Home > board game design, Horsefeathers, playtesting, print-and-play > a bit of the math behind Horsefeathers

a bit of the math behind Horsefeathers

I’ve been playtesting Horsefeathers quite a bit lately, and it’s just about finished. The idea of including special tokens that you get only when you challenge someone successfully was that little extra something the game needed. Now I just need to write up the update to the rules, upload the file, and apply to get the game in the BGG database. :-)

I thought it would be interesting, however, to share some of the math behind the game. Even a game as simple as Horsefeathers has quite a bit of math behind it, and a spreadsheet can really help keep track of it all. Spreadsheets help designers tweak all the different values in a game, and they also help designers understand what parameters are most important.

So here’s a snapshot of the spreadsheet for Horsefeathers. It’s pretty simple, but amazingly useful.

Most of it is pretty self-explanatory, but the question I was ultimately trying to answer was this:  how should I handle all the various values to make it so that the final payout due to the special tokens was worth roughly the same amount (or a little more) than a typical payout for a single round?  I wanted it, ideally, in the range of 1-2. I’m pretty happy with 0.96 to 1.87.  :-)

There are some fudgy numbers, here, but we are, after all, dealing with probabilities and statistics.  No two games will ever be the same, and there will always be that occasional game that defies the odds and comes in at 4 sigma. Outliers, however, just give a game like this texture.

The top two rows are just the number of players in the game and the number that get eliminated in a given round (always one less than the total, since rounds are always played until just one player remains).

The initial ante (10 in the sheet above) is the amount that every player must pay at the beginning of the round.  By changing just that one cell, I can change it from 10 to 5 to 20 to 0.  And each of those numbers will produce a vastly different gaming experience.

Then the domes (read: special tokens) that are taken per round.  This should be equal to the number of players eliminated divided by two.  Why two?  Well, any given challenge might prove true or false, but on average people will be right about half the time.   One player is eliminated either way, however, so a token is likely to be taken from the stack every other time.

The number of rounds is entered.  I chose 6.  This causes the game to last roughly thirty to forty-five minutes, which seemed reasonable for a game of this depth.  One neat thing about the game is that, because the number of rounds is the same for any number of players, the length of the game won’t change much.  And in that sense, it’s scale-invariant.

The number of domes (special tokens) needed to produce a game of six rounds:  since the game is over when the last token is taken from the stack, varying the number of tokens varies the (likely) number of rounds. It’s a fairly accurate predictor, though of course not infallible.

Does everyone pay to play?  This refers to a design option I considered for a while wherein every player would need to pay an additional chip anytime someone rolled.  It felt fussy, but I wanted to throw it in here just to see what it would do.  I had thought that it would cause the game to scale more reliably from 3-8 players, but it didn’t really help much.  Right now it’s off, but entering “1” will turn it on (and change all the formulas appearing later in the sheet).

How much do players pay to play?  I toyed with 0, 5, and 10, and I like the results of 5 best.

Average number of dice rolled in a round.  Here’s one of my educated guesses.  A round begins with a number of dice already on the table:  roll until a pair shows up in the center of the board, or until there are 5 dice.  In my experience, this usually means 3 or 4 dice to begin with.  But when does a round end?  Again, in my experience a round typically ends about the time there are 10 dice on the table — sometimes more, sometimes less.  So I chose 6.5 for the number here.  But there are other factors:  when a bluff is unsuccessful, the die is not added to the center (though the player is still eliminated and their “pay to play” chip is still added to the pot).  Again, it’s not perfect.

Size of pot — the size of a typical pot won by one of the players at the end of a round.  This is one of the two numbers I was wanting to compare.  It’s calculated by taking the size of the initial ante times the number of players plus the size of the subsequent ante (the amount players must “pay to play”) times the average number of rolls.

The next three rows look at the starting amount of money, the average amount a (losing) player will lose each round, and the number of rounds they’ll be able to play without winning a pot.  I wanted it to be greater than 6 (the chosen number of rounds), since I wanted players to have at least some money at the end of the game for making payments based on special token disparity.

After the first red line, I’m trying to get at the other of the two numbers I want to compare:  the size of the payouts based on token disparity.  If a player has 0 tokens, they pay everyone at the table.  How much they pay depends only on the number of tokens in play, so it is equal to the number of tokens originally in the stack.  Of course this says nothing about who gets paid, just how much is paid out.

The multiplier:  how much is paid per token?  I toyed with 5 and 10, but got better results with 10.  I thought for a while that I would have to use 5 for some numbers of players and 10 for other numbers of players, but thankfully that wasn’t necessary.  I hate fussy rules like that, and in a game like this it would be anathema.

Chips lost due to the special tokens and likely max chips received.  Chips lost is simple:  the number lost due to token disparity times the value multiplier.  Likely max chips received is another fudge factor: I’m saying here that the likely max won is comparable to the likely max lost.  And, on the face of it, this is the biggest fudge of all — there’s no real reason to assume this is the case.  But I’m guessing it’ll be close.  For one thing, when you run numbers of a typical spread (0 tokens, 2 tokens, 4 tokens, 6 tokens) the player with 0 loses 2+4+6=12 and the player with 6 gains 6+4+2=12.  There’s a bit of symmetry.

Also, I played with the numbers quite a bit and found that, after calculating the amount received by the player with the most tokens for likely scenarios, and then averaging the results, I came startlingly close to this simpler number.  I chose to go with it, though of course there will always be outliers.

So finally I calculated a ratio:  the number of chips a player is likely to receive when having the most tokens versus the number of chips a player is likely to receive when winning the pot.  And really, it all came out pretty well.

Of course none of this makes the slightest bit of difference if the game doesn’t have legs, but thankfully this one is quite a bit of fun to play.  And when playing the game you wouldn’t know there was any of this math in there at all.  When playing the game, you’re just trying to figure out if Joe is bluffing or telling the truth — did he really roll a six, or is he a low-down lying skunk?

I haven’t played the game for money, yet, but it would make a positively wicked betting game.  :-)

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