# Counterintuitive probability event

Mathematics is interesting and wonderful, but sometimes it can make you lose confidence and even make a fool of yourself in front of others. This is because behind some seemingly intuitive mathematical problems, there are often simple logics that you can easily overlook, causing you to “stumble” when calculating. It’s not because you don’t know how to make a calculation wrong, but because mathematics is actually a very clever “pretender”, it often makes you forget that it is actually wearing a veil. This article introduces 3 cases of counterintuitive probability events.
Three questions
first event is three questions. In an American TV game show “Let’s Make a Deal” in 1963, a host named Monty Hall asked this question. The three-door problem is described like this: you as a player are playing a game-there are three doors in front of you, left, center, and right, one door behind a car, and the other two doors behind each one is a goat . You have a chance to open the door. If you open the door of a parked car behind, you can win the car. If you open other doors, you will get nothing.
You certainly don’t know what is behind each door, and the dealer knows it all. When you select a door (assuming this door is No. 1), the dealer will choose another door (assuming it is No. 3), behind the No. 3 door is a goat. Then the dealer asks you: Do you want to change your choice? Shall we choose door 2 instead? At this time, you naturally think that there are only two doors left, and one is a goat and the other is a car. No matter whether you choose No. 1 or No. 2, the probability of getting a car is 1/2, so you don’t need to change your choice. However, if you think this way, you are wrong! In fact, if you choose door 2 instead, your probability of winning the car is 2/3. If you stick to gate 1, your probability of winning the car is only 1/3. Why is that?
There are three doors on the left, center, and right. There are three possible arrangements for the objects behind the door-1: cart, sheep, sheep; 2: sheep, cart, sheep; 3: sheep, sheep, cart. The position of the door remains the same, but the number can be changed-the door you pick is door 1 and the door the dealer opens to you is door 3. We assume that you choose the left door, that is, the left door is door 1 (which door is the same as door 1). For the first case, the result of whether the middle door or the right door is used as door 2 (door 3) is the same, that is, you can only win the car if you stick to the original choice; for the second case, the dealer will Open the right door as door 3 to show you, because behind this door is a sheep, and you can only win the car by changing your choice and choosing the middle door of door 2. For the third case, the dealer will give you Fancy the door, and you can only win if you change your choice and choose the right door for door 2. Therefore, of the above three situations, there are two situations where you can change your choice to win, and only one situation is to insist on choosing to win. In summary, the probability of changing the choice to win is 2/3, and the probability of sticking to the choice to win is 1/3. So, in order to be more likely to win the car, you should choose Gate 2 instead.
Question 8 prisoners of
the second event is such that eight prisoners were numbered 1 to 8 to get the chance to be a collective release, but they need a game: they will have moved into a room in numerical order. There are 8 drawers in the room, and each drawer has a small note with a number on the note. The number is one of 1 to 8 and does not repeat, and the note is randomly placed in the drawer. Each prisoner enters the room in turn, and the prisoner who enters is allowed to open any 4 drawers at most. If there is a note with the person’s serial number in these 4 drawers, the person wins. After the previous prisoner restores everything and goes out, the next prisoner can come in. Once the game starts, that is, the first person enters the room, all prisoners will no longer be able to communicate. All 8 prisoners must win before they can be released. Once a prisoner loses, the group will all be shot.
This game is much crueler than the last one! Because for a prisoner, the probability that the note with his number is in the 4 drawers he chooses is 4 divided by 8, or 1/2. Each prisoner does the selection work independently and does not interfere with each other, so the probability that all 8 prisoners will win is (1/2) 8≈0.0039, which is even less than four thousandths. This probability is too small. Some students will think it is very difficult for this group of prisoners to live here. This game is extremely unfair! But can it really be the case?
In fact, if the prisoners adopt a common strategy, the chances of everyone winning can be greatly increased, the probability even exceeds 30%. So what is this strategy? We number the drawers from left to right in the order of 1-8. We stipulate that the prisoner with serial number X must first open the drawer with serial number X, and then determine the serial number of the next opened drawer according to the number Y in the drawer with serial number X. According to this model, you will find your number until you find your own number, or use up the number of drawers you can draw. We use a layout to illustrate this strategy.
Drawer serial number 1 2 3 4 5 6 7 8
The number 4 6 1 3 7 5 2 8
in the drawer starts with prisoner 1. He opened drawer 1 and saw the number 4. Then open drawer 4, see number 3, then open drawer 3, see number 1, he won.
For 2 prisoner: drawer 2 → drawer 6 → drawer 5 → drawer 7, won
for 3 prisoners: drawer 3 → drawer 1 → drawer 4, won
for 4 inmates: drawer 4 → drawer 3 → drawer 1, win up
to 5 prisoners: drawer 5 → drawer 7 → drawer 2 → drawer 6, won
for 6 prisoners: drawer 6 → drawer 5 → drawer 7 → drawer 2, won
for 7 prisoners: drawer 7 → drawer 2 → 5 6 → drawer drawer, won
for prisoners 8: 8 drawer, won. At this point, everyone will be released.
We found that for this layout, no matter who opens the drawer, he will fall into one of the three cycles (below). The reason why everyone can win is because these three loops contain no more than 4 drawers, which means that anyone can find their number before 4 choices.
A: drawer drawer 1 → 3 → 4 → drawer 1 Drawer
II: drawer drawer 2 → 5 → 6 → drawer drawer drawer 7 → 2
Three: 8 Drawer cycle itself
, however, if in a different layout, the case can not so wonderful . For example, the following one:
drawer serial number 1 2 3 4 5 6 7 8
drawer number 4 6 8 3 7 5 1 2
There is only one cycle: drawer 1 → drawer 4 → drawer 3 → drawer 8 → drawer 2 → drawer 6 → Drawer 5 → Drawer 7 → Drawer 1. This loop contains all 8 drawers. Anyone who wants to find his own number has to open all 8 drawers. It seems that this strategy sometimes faces good luck and sometimes bad luck.
But what we care about is the probability of success. First of all, I have to mention the objective fact that there is only one loop (may contain 5, 6, 7 or 8 drawers) with more than half the number-4 drawers, because the total number of drawers is 8. And once there is a loop containing more than 4 drawers, it means that someone must fail. After calculation, the probability of occurrence of a layout that does not contain more than 4 cycles is about 0.365 (students who have studied permutation and combination can calculate it by themselves), which is also the success rate of this strategy, and this number is far greater than four per thousand.