Solution for the consecutive numbers problem

consecutive.js

// It's a bit messy, but it seems to work

function player(name) {
    this.name = name;
    this.questionCount = 0;
    this.number;
}

player.prototype.setOther = function(other, otherNumber) {
    this.other = other;
    this.otherNumber = otherNumber;
    this.possibleNumbers = [otherNumber - 1, otherNumber + 1];
}

player.prototype.knowsNumber = function() {
    if (this.number) {
        this.end();
        return true;
    }
    
    // If more questions have been asked than the other person's number,
    // and they still don't know what they are, we're the higher number
    if (this.questionCount >= this.otherNumber) {
        this.number = this.possibleNumbers[1];
        
        setTimeout(function() {
            this.other.knowsNumber();
        }.bind(this));

        this.end();

        return true;
    } else {
        this.questionCount++;
        setTimeout(this.ask.bind(this));
    }
}

player.prototype.ask = function() {
    if (this.other.knowsNumber()) {
        // If they figured theirs out first, that means we're the lower number
        this.number = this.possibleNumbers[0];
    } else {
        this.questionCount++;
    }
}

player.prototype.end = function() {
    console.log(this.name, this.number);
}

var i = 0;
var skew = 1;
var inter = setInterval(function() {
    skew = (skew+1)%2;
    if (skew === 0) {
        i++;
    }
    if (i >= 10) {
        clearInterval(inter);
    }
    var player1 = new player("player1");
    var player2 = new player("player2");

    // Associate players
    player1.setOther(player2, i+1-skew);
    player2.setOther(player1, i+skew);

    // Kick off the questions
    console.log("--");
    console.log("Should be player1:" + (i+skew) + " player2:" + (i+1-skew));
    player1.ask();
}, 100);