Bene Factum

2012/05/01

Fun with Probability – conclusions

Filed under: Gaming Blog — Tags: , , , — AlexWeldon @ 4:07 pm

I’ve been busy the past couple of weeks, and failed to summarize the general conclusions I think can be drawn from my analysis of the simple dice game I proposed. But I’d like to do so before I forget about it entirely.

As a refresher, the basic premise of the game is that players, in turn, get to pick what number they’re going to try to roll on a die. If only one player gets their number, they win, but if two or more do, the one who picked the harder number to roll is the winner.

The conclusions that I reached from my analysis, and which I think have general application to most games in which players get to choose their level of risk, are as follows:

  1. It’s never correct in this game for anyone but the last player to choose a very safe (better than 50% shot) number. The reason is that someone choosing afterwards can always take just a slightly bigger gamble (but still better than 50%) and win most of the time. What this means in general for games of this sort is that you don’t want to play it too safe until you know what your opponents are doing – and even then, only play it safe if you feel they’re all likely to fail. Games are not for the risk-averse!
  2. In a two-player game, you always want to either push just a little harder than your opponent, or else play it as safe as possible if you think he’ll fail. This kind of brinksmanship is intuitive, but the key strategy in most games of this sort would be in determining exactly where that brink lies. In the simple case of a game where the bigger gambler wins if both succeed and you keep going if both fail, you’re shooting for around a 40% chance of success.
  3. When playing with more than two players, you don’t want to match anyone else’s strategy too closely if others still have a chance to adjust theirs. This is the least intuitive results, as gamers often fall into “groupthink” patterns, wherein everyone plays a similar strategy. But it makes sense when you think about it; if two people are doing the same thing, the third player is effectively playing against a single opponent (albeit one who gets two shots at succeeding), and it’s thus easier for him to pick a winning counter-strategy. If the opponents vary their strategies, it’s hard for the remaining player to find a single counter-strategy that works against both.
  4. When your opponent gets a chance to react to your strategy, your best move is generally the one which puts him in a position where all choices are equally attractive. When there’s little advantage to choosing one strategy over another, you minimize the advantage of having that choice.

These are interesting conclusions, and intuitively correct once they’re pointed out. The third – about adopting different strategies than your opponents rather than imitating – is the most interesting of them, and will probably merit additional investigation another day.

Related: Fun with Probability – Part I

Related: Fun with Probability – Part II

Related: Fun with Probability – Part III

2012/04/19

Fun with probability – Part III

Filed under: Gaming Blog — Tags: , , , , — AlexWeldon @ 9:28 pm

Over last couple of days, I’ve been working on analyzing a simplistic game that I came up with to talk about risk-reward decisions in multiplayer games. What I thought would lend itself to easy analysis in order to prove a point, however, turned out to be a pretty complex and interesting math problem. Yesterday, I presented my findings for the two-player case. As you’d expect, it gets a lot more complicated when you add a third player; so much so that I didn’t even bother trying to work anything out for a four-player situation.

The first interesting thing to notice is an elaboration on what I said previously, about larger die sizes (and thus a larger range of choices) favoring the player who gets to pick last. When we think about multiplayer games, we can see that the actual concern has to do with the number of choices relative to the number of players; the extreme case would be that in which we have as many players as there are sides on the die. In that case, we know that all numbers will be chosen in the end. Thus, the first player has just as much information as the last, and can therefore choose the best number for himself, meaning that the last player is at the greatest disadvantage.

In the three-player case, it (perhaps surprisingly) turns out that the break-even point is once again that of the standard six-sided die. The first player should choose 4, the second should choose 5 (just as in the two-player game) and the third is now left with no better choice than to pick 1 and hope the other two fail. Thus, the second player has a 1/3 chance of winning outright, the first player will win 1/2 of the 2/3 of the remaining times, thus 1/3 as well… leaving 1/3 for the third player.

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