The holiday season means spending time with family. Spending time with my family typically means drinking, and of course, gambling. We have developed a family tradition of playing LCR. LCR is a game of chance involving dice that can accommodate an unlimited number of players. The six sides of an LCR die consist of an L, a C, an R, and three blanks. You roll a maximum of three dice, and the roll of each dictates what you do with your tokens. A blank means you get to keep it, an L means you give one token to the player to your left, an R means right, and C means center i.e., you throw a token into the pot and it is gone from play forever. You then pass the dice to your right, and so on. The game ends when only one player owns any tokens. We typically play with dollars in lieu of tokens, and the winner brings home a cool $3*n. It is probably worth it to note that, although you may lose all of your tokens, you are never ‘out’ of the game because the player to your left or right may be forced to pass a token to you. Therefore, the only way a game ends is when every token on the table is either in the center or owned by only one person.
As you can imagine, this game is low effort but full of suspense. Although fun, I hate this game because it is 100% chance; we might as well just put all of our names in a hat and draw a ‘winner’… but where’s the fun in that? After all, the fun in going to Vegas is gradually losing your money over hours of entertainment rather than just flushing it down the toilet.
Naturally, despite this being a game of chance, I was curious if there was any way to gain an advantage. My mind immediately jumped to the player sequence. My thought was that a game would almost always end with a ‘C’, so having fewer rolls (and thus, fewer opportunities to roll a C) would be optimal. I casually suggested that the person to my right start the game by rolling first, since the dice are always passed to the right after a turn. But I was curious: just how much of an advantage is the player sequence to the game?
I wrote some Matlab code that would simulate 100,000 LCR games for a minimum of 2 participants and a maximum of 16 participants. The code would note the number of rolls to finish each game, the winner of each game, and the deciding roll of each game.
The first thing we can see is the average number of rolls that it takes for the game to finish, given the number of participants (Figure 1). For example, if you have 16 participants, you can expect the game to last 149 ± 22 rolls. The relation appears to be linear, good for planning how many beers you’ll need to stock up on before the game?
Figure 1. Number of rolls (± SD) to complete a game of LCR vs. number of participants
As the number of participants increases, the likelihood of the game ending roll being a C increases, up to 91.4% for a game of 16 individuals (Figure 2). The chances of a game ending in an ‘R’ is always about twice that of ending in an ‘L.’ The only way to end the game on a non-‘C’ is to roll either an L or an R when the person immediately to your left or right respectively is the only person left with any tokens. Any individual to your right is going to have had fewer rolls than anyone to your left (assuming the order is counter clock wise, but of course, these results can be generalized because the direction is arbitrary). Therefore, the odds are better that a person rolling after you will have a token than a person rolling behind you, because the person after you will have had fewer rolls. Thus, the odds of ending the game on an R is more likely than ending the game on an L.
Figure 2. Percentage likelihood of game ending in an L (Red), C (Black), R (Blue) vs. number of participants
Then, I summed the number of games (out of 100,000) that each player won, counting sequentially from the first person. As we can see in a game of only two players, rolling first only results in a victory 38.3% of the time (Table 1). In games of 2, 3, and 4 players, the first to act has the smallest chance of winning compared to the others. As the number of participants increases, the most disadvantaged player shifts to the second or third to act.
Table 1. Win percentage of 100,000 games for each player, based on total number of participants. Green indicates highest win probability, Red lowest.
We see a similar trend, but in the opposite direction, for the last player. In a game of two players, the odds of winning as the second (i.e., last) player to act is about 2:1 (67.7%). The last roller has the highest chance of winning in games of 2-5 players. If the game consists of more than 5 players, the highest odds of winning shift to either second-to-last or third-to-last, although this advantage becomes quite small. For example, in a game of 16, rolling 14th would result in a highest odds of winning (6.68%) compared to the lowest odds of winning by rolling 4th (5.78%) – a fairly small differential.
In conclusion, there does appear to be an order effect on the outcome of the game. There appears to be a slightly higher chance of winning by being in the second-half of the rolling order compared to the first half. However, depending on the size of your family, I am guessing you won’t ever play this game enough to capitalize on these extremely small deltas in win percentage.