We calculate the probabilities for the article "Dishonest game, or how the organizers of the drawings deceive us"
Article with the analysis of the game of the well-known trading network has caused us in Cloud4Y lively interest. Here are a few excerpts to introduce you to the course:
One day, on a sunny spring morning, reading the city forum, I came across a link with a simple game from a well-known trading network. The game (action), dedicated to the World Cup, was an unpretentious field three to three, filled with soccer balls. Clicking on the ball, we opened the picture with this or that product. When opening three identical pictures, the participant was guaranteed a free receipt of the goods in one of the stores of the network. Also under one of the balls was a picture of a red card, the opening of which meant the end of the game.
The author of the article began to investigate the reasons for his loss and found out from the results of calculations the following:
A quick sketch of formulas on a napkin, and it turned out that the probability of winning is 1/4. For 5 fields it was necessary to tinker, but the calculated probability was also 25%.
Running the script, I got an unexpected result - 25% of the winnings. After playing with the number of winning elements and the total number of fields, I found out that the probability of winning in such a game does not depend on the number of fields and is equal to one divided by the number of winning elements increased by one.
We were interested in the correctness of this calculation and, replacing the napkin with Excel, we set about finding a mathematical truth. Readers, keen on probability theory, are invited under the rules to verify the correctness of our calculations.
Stecenko in his comments . He also writes:
If you look at the script of the author, it is written from the assumption that there are three cards of the same product on the box, one more card of five other goods, and one red card, while from the description of the game it absolutely does not follow - the rules do not say , that on the field there must be a winning combination.
The rules, however, say that only 26 goods are involved. It turns out that the issuance of 9 cards is generated: 8 cards are combinations of 26 products with repetitions, and one card is red.
In such conditions, the calculation of the mathematical probability of winning is somewhat more complicated than the author of the original post suggested. The game consists of several layers:
The probability of opening N cards without opening the red one is the probability of a different duration of the game without taking into account whether there will be a prize.
Probability to collect a combination of 3 cards with the same goods. This probability varies with different playing times. It is important to understand that a set of 8 cards with goods and always one red does not necessarily contain at least one pair of identical cards, not to mention three at once.
Let's start with a simple one - with an understanding of how the number of participants will decrease as the number of cards opened by them grows because of the rule of the red card.
Chances of different game duration are
For problems with a fixed number of tests or tests, if the result of any test can only be success or failure, the tests are independent, and the probability of success remains constant throughout the experiment, we use Bernoulli's formula-in Excel function BINOM.RESP.
For example, using the BINOM.DIST function, you can calculate, for example, the probability that two of the next three newborns will be boys. We calculate how likely it is that the N tests (open cards) will open red.
The last column shows how many players out of a hundred remain in the game with such a duration. The remaining players are eliminated because of the red card, and not knowing if there was a winning combination in the issued set of cards.
Now calculate the probability of collecting a combination of three cards with the same goods. This is a prerequisite for receiving the prize.
Chances for prizes
We decompose the game logically. We open the first card, and then we select a pair for this card, we select a triple for a couple. With this approach, we can calculate the probability of obtaining 3 identical cards for three attempts, proceeding from the fact that all the goods in the game are 26.
= BINOM.DIST (3; 3; 1/26; 0) by the mask
= BINOM.DIST (number of successes, number of tests, success_character, integral)
Or here's a formula for calculating the probability of obtaining 3 identical products in 8 attempts.
= BINOM.DIST (3; 8; 1/26; 0), right?
Not really. When we reached the duration of the game in 4 trials, not allowing the opening of the red and continuing to play, we get a situation with the possibility of two pairs to select a winning three.
Conditionally, two products out of 26 are A and B. Our open cards are A-B-A-B. The probability is no longer 1/2? but 1/26 + (1/26) * "The probability of two pairs for a given duration of the game" .
The probability of two pairs for a given game duration is BINOM.DIST (2; 5; 1/26; 0) ^ 2
When the game reaches a length of ? we have a combination of A-B-A-B-V-B type. This means that the probability is now
(2; 7; 1/26; 0) ^ 3) and we are looking for a third map for ? 2 or three pairs.
Knowing the probability, we build the matrix:
In it, we find the probability of each maximum number of repetitions of a product for each variant of the game's duration. We remember that the probability of winning changes on 4 attempts and at ? and therefore we take this into account in the formula BINOM.RASP
We need variants with the maximum number of repetitions of any product of 3 or more. As the game stops, as soon as we collect the triple, we add the probabilities for the columns in the area in bold.
Next, we multiply the chances of each game's duration by the chances of winning a prize at such a length. Summarizing these works, we get a probability of ??? or 11 winners per 1?000 players.
Yes, indeed this is a very low probability. Not at all 25%. Again, go back to the rules of the game:
8. The prize fund:
8.1. Available prizes: names and quantities
116000 divided by the probability of winning and get ??? participants in order to play all the prizes. Recall that on January ? 201? according to the Federal State Statistics Service in Russia, there were 14?93?921 permanent residents. Apparently, this is the idea of the organizers of the game - to give a chance to win every citizen of Russia.
The code analysis showed the author of the original publication that the script "knows" its outcome even before the game starts, but nobody knows what a predetermined result will give the script to a specific user. Knowing the mathematical side of the question, you can make a conclusion about the honesty of the organizers.
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