Odds of picking 2 specific retorts for Master of Magic

Question

In the old Dos Computer Game, Master of Magic, a wizard can gain access to extra 'retorts' when the successfully defeating defending units at Magic Nodes. If my wizard defeats the units at one of the nodes, he gains access to retorts, worth 2 picks.

(For anyone familiar with the game, my wizard has already got all the other retorts [Alchemy, Warlord, Mana Focusing, and Node Mastery]. He's also maxed out the spell books [11 Life + 2 Sorcery]).

These are the remaining possible retorts, grouped by 'number of picks'.

1 pick

1. Charismatic
2. Nature Mastery
3. Chaos Mastery
4. Sorcery Mastery
5. Sage Master
6. Archmage
7. Runemaster
8. Artificier
9. Conjuror

2 picks

10. Divine Power
11. Infernal Power
12. Channellor
13. Famous

So the wizard can get either:

• one of the 2 pick choices
• a combo of two 1 pick choices.

If I were to annotate a series of random battles as:

• 10,
• 1+9,
• 9+1,

for Divine Power, Charismatic plus Conjuror, Conjuror plus Charismatic.

The annotation of 1+9 as an outcome should be treated the same as a 9+1.

So, assuming:

• a wizard can't have the same retort twice,
• the wizard always wins the battle,
• the game is truly random

I'd like to know, how many attempts at re-running the battle at the 'Magic Node' should it take if to yield a result of 'Runemaster' as well as 'Artificier' (7+8)?

Can someone elaborate with a formula and explanation?

Many thanks.

FYI: This is where I get confused:

• Assuming you can role 1 or 2 dice.
• The first dice is a 13 sided dice.
• If the result is 10-13, you don't roll second dice.
• If it's 1-9 you roll again, and it would have to be a dice in the range of 1-8 - since first dice is one of the 9, leaving 8.
• Somewhere you effectively have to half out a result for the 1+9, 9+1 concept.

The reality:

I ran this several times before I started to note attempts.. Possibly 50 times.. I started annotating attempts to do this. It took me 102 further attempts. During that time Famous came out 13 times, Divine Power came out 10 times, Runemaster + Sagemaster came out 4 times. Probably some high standard deviations in there!

Answer 1

NOTE: I am not sure if my attempt is right or wrong but its been fun doing it

The whole picking scenario depends on what kind of probability distribution the game company chooses. You can have different answer depending on what kind of distribution you choose.

My attempt:

As you win the node, you are given with 13 choices.

Total number of combinations present 76 (4+72) where the 4 belongs to options that required 2 picks and 72 belongs to the combinations of 2 choices.

Assumption

1. all 72 options are equally likely to be occurred
2. when rolling for the first pick, all 13 choices are equally likely to be occurred implying option(10,11,12,13) have probability $\frac{1}{13}$

let x be the probability of one of the options from 72.

we have

72 x + 4*$\frac{1}{13}$= 1

x=$\frac{1}{13*8}$

Event (S) : getting (runemaster + artificer) or (artificer + runemaster)= p = 2x

Event (F): not getting (runemaster + artificer) or (artificer + runemaster)= 1-p

here is straightforward you can either get it on first turn or second turn or third turn and so on...

= S + FS + FFS + FFFS +......to infinity

Expected value = $\frac{1}{p}$ = $\frac{13*8}{2}$ = 52

On an average you need to attempt 52 to get your desired result