Bribe probabilities
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I’m in the process of updating Dan Tascau’s FLStat again. One of the stats it provides is the probability of obtaining a bribe. Anyone have any idea how he determined it?
As an example, look at Corsairs - Freeport-5 has a 19.39% probability. Looking at mbases.ini, there are seven NPCs, three of which offer Corsair bribes, as does the bartender. The bar has 5 characters, so that’s four NPCs plus the bartender. The NPCs have a choice of two bribes; the bartender four. There’s a three out of seven chance of choosing a Corsair briber, with a one out of two chance of getting the Corsair bribe. There’s a four out of seven chance of being in the bar. That would make the probability:
3/7 * 1/2 * 4/7 + 1/4 = 37.24%.
If we look at the other extreme, Planet New London is only 0.56%. That has 15 NPCs, 6 of which can be in the bar. Only one offers the Corsairs, along with four other bribes. That makes it:
1/15 * 1/5 * 6/15 = 0.53%.
Now, I’m pretty sure those calculations are wrong, but I’m also pretty sure the + 1/4 is right, which means Dan got it wrong, too.
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To tired to type atm.
Hope it helps.
Edit:
Did a bit of research.
For example Luxury liner Shetland br_06_01.
There are 4 people offering bribes. The bartender, 1 policeman br_p_grp an 2 bountyhunters gd_bh_grp.
The bartender offers 3 bribes and 5 rumors. He’s always there so the chance offering a specific bribe or rumor is 12.5%.
The Bretonian Police has a faction weight of 6%. The policeman is offering 4 bribes and 3 rumors.
6%/(4+3)=0.85714%
One of the Bountyhunters offers 4 bribes and 6 rumors, the other one 3 bribes and the same rumors. They have a faction weight of 12% for two people means 6% for one.
So for the first bountyhunter 6%/(4+6)=0.6%
and the second 6%/(3+6)=0.66667%Looking at mBases.ini, the bartender, the policeman and the first bountyhunter offer bribes for the Bretonian Police br_p_grp 12.5%+0.85714%+0.6%=13.95714 or 13.96 in FLStat.
For Spa and Cruises co_os and for the Bountyhunters gd_bh everyone offers a bribe 12.5%+0.85714%+0.6%+0.66667%=14,62381% or 14,62% in FLStat
For Planetform Inc., the only one not offering a bribe is the bartender so 0.85714%+0.6%+0.66667%=2,12384% or 2.12% in FLStat
Dan’s Calculation for 1 NPC offering a specific bribe would be:
Faction weight / number of NPC of that faction / (number of bribes offerd by NPC + number of Rumors offerd by NPC)
Writing wit a Smartphone is crap!!!
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Yep, that looks like it. Amazing! Pity it doesn’t work as a probability.
I neglected to consider rumors (which would also include the know ones, as well as misn) as a possibility. However, I don’t think it matters - you need to have a reputation of 0.2 (for lawful) or 0.4 (for unlawful) to see them, in which case you’re probably not that interested in a bribe.
I didn’t account for weighting (thought it only applied to the job board, but that is not the case). However, again, I don’t think it matters, as it’s mostly evenly distributed, anyway.
So, let’s have another look at Freeport 5. Seven NPCs in total, four spots at the bar, 7C4 = 35 combinations. Three NPCs offer a Corsairs bribe. 7 - 3 = 4, 4C4 = 1 combination without a briber. In this case all the bribers have a 1/2 chance, so regardless of the bribers present, it’s always 1/2. The bartender is always present, making a constant 1/4. That makes the probability 1/4 + 34/35 * 1/2 = 73.57%.
Planet New London has one briber for the Corsairs, who offers five bribes in all. There are fifteen NPCs, six spots. Working the combinations, it’s 15C6 total, less 14C6 without the briber, making (15C6 - 14C6)/15C6 or 1 - 14C6/15C6. The combination expands to
14131211109 = 9
151413121110 151 = 15/15, 15 - 9 = 6, so it’s 6/15. Makes sense.
The probability is then 6/15 * 1/5 = 8%.Leon Base is even more different. FLStat gives this as 16.67%. Thanks to MeisterLampe, that was worked out as the bartender being the only one to offer the bribe, so one bribe, five rumors (and two rumorknowdbs) makes 1/6 = 16.67%. As I’m choosing to disregard the rumors, it is actually 100%.
Now, something like Luxury Liner Shetland is where it gets complicated. Let’s consider a bribe for the Bounty Hunters. The bartender is fixed, so that’s 1/3. There’s three NPC bribers, twelve NPCs in total, three spots. Two NPCs are 1/4, the other is 1/3; that’s what I don’t know how to handle. Maybe it’s simply (1+1+1)/(4+4+3) = 3/11? Then we have 1 - 9C3/12C3.
987 = 347 = 21
121110 4115 551/3 + 34/55 * 3/11 = 50.19%.
The general rule would be: n NPCs, k spots at the bar, b bribers with f total bribes and bartender with t bribes:
(1 - (n-b) (n-b-1) … (n-b-k+1) ) * b/f + 1/t
n (n-1) … (n-k+1) -
In this case all the bribers have a 1/2 chance, so regardless of the bribers present, it’s always 1/2.
No, that’s not right. I was thinking like 1/2, (1+1)/(2+2) = 2/4 and (1+1+1)/(2+2+2) = 3/6, but it’s actually 1/2 (50%) if there’s one person, 3/4 (75%) if there’s two and 7/8 (87.5%) if there’s three (i.e. there’s only one possibility out of all permutations where no one offers the bribe). … Actually, that’s not right either, because the bartender should be included. So it becomes: no briber = 25% (1/4 from the bartender alone); one briber = 62.5% (5/8 = 1 - 13/(24)); two bribers = 81.25% (13/16 = 1 - 113/(224)); and three bribers = 90.63% (29/32 = 1 - 1113/(2224)). Using the combinations of each group gives us:
1 * 1 + 5 * 12 + 13 * 18 + 29 * 4 = 74.29%
4 35 8 35 16 35 32 35I’ve put it on Mathematics StackExchange. But now that I wrote it out above, I’m pretty confident that’s right.
Planet New London and Leon Base are right.
Luxury Liner Shetland is:
1 * 84 + 2 * 1 * 36 + 5 * 36 + 5 * 9 + 2 * 2 * 9 + 3 * 1 = 46.53%
3 220 2 220 9 220 8 220 3 220 4 220 -
I neglected to consider rumors (which would also include the know ones, as well as misn) as a possibility. However, I don’t think it matters - you need to have a reputation of 0.2 (for lawful) or 0.4 (for unlawful) to see them, in which case you’re probably not that interested in a bribe.
Another wrong assumption - you need to have that reputation with the faction offering the bribe, not the faction being bribed (duh!). However, it seems the bartender will always offer bribes; NPCs won’t talk to you at all if you’re below 0, but they’ll offer bribes & jobs if you’re below the above.
I didn’t account for weighting (thought it only applied to the job board, but that is not the case). However, again, I don’t think it matters, as it’s mostly evenly distributed, anyway.
After looking at some other bases, I believe weighting is a factor. Unfortunately, I don’t know how to take that into account.
And just to be really thorough, Dan also didn’t take into account the Liberty capitals, where you can also get some bribes at each Commodity Dealer.
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Well, in the end I just included misn and know and let it be (it should be relatively correct, even if it’s not mathematically correct). I think what it should really do is list each individual offering the bribe (since faction matters), with the individual’s probability range (faction weight / NPCs in faction / (misn + bribes + rumors + knows) to / (misn + bribes), which still isn’t quite right as it doesn’t take the number of spots available into account, but character_density itself isn’t reliable, either), but that’s more than I can do via patching.
