# Tricked videoslot

(This article applies to Italy, but I'm pretty sure it's the same in all other countries)

Let's talk a little bit about those new slot machines, that have appeared in a lot of tobacconists for a few days ...
To be exact, those new slot machines, authorized by the State to provide cash winnings.
I immediately state, to avoid misunderstandings, that I'm NOT gambling-addicted, and that my first and always valid advice is NOT TO PLAY THESE GAMES. This "article" comes from my passion for computer science and statistics, reasons for which I know the game of Blackjack.

I immediately became "suspicious" for the choice of the game on which to base this new type of slot machine, and I wanted to deepen. And let's see what I discovered ...
The Games depicted are two, a real slot machine, and a game based on blackjack, and this is what made me "raise my own antennas" ...
This blackjack game works like this: two cards are given, from a (supposedly) 52-card deck, and placed one on the right and one on the left, then a card is dealt, one at a time, and you have to decide whether to place that card that came out on the right or left piles.
The aim of the game is to approach, or, to the limit, do, 21, without overcoming it, in neither of the two contemporary games (the right and the left).
The payments, against a play of 0.50 Euro, are as follows:

Sum of the two games = 35 -> € 0.50 payout
Sum of two games = 36 -> € 1 payout
Sum of two games = 37 -> € 1.50 payout
Sum of two games = 38 -> € 2.50 payout
Sum of two games = 39 -> € 5.00 payout
Sum of two games = 40 -> € 20.00 payout
If just one of the two games makes 21 points EXACT -> € 50 payout

This game immediately seemed very strange to me, and I wanted to go into this question ...

PREMISE:
Blackjack is played with a certain number of 52 card decks, generally 6 decks are used in casinos, for a total of 312 cards.
The cards from 2 to 10 have their face value, the figures are worth 10 points, the ace can be worth 1 or 11, according to convenience. The aim of the game is to make 21 exact points (or to get as close as possible) without exceeding this score (if you exceed then you lose: the term is "bust")
For now I simplify the reasoning, supposing to use a single deck, we will see later that the thing does not change much, given the result that I intend to demonstrate.

So, whoever plays that game expects the cards to be drawn REGULAR, ie EVERY CARD is drawn with probability 1/52 if it is the first, 1/51 if it is the second etc ...
Let's see what would happen if this were true: Suppose we concentrate ONLY on the FIRST SEQUENCE, that is, the two initial cards plus the first one drawn, that we can choose WHERE to place.
Well, WHAT PROBABILITY WE WILL HAVE TO DO plain 21 JUST WITH THE FIRST SEQUENCE OF THREE CARDS?
The total possible combinations, as known, are 52 x 51 x 50 = 132,600 (I have assumed that it is only one deck of 52 cards, so there are NO repetitions)
The winning combinations are obviously those made like this:

First two: at least a 10-point card -> Third: an Ace
First two: at least one Ace -> Third: a 10-point card
This is because, as mentioned, I am free to put the third card to the right or left.

Do you want to know how many combinations are winning? They are 11.648 !!
11.648 / 132.600 is equivalent to say a probability of 8.78%, which translates into 1 out of 11.4 !!

Do you want a strictly statistical proof? Here she is:

N = number of decks
CA = number of cards = 52 * N
D = number of cards worth 10 (which are 10, J, Q, K) = 4/13 * CA
A = number of aces = 1/13 * CA

PD = Probability to extract a card from ten = 4/13 (independent of N)
PA = Probability of extracting an Ace = 1/13 (independent of N)

P2XD = Probability to extract two cards of 10 in the first two = PD * (D-1) / (CA-1)
this is because it is equal to the probability of extracting the first ten, which is PD, multiplied the probability that the second is a ten, given the fact that the former is ten too, then D-1 ten-point cards have remained, out of a total of CA-1 cards, so the P that the second is ten is now (D-1) / (CA-1)

P2XA = Probability to extract two aces in the first two = PA * (A-1) / (CA-1)
same reasoning as before

Now comes a slightly harder passage, but I want everyone to understand it, so I have to explain HOW you add two probabilities: I will use set theory, which is easy, even visually, look at the figure: Each of the two ovals represents the probability of extracting an ace, so the surface of each oval is PA. The meaning of the figure is: the RED area is the probability that ONLY the first card is ace, the GREEN area is the probability that ONLY the second card is ace and the GREEN-RED area is the probability that BOTH are both Ace.
What I want to calculate now is the probability that AT LEAST ONE card is ace. What is realized if the first is ace OR the second is ace, or both.

This probability is therefore the surface of the figure composed by the UNION of the two ovals. If I sum the surfaces of the two ovals I would have counted TWO times the GREEN-RED part, then, to find the surface of the figure, I have to add the two ovals and TO REMOVE the GREEN-RED surface once.
Here we are, we have arrived: the colored surface, which corresponds to the probability that AT LEAST a card is ace is

PA + PA-P2XA

Well we can continue:

P1XA = probability that at least one of the first two cards is ace = PA + PA-P2XA
P1XD = probability that at least one of the first two cards is worth ten points = PD + PD-P2XD

PDA = probability of having an ace and a 10 in the first two = PD * A / (CA-1) + PA * D / (CA-1)
this is always a sum of probabilities, but since here there is no overlap, as before in the case of the ovals, I must not subtract anything.
I'm adding the probability that the first two cards are "D A" and the probability that the first two cards are "A D".
The prob. that the first two cards are "D A" is prob. that the first is D ie PD, multiplied by the prob. that the second one is A, since the first one was D in the deck CA-1 cards remained and all the aces remained, which were A, then the prob. that the second card is A is A / (CA-1).
Same reasoning for the other by adding ...

Courage ... we're almost out ...

P1AnoD = Probability that in the first two cards there is at least one ace and NO ten card = P1XA-PDA
P1DnoA = Probability that in the first two cards there is at least one card of ten and NO ace = P1XD-PDA

Well: the COMBINATIONS and PROBABILITIES of the first two cards that can give rise to a win (make the 21 with the THIRD card, which I can decide where to put it) are just:

1) at least one from ten without ace: P1DnoA
2) at least one ace without ten cards: P1AnoD
3) both a ten and an ace: PDA
There are not any others!

Results:

case 1)
if an ace comes out you win. There are still all the aces, which are A and CA-2 cards are left.
The prob. that an ace is A / (CA-2) and the probability of winning in case 1 is P1DnoA * A / (CA-2)

case 2)
if a ten-card comes out, you win. There are still all and they are D and CA-2 cards are left.
The prob. that a ten card comes out is D / (CA-2) and the probability of winning in case 2 is P1AnoD * D / (CA-2)

case 3)
if an Ace OR a 10 come out, you win. There are still (A-1) aces and (D-1) cards of ten and CA-2 cards remained.
The prob. that a winning card (ten or ace) is [(A-1) + (D-1)] / (CA-2) and the probability of winning in case 3 is PDA * [(A-1) + (D -1)] / (CA-2)

Well ... the substitutions we do to Excel ... and here are the results, with one deck on the left, and with 1000 decks on the right: (sorry, it's an old image of mine, and it's still in Italian)

So we have the following fact, which - you will agree with me - is quite STRANGE ...
In the worst case it is SHOULD win the MAXIMUM (points 21 = 50 €) already at the THIRD card extracted, with a probability of 8.557% !!
Believe me if I tell you this does NOT happen!
Waste some of your time, if you want: watch someone play and count how many times you see a 21 at all ...
The games that I did (a few ...) and those that I looked to, are over a hundred, and I have not yet seen a single 21 !!!

Do you want more evidence? Here is the same game, but with REGULAR card distibution, as you can see yourself (a little bit of code in excel, nothing transcendent)
Download it by clicking HERE (do an antivirus check, for safety, ALWAYS do it, even if you trust me ...), I inform you that, of course, the excel sheet contains "macros" of which you must authorize the execution. (P.S. I have found that on Excel 97 it does not work well, you have to have Excel 2000, at least)
Take a look at it, make some games ... YOU WILL WIN A BIG BAG of POINTS !! (100 points = € 1)

Counter-proof: we said that the MINIMUM probability of making 21 already appearing on the third card is 8.5571%, well, I wanted to have fun, I invested 4.5 euros in my personal survey, and I played.
I managed to do (playing 0.25 Euro a knock) 36 shots ... WITHOUT, of course, peep NEVER the fateful 21 to the third card.
I asked myself: what chance did I have to do 36 shots without pecking this "magic" 21?

You must know that when the result of an experiment can be defined as "success" or "failure", and when these have known probabilities, then there is a formula for calculating the probability that you should wait for failures before success:
in fact, if p is the probability of success and q the probability of failure, in addition to the fact that p + q = 100%, we have that:

Probability of waiting for failures before success = p * qx
We know that p = 8.5571% and therefore q = 91.4429%, x = 36 so we can calculate it!

Do you know how much it does ?? It is 0.34% !!!!

Yes, sir! I had only 3 chance out of 1000 to do 36 shots without the 21 ... and I did it !! I'm lucky, am not I ???

But that is not all !! During the game it happened to me to have immediately an ace. I thought: "How nice! Now if you gave me a ten I would finally win ..."
Well, IT HAS NOT GIVEN ME !! It happened three times ... he extracted, as if by magic, 4 or even 5 cards ALL OF LESS THAN 10, as long as ten has become useless ...

You will have understood by now that I am curious ...
HOW MANY PROBABILITY IS THERE TO SEE 5 CARDS AT A ROW , ALL LOWER THAN TEN?
p = probability of success, ie a ten-point card = 4/13
q = probability of failure, ie a card NOT of ten = 9/13
x = 5

The probability of extracting 5 cards all less than ten is p * q5 = 4%

So here's another strange thing: THREE TIMES occurred an event that only had a 4% chance of occurring !!

Conclusions:

The game object of this analysis is presented in a certain way. The only indication in this case is that the game itself returns 75% of the bets as MINIMUM. There are no other indications.
So the player can legitimately expect simply what he sees: a game in which cards from a certain number of decks are dealt. These cards should be those that are normally in a 52-card deck, and should therefore have the normal odds of these decks.

The game advertises some winning combinations, which combinations have a theoretical probability that you could legitimately expect.
Probability that IN PRACTICE not only does not occur, but CONSTANTLY events occur that should have VERY LOW odds !!
Which leads me to formulate the following conjecture:

PROBABILITIES ARE FALSIFIED !!

The FINAL test:
I put my hands on a commercial invoice: that is the bill issued by the company installing the game, to the tobacconist. Meanwhile, I let you know that the slot machines are produced, programmed and distributed by private companies. CERTAINLY not beneficial institutions! Then let's know how the economic mechanism works: the machine keeps track of all the bets and winnings, and therefore makes the difference between the two figures.
The bill I saw said: played € 1800, € 1404 won, difference ("CASH") € 396.
Now, pay attention to the following invoice's item: TAXES = 13.50%, but not on the difference !!! 13.5% OF THE BETS !!! Do you got it ???? TAX = 13.5% out of 1800 = 243 Euro !!
More some other percentage point for other taxes ... And only at this point the difference between Bet and won, MINUS TAXES, become gain, which is divided between the owner of the machines and merchant.
So let's recap: the machine must return at least 75% of what it collects (the rule, I believe the law, for all the slot machines we are talking about is that they return 75% measured on cycles of 14,000 played), but 15% of what you collect goes into taxes. 75% + 15% = 90% !! The operator of the machines remains 10% of the bets, to be shared with the operator.
According to you REALLY who program the machines RISK not to cash anything? In your opinion, is the fact that, of the first bill, € 1404 and 78% of € 1800 is ALL RANDOM ??

Do you think REALLY donkeys fly ??? (maybe they do not fly ... certainly they play these games !!)

Note
This article was written at the appearance of the first video slots, in the year 2005, more or less. Since then I think it has become clear to all that the video slots are electronic devices specially calibrated so that they return at least 75% of what is played, in the form of winnings for the players. Given also for SURE that even taxes are applied to the volume of money played, and given that, as rightly so, the operator of the equipment and the merchant who hosts them, I think it is correct to conclude that there is no way to win, if not playing very little, hope for luck and leave immediately.

This document must be considered protected by Article 21 of the Italian Constitution, which protects the freedom to express one's own thoughts.
The information and deductions presented are supported by the theory of probability calculation, are presented objectively, and can therefore be challenged only by bringing new elements, which should not have been considered.
I am available to receive any kind of report, and to give my justifications. To contact me, use the appropriate contact form.

I reported this analysis to: Altroconsumo, Mi Manda Raitre, Striscia la notizia. I'll keep you up-to-date...