Overthinking Rhino Hero: Super Battle

I love to overthink simple mechanics, especially if they're a little outside of the typical die roll or card draw. Rhino Hero: Super Battle has an interesting little die roll-off system that I hadn't seen before. This is a quick and easy kids/family game, right, so why not ruin it by overthinking the hell out of it and doing some probability calculations.

In RHSB, players are taking turns adding walls and floors to an ever-growing tower. Each player has a playing piece and they balance that piece on the floors as they build. When the tower falls, whoever was highest wins (unless that person is the one who trashed the place, in which case everyone else wins).

Two players cannot have their piece on the same level because someone always has to be highest. So if a player moves their piece up to the same level as someone else, those players roll off and the higher roller stays while the loser moves down a floor. If there's someone on that floor then those two players roll off.

There are two dice included with the game. The red die is used by the player who just moved and has sides of 2, 2, 4, 4, 6, 6. The blue die is used by the player who was already there, and has sides 1, 1, 3, 3, 5, 5.

This has the nice effect, especially when playing with young'uns, that there are no ties. You both roll and the result is indisputable. It has a downside that you can't just grab a die and roll, you have to make sure the right person is using the correctly colored die. But does it give one player an advantage? And how much?

Looking at the numbers, it's pretty obvious the red die is better. The expected value (the mean result you'd roll over infinite rolls) is 4 for the red die and 3 for the blue die (a normal d6 has EV of 3.5). But how much advantage does that give?

Let's look at the red die. One-third of the time (when it rolls a 2) it will win one-third of the time (when the blue rolls a one). One-third of the time (when it rolls a 4) it will win two-thirds of the time (when blue rolls a 1 or 3). And one-third of the time it will always win (when you roll a six).

You'd write that out as such:

((1/3)*(1/3)) + ((1/3)*(2/3)) + ((1/3)*(1)) = ?

(There are simpler ways to reduce the formula for calculating such things, but I like breaking it out so people can see what's going on.)

Each of those sets is the probability that you roll a number times the probability you win given you rolled that number. You can add these together because these events are mutually exclusive (meaning that you can't roll a 2 and a 4 at the same time).

This comes out to 2/3. So if you're rolling the red die, you will win 2/3rds of the time. That means that blue die will win 1/3 of the time. Which all means that the "attacking" player who just moved will win the roll off twice as often as the player who was there first.

I think this is perfect for this game. If you just moved then that means you just did the hard thing of adding a new floor to the structure and then moving your piece without knocking anything over. So you should get the advantage. But it's not a perfect guarantee.

It also means that if you were already in place and you get bumped, you're pretty likely to only lose one floor out of the deal. If you lose a roll-off and move down to a floor where there is already another player, then you are considered the attacker and you get to roll the red die. This means you're less likely to get knocked all the way down to the bottom, you're likely to just lose one floor, which is good. But again, it's not guaranteed.

A flat roll-off of regular d6s where you reroll ties means just even odds; each player has a 50% chance to win. I don't think that'd be good for this game, and it is one more minor rule to remember (reroll ties).

Here's another system. Each roll a regular d6, attacker wins ties. Probability-wise, the attacker will win:

((1/6)*(1/6)) + ((1/6)+(1/3)) + ((1/6)*(1/2)) + ((1/6)*(2/3)) + ((1/6)*(5/6)) + ((1/6)*(1)) = 7/12

(Again, this is broken out. I believe for an N-sided die the attacker will win with probability [(N+1)/(N*2)]. In this game you're basically both rolling a d3 with attacker winning ties. I'll leave it as an exercise to the reader to confirm the math works out.)

This has the advantage that you don't have to pass dice around the table or switch from rolling the blue to the red when you get bumped off a floor. But in this case the attacker is only going to win about 58% of the time (compared to the current 67%), and I don't think that's enough. You'll have too many turns where you pull off some crazy balancing act and then have to move down anyway, or you made it to the top only get to bumped down a floor by every other player as you plummet with failed roll after failed roll.

Let's say there's 5 floors and you're up on top with someone on every floor. Some pesky opponent threatens your floor (which isn't possible without leaving a hole mid-structure, but this is theoretical). How far down are you likely to drop?

You're on the defense at first, so you'll drop from 5 to 4 two-thirds of the time. But then you switch to attack. So you'll drop (rounding):

From 5 to 4: 67% of the time
From 5 to 3: 22% of the time
From 5 to 2: 7% of the time
From 5 to 1: 2% of the time

This seems like a good spread. You did the work to get up there, after all. But you did that work a few turns ago, and every turn gets harder. A spread like this helps to reduce the chances that any player drops too much, which is just a bad feeling for anyone playing a kids game. A lot of the time when you have a new leader, everyone will just drop a floor or two to make room, and that's easier to swallow than a higher chance of someone taking a big hit.

Plus, since there can only ever be one person per floor, the attacker will always leave a hole by moving up to you. You can only climb up to 3 floors (this is also dependent on a die roll that goes from -1 to +3), so on a taller structure the worst it can ever get for a player is that they swap positions with the player that just took their spot, so your loss is limited and you're always left with a chance to get back to where you were on your next turn (you can't fall so far that you can't get back to where you were, though there may be new floors by then so you won't always reach the top).

You could also play this where you just always bump the other player down if you're the one moving. Attacker always wins, basically. This would make playing it easier, as you wouldn't have to stop to make die rolls or make sure the right person is rolling the right color. But, given that this is a family game, I think the randomness is probably a good idea, and the weighted randomness is a good middle ground.

It's a clever way to make the roll-offs simple and quick and get back to the game. By far the only issue I have with it is that you have to swap dice around, especially if you're dropping floors and going from defender to attacker. But I think that's a fair trade for a system that has no questions to ask after the single roll happens, and that actually still weights winning towards the player who "should" win.