Pirate gold coins

**Problem:**

Five pirates are listed in the order of 5 to 1. The largest pirate has the right to propose how to share 100 gold coins. But other people will vote on this. If the majority (the majority of all people) oppose this, they will be killed. What kind of solution should he propose, so that he can get as many gold coins as possible and won't be killed?

Answer:

The allocation scheme is 98,0, 1.

Whether a fifth-level pirate will be killed depends on whether other pirates will gain more benefits after the fifth-level pirate's death. If you get more benefits, you will certainly oppose it. If you get less benefits, you will certainly support it. If the benefits do not change, you can both oppose or support it.

If a fifth-level pirate dies, four levels of piracy are distributed, and four levels of piracy face the same problem. It depends on the change in interest distribution after their death. Then there are level 3 and level 2 pirates.

No matter what a solution is proposed, no one of the second-level pirates will disagree with the majority of the people (for their own support, but the other party's opposition cannot constitute a majority of opposition ). So Level 2 pirates will certainly propose a, 0 allocation scheme, and they will enjoy all the gold coins.

After you have guessed the distribution scheme of level 2 pirates, Level 3 pirates will propose a distribution scheme of, 0, and 1. In this way, level-1 piracy will support level-3 piracy because it has obtained more gold coins than level-2 piracy solutions.

After you have guessed the distribution scheme of Level 3 pirates, Level 4 pirates will propose a distribution scheme of, 0, and. In this way, level 2 Pirates get more gold coins than the level 3 pirate solution, so they will support the level 4 pirate solution.

After you have guessed the distribution scheme of 4-level pirates, 5-level pirates will propose the allocation scheme of, 0, and 1. In this way, level 1 and level 3 pirates obtain more gold coins than the level 4 pirate solution, so they will support the Level 5 pirate solution.

The three boys fell in love with a girl at the same time.

**Problem:**

Three single programmers fall in love with a girl at the same time. To decide who can marry the girl, they decide to fight with a pistol. The hit rate of A is 30%, B is better than him, and the hit rate is 50%. The most outstanding gunman is C. He never makes a mistake and the hit rate is 100%. For the sake of fairness, they decided the order: A starts shooting first, B is second, C is last, and then the cycle is like this until they only have one person left. Who will have the greatest chance to survive the three? Who has the least chance? What policies should they adopt?

Answer:

First, let's make it simple and prepare for solving the problem.

If AB is single-handedly selected, A is shot first, and a is likely to survive.

30% + 70% × 50% × 30% + 70% × 50% × 70% × 50% × 30% + ...... = 0.3/0.65

Correspondingly, the probability of B Survival is 1-0.3/0.65 = 0.35/0.65

Then let's see how A, B, and C fight. If there are three people, we should first consider the goal. For A, when C is still alive, it will certainly not aim at B. If B is not hit, it will be a waste of opportunities, even worse, the next round is directly cracked by C. Therefore, the first major threat solution is the positive solution, so a will target c.

Similarly, B will take Objective C and Objective C will take objective B as the goal.

Let's take a look at the survival probability of A in three ways:

A killed C and B did not killed a. At this time, the problem degraded to AB dual A and shot first.

30% × 50% × 0. 3/0. 65

A did not kill C, B killed C, and the problem also degraded to AB dual a first shot.

70% × 50% × 0. 3/0. 65

A has not killed C, B has not killed C, C has killed B, and a has killed C.

70% × 50% × 30%

The total probability event probability is added, and the probability of a survival is

0.105 + 3/13 ≈ 0.336

B can survive in three ways:

A kills C and B kills.

30% × 50%

A killed C and B did not killed a. At this time, the problem degraded to AB dual A and shot first.

30% × 50% × 0. 35/0. 65

A did not kill C, B killed C, and the problem also degraded to AB dual a first shot.

70% × 50% × 0. 35/0. 65

So the possibility of B's survival is 0.15 + 3.5/13 ≈ 0.419

C has only one survival case, which is relatively simple:

A did not kill C, B did not kill C, C killed B, A did not kill C, C killed.

70% × 50% × 70% = 0.245

**Therefore, we learned the following truth: Start with strength!**

If you allow waivers or shot in the sky, the situation changes again.

First, B cannot withdraw, because B knows that he is the target of C, and C has hundreds of thousands of failures. This first-hand opportunity must not be abandoned. If C is already dead, two people will fight each other, the first-hand opportunity also has an advantage.

C is even less likely to lose. One is that he is the best, and the other is that he is the goal of both A and B.

If a waivers in the first round.

.

The survival probability of A is divided into two possibilities:

B killed C. At this time, the problem degraded to a's fight with a first shot.

50% × 0. 3/0. 65

B did not kill C, C killed B, and a killed C.

50% × 30%

The total probability event probability is added, and the probability of a survival is

0.5 (0.3/0.65 + 0.3) ≈ 0.381.

38% of the survival probability! It is much higher than the best solution of the previous 33.6% million!

There is only one possibility of B's survival probability:

B kills C, and the problem degrades to AB dual A's first shot.

50% x 0. 35/0. 65 ≈ 0.269.

The survival rate of C is also very simple:

B did not kill C, C killed B, A did not kill C, C killed.

50% × 70% ≈ 0.35

Wow, it seems that a's first round of waiving gave a and c A a greater survival rate and made a most promising surviving player!

Therefore, if a's logical reasoning is good, he must choose to drop in the first round.

**From this, we learned another truth: Yu Weng benefited from the fierce competition.**

**Welcome to my independent technical blog**

Syntaxes | sameideal.com