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Probabilities in examples and their practical use

Your neighbor from above has long been rumored to be very lucky, although the grandmothers council of the entrance calls him a gambler. He has the glory of being unbeatable in backgammon, card and dice games. But personally you often see him in the library where he reads books in the corner, but there is always a calculator next to him and he is making notes.

The fact that he floods you and you are a faience has given you the opportunity to ask him in a straight text where the key to his success lies. Everything turned out to be too simple, but not quite, because it turned out that in these games not only the risk taker wins, but more often the one who calculates the probabilities of one event or another and thus minimizes the risk to himself, respectively, increases the chances of winning.

Probably the easiest way to explain the complex concept of probabilities is to give examples of thrown coins, dice and pulling a card from a deck.

We started with barbut, a three-dice game that has a certain amount of points for different dice combinations. But I specifically assessed the risk associated with the probability of two winning dice already, throwing 1 to be eligible for a bonus roll. Or simply if we roll a single, ordinary dice, what is the probability of it falling to 1?

Imagine throwing a regular dice  with 6 sides, each of which has numbers from 1 to 6 and each of them can fall on a roll. Only one of these sides has 1.

The probability of falling 1 can be calculated as a fraction. Since there is a high probability of hitting 1 and there is only one side with 1, the probability numerator is 1. There are six possible outcomes, so the denominator is 6. The probability of hitting 1 is 1/6 = 16.7% . Or the opposite, there is 83.3%  probability to lose. The final choice is always yours, but don’t blame the neighbor for winning. For those of you who have forgotten how to switch from fractions to percentages, the numerator (the number above the fraction) is divided by the denominator (the number below the fraction) and the result is multiplied by 100.

For some of you, it may be easier to work with fractions, but the percentages have definitely kept me from throwing dice, but when I got home, I remembered the basic rules for simplifying fractions, and even forgot that there were proper and improper fractions (do you remember the difference?). I was concentrated on the proper fractions because right now we’re talking about statistical probabilities, and they can’t be more than 1 or 100% otherwise. And I remembered the comic event I came across on my daughter’s birthday when I had to cut the cake into seemingly identical pieces. I have 8 guests told me my daughter. Piece of cake I thought:

The cake cut by half –  ½ or 4/8, every half in half – ¼ or 2/8 every quarter in half 1/8.

The pieces were decent in size, but besides the guests and my daughter was there (((they needed 9 pieces.

Back then, we covered up with a box of tales about diet and food intolerance, but now, remembering, I sat down and tried to draw a chart on a piece of paper, and started from the back side – I needed 1/9 part. 9 is a number divisible by 3 without a remainder. That is, I first need to cut the cake into three equal parts 3/9, which after simplification is 1/3, then each third to three more. Here is one of my intermediate drawings when I thought of 6 pieces:

If you are interested in try with 12 pieces, but remember – they are the same size !!!

The bathroom was ready, but my encounters with my neighbor continued, as I progressed, I searched for more complicated examples. It turned out that apart from the well-known probability that the slice of bread would fall on the carpet with the smeared down, which in the Murphy Laws alone is 100%, and not the 50% standard calculated, there are many other interesting cases. The most commonly used examples of probability methods are dice and cards.

As  a beginning lets start with rolling a non-standard die six sides but  has the following numbers on its six sides: 1, 1, 3, 3, 5, 5. What is the probability of rolling a 3?

With this funky die the number 3 shows up twice, so there are 2 favorable outcomes, making the numerator 2. There are still six possible outcomes so the denominator is still 6, making the probability . We can, and should, simplify probability fractions when possible so the answer to this probability problem is 1/3 = 33,3%.

Backgammon !!!

My neighbor’s other favorite game. A game of two dice and combinations of moving pools. There are different combinations of moves at different stages of the game, but at some point we need a certain combination to win. And now comes the time when with both standard dice with 6 sides and the numbers from 1 to 6 we have to throw more than 9. What is the probability that the amount of throw will be greater than 9?

To get started, let’s simplify the process and imagine the likely results. We can make a table of the possible results of the sum of the two dice from 1 to 6.

 

Die 1
   Die 2 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9
5 6 7 8 9
6 7 8 9

 

As you can see there are 36 possible combinations, 6 of which are greater than 9, and these are shaded. So, we have 36 possible outcomes and 6 favorable ones:

6/36 = 1/6 = 16,7%

With such a percentage we will surely need a bonus luck)))!

What have I learned from these calculations? I began to calculate the probability of an opponent throwing some combinations and hitting me. I definitely changed my gameplay stereotype and started winning on the bench in front of the block, under the staring gaze from the neighbor’s balcony above. It turns out that in addition to knowing the rules of the game, there is another side of knowledge that helps you win.

 

With this our communication did not end, we switched to card games and their probabilities.

As always, we started with the simple ones. What is the probability of drawing a 10 of clubs randomly from a standard deck of 52 cards?

First, it’s important to know what is in a deck of cards.

52 cards (not including Jokers) with 26 red and 26 black ones in 4 suits: diamonds(red), clubs(black), hearts(red), and spades(black). In suit we have: A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K(did you count them) and 3 face cards in each suit: Jack, Queen, King (Ace stands for number one).

Since we have 52 cards and only one of them is 10th party, the probability is 1/52 = 1.9%. Too little chance, right? I remembered the poker channels on television and how long some of the players have been thinking, it turns out that they are calculating not only their odds, but also those of their opponents (which I as a viewer see, but not them) depending on the cards on the table. Millions of prize money are totally worth it.

What about pulling any diamond card? The probability of pulling a diamond card would be  since there are 13 cards in the suit of diamonds, and 52 cards in the total deck of cards. Simplifying the fraction, we get . So, there’s a much better chance of pulling a diamond card than pulling a 10 of clubs.

What is the probability of pulling an ace from a standard deck of 52 cards? The outcome we want is an ace, and there are 4 aces, so the probability is 4/52 = 1/13 = 7,7% .

As we discussed in the text above, we are probably talking about possible solutions in the range of 0 to 1 or Impossible and Certain events.

Let’s go back to the fractions. And why do we look only at the regular fractions. Leys take a closer look at the extreme cases – zero and one. For the first “the translation”  says that this can not happen (the probability of happening is 0), which in the above case with a deck of cards without jokers is more than clear simply because they are not in the deck or the mathematical expression is, that the probability is 0/52 = 0%  . Any event that is impossible has a probability of 0.

Now what about the 1 from probability point of view. Some things are impossible not to be done. Hate those double negatives? Let’s say it another way; some things are absolutely certain. Here’s an example. What is the probability of picking either a red or a black card from a standard deck of 52 cards? Well, all the cards are either red or black, all 52 of them. So the probability is 52/52 = 1 = 100% .

And millions of combinations between 0 and 1 …

Do you feel ready to go to a Casino?))))

These were simple situations, but sometimes there are more complex situations where, in addition to throwing dice with sum of 7, we need exactly 5 and 2 in order to get the pool out. Neither 6 and 1 nor 4 and 3 do our job. I.e. we need to fulfill two conditions at the same time. This type of probability is also called AND probability. In the other case, we need 7, regardless of the combination of the numbers 6 and 1 OR 5 and 2 OR 6 and 3 OR 4.

So we come to the definition of “And” probabilities, where the result must meet both (possibly more) conditions simultaneously.

In the case of “OR” probabilities where the result must meet only one condition, OR the other condition , OR both at the same time, i.e.  includes the “AND” result.

Let’s see one probability in two ways:

We need the amount of dice while playing backgammon 7 with only 2 and 5.

The combinations are 36 and the possible variants are only 2 (((

P = 2/36 = 1/18 = 5.5% You will probably lose this game unless your opponent goes wrong.

What is the probability of drawing a card from a deck and it being red and a face card (King, Queen and Jack)?

For this probability, we need to look at which cards are both red and face cards. There are 6 of them: Jack of Hearts, Queen of Hearts, King of Hearts, Jack of Diamonds, Queen of Diamonds, and King of Diamonds.

P = 6/52 = 3/26 = 11,5% higher but still very tricky.

Now let’s calculate the probability of throwing 7 no matter what the dice numbers are, with the following combinations of digits 5 and 2 OR 6 and 1 OR 4 and 3 OR 2 and 5 OR 1 and 6 OR 3 and 4.

6 possible combinations of 36.

6/36 = 1/6 = 16.7% still 3 times more likely, and a matter of luck.

What is the probability of drawing a card from a deck and it being red or a face card?

This time the card can be red, or a face card, or both at the same time. There are 26 red cards (6 of which are also face cards). In addition, there are 6 more face cards that are not red: Jack of Clubs, Queen of Clubs, King of Clubs, Jack of Spades, Queen of Spades, and King of Spades. That is a total of 26 + 6 = 32 cards.

P = 32/52 = 8/13 = 61,5%

Be careful not to just add up the number of face cards (12) with the number of red cards (26). That would give a total of 38 cards, but it would count the red face cards twice.

Notice how much these two probabilities differ. One little word changes the whole problem!

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