Photon 2 Can´t join random room - unity3d

Question

I´m new to Networking and can´t find the problem on my self.
I made a matchmaking system for my game and it worked well.
But after some changes in my Game it stopped working.
It should create a room if no room is there. Else join a random room.
But it always creates a new room. no errors etc.
here are the scripts that should do the work.
Edit:
broke it down to this script:
using Photon.Pun;
using Photon.Realtime;
using System.Collections;
using System.Collections.Generic;
using UnityEngine;
using UnityEngine.UI;
public class PhotonLobby2 : MonoBehaviourPunCallbacks
{
public static PhotonLobby2 lobby;
public GameObject battlebutton;
private void Awake()
{
lobby = this; // create singleton
}
void Start()
{
PhotonNetwork.ConnectUsingSettings();
}
public override void OnConnectedToMaster()
{
Debug.Log("connected to masterserver");
PhotonNetwork.AutomaticallySyncScene = true;
battlebutton.SetActive(true);
}
public void OnBattleButtonClicked()
{
PhotonNetwork.JoinRandomRoom();
}
public override void OnJoinRandomFailed(short returnCode, string message)
{
Debug.Log(message + returnCode);
Debug.Log(" failed to join random game");
CreateRoom();
}
void CreateRoom()
{
PhotonNetwork.CreateRoom("Room");
}
public override void OnCreatedRoom()
{
Debug.Log("Created Room");
base.OnCreatedRoom();
}
public override void OnPlayerEnteredRoom(Player newPlayer)
{
Debug.Log("Entered");
base.OnPlayerEnteredRoom(newPlayer);
}
}
Debug says: No match found32760

Answer 1

It is because the two players join matches in different regions... you can force this in the photon server settings Fill in Dev region and Fixed region with one particular server region that has better ping for you. Make the property Use sure name serve is also checked

Answer 2

I resolved this issue by checking the room properties which first player creates.
There is a variable of Room named "IsOpen". If it is false then second player cannot join the same room. I just set it to true and now its perfectly joining the same room.