Note that the original version of this article contains an erroneous calculation. This is the version as of 28 februari, 19:16.

As we have seen, different strategies in the game of thirties give us different expected results, even if the game is largely determined by luck. This raises the question: to what extent is the game a game of chance?

Obviously, to answer this question one needs to have a rigurous definition of what exactly is a game of chance. Some research in this area, particularly to be able to advise courts in The Netherlands, has been done by Dr. Ben van der Genugten. The method is explained in this article in the Dutch state newspaper:

The idea is that one defines the learning effect LE in a game to be the expected value of the optimal strategy minus the expected value of a strategy that may be employed by a “beginning player”, and the random effect RE to be the expected value of a player with advance knowledge (i.e., he knows what he is going to throw, should he throw again) minis the expected value of a player employing the optimal strategy. Then, the relative skill of a game is defined S=LE/(LE+RE).

On the grounds of Dutch jurisprudence, the boundary value has been set at 0.20; for example, roulette has value 0, black jack has value 0.049, and some simple forms of poker (the full game is too difficult to analyze) have values around 0.40.

To be able to calculate the random effect, I wrote a small simulator that, for random dice throws, calculates the optimal strategy. This suggests that with prior knowledge, the game has an expected value of about 32.73. As we know, the expected value given an optimal strategy is 30.1520.

So what, in this case, is the relative skill? For this, we need to define what a beginning player would do. Experience suggests that the strategy “put all fives and sixes away, and the last dice if it is 4 or higher” may be reasonable. In this case, one gets expected value 29.7232. This would suggest that the relative skill is 0.14, so the game just about a game of chance. This would seem to be reasonable.

There are some pitfalls, though. First of all: is this choice of strategy reasonable? As we know, the heuristic strategy is pretty easy to remember, and has an expected value which is just about equal to the expected value of the optimal strategy. In this case, the relative skill is only 0.0024.

Also, it is a bit difficult to precisely specify a random execution of the game, since the number of different throws may vary. In an earlier version of the simulation, I actually did this wrong. Calculating exactly (i.e., not simulating) the complete expected value with prior knowledge would probably be a large computation..

While playing a bit of the game of 30s Tuesday, I learned that the second stage of the game is actually a bit different from what I previously described. In fact, it works as follows: say one has thrown 32. Then one starts throwing again, putting away 2s while new ones are thrown. But now, the previous score (in this case, 2*6 thrown in the second stage+2) gets doubled, and one starts throwing 3s. This can go on (doubling the score each time) until one gets at 6s; in this case, if 6*6s are thrown, one just continues throwing 6s — presumably, this is already bad enough for the opponent.

Unfortunately, this makes the calculation of the expected value for the second stage game more difficult for two reasons. First, the expected value is no longer linearly dependent on the value of the dice: the lower the score on the dice, the more impact going from one dice value to a higher one is. This is not a real problem since it only increases the number of equations by, say, 6. The real problem is the doubling, which makes the score dependent in a non-linear fashion on the score obtained earlier.

I’ll spare you the analysis for now, but here’s the Maxima code I used to define P(x,n), which is the expected value when throwing xs when n points have already been scored. I do this by ignoring terms for very large n. This seems justifyable given that, letting n tend to infinity, the calculated values converge.

d(x,n):=if(x=0) then 1 else if n=0 then (5/6)^x else
   sum((1/6)^i*(5/6)^(x-i)*binomial(x,i)*d(x-i,n-i),i,1,n);
d(6,0)+d(6,1)+d(6,2)+d(6,3)+d(6,4)+d(6,5)+d(6,6);
a:sum(d(6,i)*6*i,i,0,6);
b:d(6,6);
P6(x):=if x>10000 then 0 else x+a+b*(36+x+P6(72+2*x));
P(y,x):=if y=6 then P6(x) else
   x+sum(d(6,i)*y*i,i,0,6)+b*(6*y+x+P(y+1,12*y+2*x));

This gives the following results:

Note that these values are considerably higher than the 2.88d obtained using the simpler variation of the endgame.

Small update: a reversed version of the source code is available. It now has command-line support, so that recompilation if one wants to use another strategy is no longer necessary. A Linux binary (static, gzipped) is available as well. Also, here is a slightly revised version of the paper, now with some formalization of expected values as well.

Last Thursday, I finished tidiying up the code for the strategy calculator: it can now also compare strategies, and it is pretty easy to add new ones. I will post the code in a while. Also, I started a bit on a mathematical formalization of the problem and the concepts involved. A first effort on a formalization (due to be released on Cleopatra’s “Schierkamp”), can be read here

Earlier, I described the optimal strategy in the game of 30s if one wants to maximize the number of points. I already mentioned then that there may actually be other things one wants to optimize. Also, since the optimal strategy may be a bit hard to remember, I developed a heuristic strategy.

So, I implemented some simple strategies, and calculated the optimal strategies to optimize various aspects. I calculated the following strategies:

• Maximal – Maximize number of points
• Maximize >30 – Maximize chance of getting 30 points or more
• Minimize loss – Minimize loss (defined as number of points below 30)
• Heuristic – Heuristic strategy resembling the maximal strategy, as described here
• Leave >5 – Put aside 5 or 6, or otherwise the highest dice
• Leave >6 – Put aside 6, or otherwise the highest dice

The various parameters of the different strategies are summarized in the following table:

When interpreting these numbers, one has to keep in mind that the strategies only optimize for the given varible; thus for example, the “over 30s” strategy does not favor a score of 36 over a score of 31. So, if the first five stones are sixes, then the “over 30s” strategy will not discriminate between throwing the last dice or putting it aside. This somewhat troubles the results. Still, it can be noted that for any of the three parameters, there is a somewhat different strategy to reach its optimum.

Next time: a more detailed analysis of the differences between the various strategies. For now, see the source code of the analyser.

As noted before, the game of thirties normally consists of two phases. In phase 1, the player attempts, with six dice, to throw a score higher than 30. The optimal strategy for this part of the game was calculated here. If the score is above 30 case (say x is the number of points scored above 30), then the player enters the second phase. Here, the player throws all six dice, putting aside all dice with value x. As long as a new x is put away, the player can throw again. If all dice are put aside, the player can start all over again. Finally, the player next to the throwing player gets x times (number of dice put aside plus 1) points (which is a bad thing, because points correspond to drinking).

Obviously, there is no such thing as an optimal strategy here, but we can calculate the expected value of the game. We will calculate the expected number dice we put aside; the expected value of the game will clearly be this value times x.

Thus, define f(a,b,c) to be the chance that, given a dice we throw with, we get b times the dice with value x, times b+c:

Define t₆, …, t₁ to be the expected value of the game with 6, .., 1 dice respectively. Consider for example t_{2. If we throw the dice we want 0 times, the game is done. If we throw it 1 time, then the expected return is 1+t1, since we then do a 1-dice game. If we get the dice we want 2 times, then the expected return is 2+t6, since we repeat the 6-dice game. Hence we get the expected payoff f(2,1,t1)+f(2,2,t6). In general, we get the following set of equations:}

To solve this system of equations, I used the great open-source computer algebra system Maxima with the following commands:

f(a,b,c):=(1/6)^b*(5/6)^(a-b)*binomial(a,b)*(b+c);
solve([
  t6=f(6,1,t5)+f(6,2,t4)+f(6,3,t3)+f(6,4,t2)+f(6,5,t1)+f(6,6,t6),
  t5=f(5,1,t4)+f(5,2,t3)+f(5,3,t2)+f(5,4,t1)+f(5,5,t6),
  t4=f(4,1,t3)+f(4,2,t4)+f(4,3,t3)+f(4,4,t6),
  t3=f(4,1,t2)+f(3,2,t4)+f(3,3,t6),
  t2=f(4,1,t1)+f(2,2,t6),
  t1=f(1,1,t6)], [t1,t2,t3,t4,t5,t6]);