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.)

Comments on this post are now closed. If you have particular grounds for adding a late comment, comment on a more recent post quoting the URL of this one.

Tags 1920s 1930s 1940s 1950s 1960s 1970s 1980s 1990s 2000s 2010s 3d printing action advent of code aeronautics aikakirja anecdote animation anime army astronomy audio audio tech aviation base commerce battletech beer boardgaming book of the week bookmonth chain of command children chris chronicle church of no redeeming virtues cold war comedy computing contemporary cornish smuggler cosmic encounter coup covid-19 crime cthulhu eternal cycling dead of winter doctor who documentary drama driving drone ecchi economics en garde espionage essen 2015 essen 2016 essen 2017 essen 2018 essen 2019 essen 2022 essen 2023 existential risk falklands war fandom fanfic fantasy feminism film firefly first world war flash point flight simulation food garmin drive gazebo genesys geocaching geodata gin gkp gurps gurps 101 gus harpoon historical history horror hugo 2014 hugo 2015 hugo 2016 hugo 2017 hugo 2018 hugo 2019 hugo 2020 hugo 2022 hugo-nebula reread in brief avoid instrumented life javascript julian simpson julie enfield kickstarter kotlin learn to play leaving earth linux liquor lovecraftiana lua mecha men with beards mpd museum music mystery naval noir non-fiction one for the brow opera parody paul temple perl perl weekly challenge photography podcast politics postscript powers prediction privacy project woolsack pyracantha python quantum rail raku ranting raspberry pi reading reading boardgames social real life restaurant reviews romance rpg a day rpgs ruby rust scala science fiction scythe second world war security shipwreck simutrans smartphone south atlantic war squaddies stationery steampunk stuarts suburbia superheroes suspense television the resistance the weekly challenge thirsty meeples thriller tin soldier torg toys trailers travel type 26 type 31 type 45 vietnam war war wargaming weather wives and sweethearts writing about writing x-wing young adult
Special All book reviews, All film reviews
Produced by aikakirja v0.1