Don't be put off by that 1970s advert; it's a much better game than the advert would suggest. I'd forgotten how much mathematics, statistics, and risk analysis the game involve.

The mathematics bit is simple. Adding the scores of the dice - and learning the shortcuts. 1, 2, 4, 5, 6. "That's an easy sum," I said, as CGF Sprog 3 added each score in turn. "What's 1 plus 5?" I asked? Followed by "and 2 plus 4?" Finally "and what's 3 times 6?"

There's also a bit of profit-and-loss accounting. You score an extra 35 points if you get at least 63 in the top section. 63 is three each of 1, 2, 3, 4, 5 and 6. So if you get only one 1, you can balance it out by getting four 2s - or indeed four of any other number. Getting only one 6, however, is much more difficult to mitigate.

Risk analysis? Which dice should you keep and which should you throw again? What are the chances of throwing the numbers that you need? After one throw you get two 2s. You throw the remaining three dice again, and get two 3s and a 4. Should you throw the same three dice again, hoping to get some 2s this time? Or change your mind and throw the 2s and 4 to try to get 3s? I'm pretty certain there is no difference in the odds of these options. Or would you do better to just throw the 4 and hope to get a 2 or a 3?

How easy is it to get a full house (three of one number; two of another)? Empirically this is quite common. So common it is generally not worth aiming for - you will get several full houses during a game without even trying.

How about this scenario? After two throws you have two options: throw one die for a 4, or throw two dice for a 3. The first option would give you 40 points; the second, 30. So here we have an example of risk and reward - the more risky option gives the better result. Or, put another way, there is less to lose from the less risky option. I suppose the option you take (and your justification for taking it) depends on how risk averse you are.

OK, so what are the chances of each option? Well, the first option is easy. That's 1 out of 6. Logic tells me that the second option must be 2 out of 6 - 1 out of 6 for each die. Now, although I claim to know nothing about statistics I do know that isn't correct. It's 11 out of 36. Throwing two dice (or the same die twice - it doesn't matter) gives 36 possible outcomes. Only 11 of these contain a 3 (1-3, 2-3, 3-3, 4-3, 5-3, 6-3, 3-1, 3-2, 4-3, 5-3, 6-3).

Here's a trickier problem. Are you more likely to get two (or more) 1's from four dice, or one (or more) 2 from two dice? Or is there no difference (it is, after all, twice as many successful results from twice as many dice)? I had no idea how to work this out, so I wrote a program to work out all possible permutations, and count the ones that mattered. It is 171 out of 1296, or 19 out of 144 - which is considerably less likely than the 11 out of 36 for the other option.