not possible to occur 1 and 6 at the same time when dropping cube (and other two numbers, we don`t consider “standing on a edge” and “hover in the air” cases) these events will be incompatible between each other.

I offer you the first law of prob theory, addition rule: probability of occurring in experience any one (it doesn`t matter which exactly) of results is sum of these results probabilities if each two of them are incompatibly between each other (we can write it like Р(А or В)=Р(А)+Р(В), where Р(А) is probability of event A). For example if you are interested in probability of even number occurring (that means 2, 4 or 6) we have to sum up probabilities of each of given numbers occurring, we`ll get 1/6+1/6+1/6=1/2 or 50%. Odd number occurring (1, 3 or 5) will be opposite even number occurring event. Sum of opposite events probabilities is 1, that is probability of odd number occurring is 1-1/2=1/2. Event probability is often opportune to calculate with probability of opposite event occurring. So a probability of less than 6 number occurring can be define as probability of 6 UN occurring and calculate like 1-1/6=5/6. Of course in the case of dices it is not hard to sum up 1/6 for five times but in the most of real sums it will be vastly harder.

Let`s pass on conditional probability notion. Suppose that we have two cubes: one standard with numbers from 1 to 6 and the second where 1 and 2 numbers repeated twice and 3. If we`ll drop the first cube, 1 will occur with 1/6 probability, and if we`ll drop the second we`ll have the 1/3 probability already. And 6 will occur with 1/6 and 0 probability (event with 0 probability calls impossible, and 1 – reliable) respectively. Probability of dropping some number on the first dice, for example 1, calls unconditional as it doesn`t depend on other events and probability to drop 1 on both cubes at the same time will be conditional already (some conditions have to happen at the same time – 1 on the first and the second cube). How can we detect this probability? Probabilities multiplication rule will help us: Product of two events probability is product of probability of one of them and conditional probability of other one, calculated with the condition that first event have happened already (Р(А and В)=Р(А)*РА(В), where РА(В) is B event probability when A performs). That is in our example with cubes probability of event A – 1 occurring on the first cube – is 1/6, probability of 1 occurring on the second cube, with the condition of 1 occurring on the first cube too, is 1/3. In all 1 occurring on both cubes is 1/6*1/3=1/18.

Indeed probability of 1 occurring on a second cube is 1/3 independently of what will occur on a first cube. Thereby occurring of 1 on each cube are independent events. In that case multiplication rule simplifies to Р(А and В)=Р(А)*Р(В) – probability of combined occurring any number of independent of each other events is product of probabilities of these events. Probability of two 6 or 1simultaneous occurring on both ordinary cubes is 1/6*1/6=1/36. And what is the probability of 4 or 3 occurring? Is it 1/36 too? No! For 6-6 occurring 6 have to occur on a first cube and 6 on a second. And for 3-4 it can be 3 on the first and 4 on the second or 4 on the first and 3 on the second. It happens that we have to find two conditional probabilities and then sum up them, we`ll get 1/18 probability. Armed with this knowledge you can become like Guigens and answer the question he was asked for. Find all combination which give 11 and 12 points (for example 11=1+5+5, 11=2+4+5 and so on), take into account possibility (or impossibility) of combinations occurring in different order (that is 1,5,5 or 5,1,5 or 5,5,1 – three possible combination in all). Then you multiply and sum up everything. You have to get 25 combinations (each has 1/6*1/6*1/6=1/216 probability) for 12 and 27 for 11. I`ll repeat once again – cubes droppings or numbers occurring in roulette are independent events, probabilities of each are the same and occurring of cards from one batch are dependent already. For example probability of pulling ace first from a 52-cards batch is 4/52. And probability of second ace is 3/51 already (as there are only three aces and 51 cards remained in batch). As a result probability of pulling two aces one by one is 4/52*3/51. And if the first card was not ace the probability of pulling ace with a second card was 4/51 and total situation probability was 48/52*4/51.

Indeed this given higher theory is enough for calculation of most situations in casino games. In the same roulette or dices it is more than enough for calculating of different probabilities. Let`s take “the favorite” roulette-men`s example – 10 reds one by one and calculate it`s probability. There are 37 numbers in total, 18 red among them, so the probability of red occurring is 18/37. Probability of two reds one by one is 18/37*18/37 already, and 10 one by one is (18/37)^10=0,074% or 1 / 1347 (^ – involution). If calculate not only colour but colour plus zero which is dangerous for martingale-lovers we`ll get (19/37)^10=0,13% already or 1 / 784. That is during 1000 spins at the average (one day of playing) you expect occurring of series from 10 one by one not of “your colour”. And if there are 10 of any colour one by one (we`ve counted only the one colour) a probability doubles as series can be both red and black. Let`s look for probability of not occurring of dozen 15 times running. First dozen not occurring means that only second and third dozens occur, plus zero. Probability of their occurring is 1-12/37. Probability of their 15 times running occurring is (25/37)^15=0,28% or 1 for 358 spins. But again if we are interested in not occurring of any dozens but not exactly the first one, we multiply our total by three and approach to 1 case for 100 spins. One who wishes can calculate other probabilities too as number occurring for three times running or not occurring of number during 100 spins.

So, let`s back to order holder d`Mere`s sum. We know that probability of two 6 occurring is 1/36. Probability of two 6 not occurring (opposite event) is 1-1/36=35/56. Probability of two 6 never occurring during 24 droppings is (35/36)^24=50,8%. So, two 6 occurring probability once at least is opposite event with the probability 1-50,8%=49,2%. As we can see d`Mare`s supposition was wrong, you can be convinced after Paskal that it is 25 droppings necessary for 6-6 occurring probability exceed 50%. Turn your attention that sums of “at least once happened event” optimally solve with probability of opposite events.

One more very useful function is Bernoulli formula which makes it possible to estimate probability that during N tests with probable success p amount of successes will be P=p^n*(1-p)^(N-n)*N!/n!/(N-n)! If you use Excel there is function BINOMRASP which enables to calculate respective probabilities quickly. Generally given formula enables to calculate binomial distribution coefficient. This formula is most urgent for “chances” check of one or other events. For example 5 zeros (n successes) occurred during 100 spins (N tests), occurring probability is 1/37 (success probability p). We substitute to the formula or better use BINOMRASP in Excel (“Integral” parameter set as “lie”), we`ll get 8% probability of 5 zeros occurring during 100 spins, that is quite a lot. Pay your attention that is the probability that zero will occur 5 times running! But it can occur 6 times with 3% probability and 7 times with 1,3% probability etc.  If we want to know probability of 5 and more times occurring we have to sum up these probabilities or it is handier to calculate in Excel by 1-BINOMRASP (4;100;1/37;ИСТИНА)=13,5% formula (the event opposite to zero occurring 4 or less times).

When solving different sums of applied probability theory you can need main combinatorial analysis formulas:

• Permutations – with how many ways you can arrange in order available n elements: P=n! (I hope, you`ve got acquainted with factorial notion – that is product of all natural numbers from 1 to n. n!=1*2*3*…*(n-1)*n). When we talked about two or three dropping of cubes, we were interested exactly in permutations for two or three different numbers. Actually we multiplied probability of defined combination occurring by permutations quantity, for three cubes we have Р=3!=6 variants of permutations. If we have  some (m) same elements the formula changes to P=n!/m! (when two cubes of three are the same as 1-5-5, there will be 3!/2!=3 permutations, and if all three cubes are the same, we`ll have 3!/3!=1 permutations).
• Placements – with how many ways you can rearrange m elements from available array to n: A=n!/(n-m)! For example how many different numbers can we compose using only three numbers of ten? 10!/(10-3)!=8*9*10=720 (the digit is less than thousand because digits like 111 or 505 use one figure more than once, for placement with reiteration the formula will be n^m, that will give us 1000 of three-digit figures). Formula of placement often uses for possible poker combinations detection. For example what is a probability to collect royal flesh with five given cards? Only 5 cards of 52 in batch can be deal with 52!/47!=311.875.200 ways (we`ll use placement formula right). Royal flesh of defined suit can be collect with 5!=120 ways (the amount of 5 cards placements which form royal). As there are only 4 suits, the probability to collect royal flesh will be 4*120/311.875.200=1/649740 or about every 650-thousand dealing.
• Combinations – with how many ways m elements can be choose from n elements: C=n!/m!/(n-m)! For example with how many ways two preflop-cards can be deal in Texas holdem? 52!/2!/50!=51*52/2=1326 ways, when 9♣8♠ and 8♠9♣ consider as the one way, as order is not important for us. Further each pair can be compose with 4!/2!/2!=6 ways already (the quantity of combinations is in twos from four), each unpaired combination with 4^2=16 ways (4 ways to choose first card and 4 for the second, multiply analogous to probabilities), one-suit combination of them with 4 ways (by suit quantity) and each different-suit with 16-4=12 ways.

I hope, you`ll understand what is written here. For practice you can train, calculate probability of different combinations occurring in dices and sic bo and compare them with payments; calculate probability of getting different “series” in roulette; calculate probability of getting different bonus payments in different blackjack versions, and compare results for different batches quantity – 1, 2, 3, 4, 8 etc. at that. I wish you success in mathematics mastering and game!