RogerBW's Blog

Leaving Earth outcome odds 27 June 2021

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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