I did some analysis of card draws in Leaving Earth to try to work
out the odds in a particular situation.
For most of the technological advances one can acquire during the
game, one draws a small deck of three random outcome cards without
looking at them. Of the 90-card supply, 60 are successes, while the
other 30 are failures (15 each minor and major).
Each time one uses that technology (firing a rocket, docking or
separating spacecraft, etc.) one draws a card from that deck, thus
generating a success or failure outcome. At that point one can either
pay to remove the card (more for a success than for a failure, because
one learns more easily from failures) or shuffle it back into to the
deck.
Opinions differ as to whether one should pay the higher cost to remove
success cards. If one draws (and replaces) several cards in a row all
of which are successes, maybe there are one or even two failure cards
that just haven't come up yet, which may show up to wreck a later
mission… or maybe they're all successes and resources spent on
removing them would be wasted. What are the odds that they are all
successes?
This seems like a case for Bayesian probability analysis.
There are four possible combinations of cards, from 0 to 3 failures.
Let's assume that the probability of a single random card being a
success is always ⅔. (Which it won't be exactly, because the supply of
outcome cards isn't infinitely large – for example if when I start the
game the first two cards I draw are successes, the chance that the
third one will be too is 55/88 or about 0.659. But I'll ignore this
complication, in large part because there's no way to know the state
of the supply of outcome cards.)
Then the possible combinations are:
Combination |
P(combination) |
P(N / combination) |
SSS |
⅔^3 = 8/27 |
1 |
SSF |
⅔^2 × ⅓ × 3 = 12/27 |
⅔^N |
SFF |
⅔^1 × ⅓^2 × 3 = 6/27 |
⅓^N |
FFF |
⅓^3 = 1/27 |
0 |
where P(N | combination) is the probability of getting N successes in
a row if that's the combination we have.
Bayes' theorem tells us that
P(SSS | N) = P(N | SSS) × P(SSS) ÷ P(N)
So we know P(SSS) (8/27); we know P(N | SSS) (1); and we can work out
P(N), because it's the sum of P(combination) × P(N | combination) for
each combination, which will be
8/27 + 12/27 × ⅔^N + 6/27 × ⅓^N
(I'm ignoring the case where N=0 because I don't care about it.)
In other words, simplifying,
P(SSS | N) = 4 ÷ (4 + 6 × ⅔^N + 3 × ⅓^N)
which, as a quick sanity-check, is asymptotic to 1 (the more success
cards you draw, the greater the chance the configuration is SSS, but
you can never be completely certain.)
N |
P(N) |
P(SSS / N) |
1 |
0.666666666666667 |
0.444444444444444 |
2 |
0.518518518518518 |
0.571428571428571 |
3 |
0.436213991769547 |
0.679245283018868 |
4 |
0.386831275720165 |
0.765957446808511 |
5 |
0.355738454503887 |
0.832904884318766 |
6 |
0.335619570187471 |
0.88283378746594 |
7 |
0.322410201696896 |
0.91900409706902 |
8 |
0.313671696387746 |
0.944606413994169 |
9 |
0.307868606298724 |
0.962411529575709 |
10 |
0.304007406278402 |
0.97463512459613 |
11 |
0.301435781833418 |
0.982949981897101 |
12 |
0.299722201837394 |
0.988569730503459 |
13 |
0.298580093940256 |
0.992351138973059 |
14 |
0.29781878159779 |
0.994887880162139 |
15 |
0.297311271010243 |
0.996586154603227 |
16 |
0.296972940943245 |
0.997721527608545 |
17 |
0.296747391006812 |
0.9984798696663 |
18 |
0.296597025529712 |
0.998986067938885 |
19 |
0.296496782260709 |
0.999323817402388 |
20 |
0.296429953542172 |
0.9995491101885 |
If you draw 7 successes in a row, there's a ~92% chance that all the
cards are successes. 9, ~96%. 12, ~99%.
(Thanks to Lieven and Gus for double-checking this; it's over 30 years
since I did Bayesian analysis.)
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