Q1. You’re playing Stone Age. You place 2 workers on gold, 2 on brick, and 1 on food. What is the expected amount of each resource you’ll collect?

A1. Let G, B, and F be, respectively, the number of resources we get. Let X, Y, and Z be the rolls.

- P(G=0) = P(X in {2,3,4,5}) = 10/36
- P(G=1) = P(X in {6,7,8,9,10,11}) = 25/36
- P(G=2) = P(X=12) = 1/36
- (AN: Sorry, can’t use summation notation in wordpress) E[G] = 0 P(G=0) + 1 P(G=1) + 2 P(G = 2) = 0(10/36) + 1(25/36) + 2(1/36) = 27/36 = 0.75 gold

Similarly, we find E[B] = 49/36 = 1.36 brick and E[F] = 9/6 = 1.5 food. This a simple exercise in **expectation **for discrete random variables.

Q2. You have the same situation as before, but now you have a single tool. Assume you roll for resources in order: gold, brick, food. Also assume you use the tool if it will allow you to gain an additional resource but otherwise not.

A2. Let event TG be true if we have the tool before rolling for gold, and similarly for TB and TF. (One of the problems with probability notation is using capital letter for both events and random variables. Have to think of an alternative. Maybe Roman and Greek letters?) Furthermore, let TGc be the complement of event TG, and similarly for the others. We have already calculated:

- E[G|TGc] = 3/4
- E[B|TBc] = 49/36
- E[F|TFc] = 3/2

We should recalculate the expectations given the tool (the calculations are very similar to the above, so they’re omitted):

- E[G|TG] = 33/36 = 0.92
- E[B|TB] = 59/36 = 1.63
- E[F|TF] = 2

Also, we should calculate the probability that we retain the tool after rolling for each resource:

- P(TB) = P(X not in {5,11}) = 30/36
- P(TF) = P(TF|TB) P(TB) + P(TF|TBc) P(TBc) = P(Y not in {3,7,11}) P(TB) + 0 P(TBc) = (26/36)(30/36) + 0 = 65/108

That last line comes from the **law of total probability** in **conditional probability** since {TB,TBc} is a partition of the event space. Finally, we can use the analog of the law of total probability for **condition expectation**:

- E[B] = E[B|TB] P(TB) + E[B|TBc] P(TBc) = (59/36)(5/6) + (49/36)(1/6) = 344/216 = 1.59 brick
- E[F] = E[F|TF] P(TF) + E[F|TFc] P(TFc) = (2)(65/108) + (3/2)(43/108) = 389/216 = 1.80 food

In summary, by adding 1 tool, a player in this situation can expect 0.17 extra gold, 0.23 brick, 0.30 food, which I estimate to be worth 2.6 pips that turn, and you get to use the tool every turn. This is why tools are a good investment.

Q3. Instead of having 1 tool, say you now have 5 and then you roll a 7 for gold. Do you use all 5 tools to get an extra gold, or do you save the tools for the other rolls? Because it possible to get as many as 8 pips (1-2 tools for one brick, 3-4 tools for two food). This is a trickier problem which I’ll save for Part 2.