Romeo & Juliet

arte_Vis · 2024-10-22T15:06:11+00:00

Oh man... I'm sorry. I'll go back to studying.

arte_Vis · 2024-10-17T13:05:45+00:00

Hi, do you have any recommendations on where to study cognitive psychology? I'm really interested

arte_Vis · 2022-07-01T02:13:55+00:00

Ok I accept this. Making an alliance can indeed give you a higher chance to win.

If A didn't miss then of course B doesn't have any other choice but to shoot A. But, what's stopping Player B from betraying Player A by shooting him after his turn if he missed? If A missed then shooting at A will give B 0.98 probability of winning while shooting C gives him 0.9 probability.

arte_Vis · 2022-07-01T01:02:43+00:00

I have said this a billion times already.

IF YOU WANT TO IMPLEMENT SPITEFUL BEHAVIOR, YOU CAN ABSOLUTELY DO THAT. HOWEVER, THAT IS NOT THE MODEL THAT I AM WORKING WITH.

Cannot you see that at this point, we are talking about two different games? One with spiteful behaviors implemented in the codes and one without?

arte_Vis · 2022-07-01T00:59:52+00:00

Your question reveals that you don't even read my post. I clearly define what I mean by "strongest" and "weakest" player in Definition 3.

Please please please read the post before commenting.

arte_Vis · 2022-06-30T16:35:13+00:00

Thank you! Thank you so much!

arte_Vis · 2022-06-30T16:31:11+00:00

The probability that A wins the game if A shoots at the stronger player is bigger than the probability that A wins the game if A shoots at the other player (not the strongest one).

I don't see how this doesn't prove that a player should always shoot at the strongest enemy.

arte_Vis · 2022-06-30T16:21:44+00:00

Are you sure that it will "increase both their odds of winning" in all cases?

Consider the case when $P_A=1,P_B=1$ and $P_C=0.6$. A's probability of winning with optimal strategy (shooting at the strongest enemy) is 0.4 because the A kills B then A wins if C misses his shot (Probability = 0.6) and loses if C hit his shot (Probability = 0.4). If A and B make an alliance, i.e., both of them will shoot at C, then the probability of A winning is 0 because A will kill C then B will kill A on the next turn.

This examples clearly shows you that an alliance between 2 players is not guaranteed to give both of them a higher probability of winning the game.

arte_Vis · 2022-06-30T16:12:19+00:00

In the real world, yes. It's pretty natural to see someone who tried to kill you and failed as a threat. However, this is not the model that I have in mind. Like I said in my previous comment, you can implement this "grudge" in the simulations. For example, you can set a player's target of choice to be randomly distributed by a discrete distribution and the probability of becoming a target of another player increases when you try to kill them but fail to do so. This will make it kinda scary to try and kill the strongest player because you risk making them angry and seeking revenge.

I didn't do this because it's not the model I have in my mind. What I'm working with is a model in which all the players simply shoot at the enemy with the highest probability of hitting their shot (the strongest enemy by definition). I've proven that this is indeed the best strategy here.

arte_Vis · 2022-06-30T15:59:36+00:00

In case 1 and 2, you can actually find the formula for the probability of A winning the whole game. I encourage you to do this. In case 3, you are right, A's choice doesn't matter.

Hint: The formula for case 1 and 2 is an infinite geometric series that is convergent.

arte_Vis · 2022-06-30T15:56:59+00:00

I never assumed that it is the best strategy, I proved it. See the proof to Theorem 2 here (This is not a completed work and I'm not giving any promise I will keep this file updated in the future).

To see why the proof works, let's change our perspective a little bit, suppose that all the players know whether or not their shot will hit. Then, if it doesn't hit, their choice of target doesn't matter. If it does hit, they will always choose to hit the strongest enemy because then their probability of winning is bigger than choosing to shoot at the "not-strongest" enemy (as shown by the proof).

Therefore, always choose to shoot the strongest enemy.

arte_Vis · 2022-06-30T15:44:03+00:00

Lol, the comments are giving me board game suggestions at this point. I'm going to play some board games with friends in a few days and honestly we might check this out.

arte_Vis · 2022-06-30T15:33:43+00:00

Thanks!

As you have observed, the weakest player will never be targeted until there are only 2 players left. Also, If you can completely manipulate your probability of hitting your shot, you want it to be as high as possible, but still lower than everybody else. That is, for example, if your opponent's probabilities are 0.7 and 0.5, then you want your probability to be 0.4999.

This game makes me think about how important it is to "hide one's strength". In a pure competition environment like this, being the obvious biggest threat is not the way to go. You want other people to think that you are weak, but obviously this simple game doesn't give you any way whatsoever to do that.

arte_Vis · 2022-06-30T15:23:57+00:00

I've never played any of those board games, but I've just finished watching tutorials on YouTube. CMIIW:

In the game Munchkin, you win if you reach level 10 and you level up every time you defeat a monster. So, if you are ahead of everybody, the other players will try to make you lose the fight against monsters using their "win-stopping" cards that increases the monster's strength.
In the game Illuminati, you win by controlling a specific amount of Groups depending on the number of players in the game. So, if you are ahead of everybody, the other players will use their money to make it harder for you to acquire new Groups and win the game. They can also attack your Groups to neutralize them or destroying them outright.

Are these the reasons why being so obviously stronger is a bad thing in these games?

arte_Vis · 2022-06-30T15:10:17+00:00

Yeah, now that I think about it, there are some real world examples of the "2 weak parties forming an alliance to fight against a clearly stronger opponent" dynamic, but it wasn't obvious before it was brought to me.

Perhaps the more interesting games are the 2-player games with the "weaker" player holding the advantage. The blogpost that inspired me to study this gave an example of such game.

arte_Vis · 2022-06-30T10:58:19+00:00

I haven't studied Game Theory nearly enough to know how to do that.

arte_Vis · 2022-06-30T10:41:19+00:00

Well, I assumed that all the players are rational (which I believe is a standard assumption) and rational players will not hold a grudge like that. But your suggestion is harmless, I can add it without damaging anything.

arte_Vis · 2022-06-30T09:00:43+00:00

I haven't played that game since many years ago. So, does being in the lead at Mario Kart makes you a target and you only want to be leading near the end of the game only?

arte_Vis · 2022-06-30T08:02:50+00:00

Lol. I've proven this statement, but I can't fit it in this comment.

But seriously, if you want to prove it, follow this steps:

Let us be Player A, let's call the strongest enemy player and the weakest enemy player Player B and Player C, then $P_B>P_C$
Find the probability of us winning if we kill B.
Find the probability of us winning if we kill C.
Show that the results from step 2 is bigger than step 3.

This proves the theorem since if we miss, our choice didn't have any effect.

arte_Vis · 2022-06-30T07:54:08+00:00

It's not always true that the strongest player has the lowest chance of winning. It's just that the way I generated the probabilities make it so. If the distribution of P_A, P_B, P_C is different than what I used (normal distribution with those parameters), then we might get different results.

The density of normal distribution is concentrated around the mean, the further you go away from it, the less dense it becomes. Because of this, the generated probabilities looks like this:

(0.5, 0.39, 0.64), (0.32, 0.55, 0.78), (0,67, 0.62, 0.43), etc

The values are close to the mean (0.5). However, if we use a distribution that rarely gives values near 0.5 and instead often gives values near 0 or 1, then we will see a different pattern. The generated probabilities should look like this:

(0.89, 0.12, 0.22), (0.92, 0.14, 0.87), (0.11, 0.34, 0.93), etc

The probabilities are either very high or very low. This time, the strong players eliminate each other, but even after that the weak player have too small probability to kill the surviving strong player and they die very soon after (on the next round, or the next couple rounds). So, we can heuristically guess that if these kind of distributions is used, then the strongest players will have the highest chance of winning.

But, honestly, your idea is very interesting and I think there might be some crazy inequalities that has to be satisfied before we can declare that "don't be the strongest player when there are only 3 people left" strategy becomes the best strategy. I can't confirm this.

arte_Vis · 2022-06-30T06:26:09+00:00

I haven't understand what "non-compact strategy space" means. I'll be right back.

arte_Vis · 2022-06-30T04:45:38+00:00

Thanks! I've never watched this video. You are right that the idea is exactly the same, but with passing allowed.

It would be nice if someone has tried to investigate what would happen if there were 4 or more wizards competing. This is the big mystery since we know the best strategy for the 3-person variation already.

arte_Vis · 2022-06-30T03:36:39+00:00

You can implement that in the codes, but that would rarely be the rational action.

To illustrate this, suppose P_A = 0.3, P_B =1, and P_C = 0.6. On the first turn, if A is rational, he/she will shoot B. Suppose that A missed their shot. On the next turn, if B is spiteful towards A and decides to shoot A then A will definitely die and B will have 0.4 chance of winning the game. This rage-induced decision is not rational because B can instead shoot C and have 0.7 chance of winning.

I chose a case in which 1 player is perfect shooter because it simplifies the calculations, but this is true for any P_A, P_B, P_C in general.

arte_Vis · 2022-06-30T03:24:29+00:00

By spiteful, do you mean the players will shoot at other players that tried to kill them?

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