EPTAll percentages will be approximated to 2 significant digits.
The players will be designated the names a, b, c, d, e, f, g, h, i.
Most of the text containing math and numbers are hidden in spoilers to allow for easier reading, I recommend opening each spoiler, read the text but ignore the numbers unless you are particularly interested in them.
In the game of resistance using 9 players, 3 of which are spies and uses a 3 4 4 5(2) 5 setup, the following is true:
If teams are chosen completely randomly and spies always sabotage missions (which is not true, but if it was then this would be true):
The chances of picking a team where the mission is succesful on turn x is (paranthesis is chance to be sabotaged):
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1: (6!/(3!3!))/(9!/(3!6!)) = 5/21 approx 23% (16/21 77%)
2: (6!/(4!2!))/(9!/(4!5!)) = 5/42 approx 12% (37/42 88%)
3: (6!/(4!2!))/(9!/(4!5!)) = 5/42 approx 12% (37/42 88%)
4: ((6!/(4!2!))+(6!/(5!1!)))/(9!/(5!4!)) = 1/6 approx 17% (5/6 83%)
5: (6!/(5!1!))/(9!/(5!4!)) = 1/21 approx 5% (20/21 95%)
2: (6!/(4!2!))/(9!/(4!5!)) = 5/42 approx 12% (37/42 88%)
3: (6!/(4!2!))/(9!/(4!5!)) = 5/42 approx 12% (37/42 88%)
4: ((6!/(4!2!))+(6!/(5!1!)))/(9!/(5!4!)) = 1/6 approx 17% (5/6 83%)
5: (6!/(5!1!))/(9!/(5!4!)) = 1/21 approx 5% (20/21 95%)
There are only 20 possible scenarios, given that teams are chosen completely randomly and spies always sabotage, the chances for each scenario looks roughly as follows:
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s = success, f = fail
1: sss 125/37044 0.34%
2: ssfs 925/222264 0.42%
3: ssffs 4625/4667544 0.099%
4: ssfff 23125/1166886 1.2%
5: sfss 925/222264 0.42%
6: sfsfs 4625/4667544 0.099%
7: sfsff 23125/1166886 1.2%
8: sffss 6845/4667544 0.15%
9: sffsf 34225/1166886 2.9%
10: sfff 34225/222264 15%
11: fsss 50/27783 0.18%
12: fssfs 250/583443 0.043%
13: fssff 5000/583443 0.86%
14: fsfss 370/583443 0.063%
15: fsfsf 7400/583443 1.3%
16: fsff 1850/27783 6.7%
17: ffsss 370/583443 0.063%
18: ffssf 7400/583443 1.3%
19: ffsf 1850/27783 6.7%
20: fff 5476/9261 59%
1: sss 125/37044 0.34%
2: ssfs 925/222264 0.42%
3: ssffs 4625/4667544 0.099%
4: ssfff 23125/1166886 1.2%
5: sfss 925/222264 0.42%
6: sfsfs 4625/4667544 0.099%
7: sfsff 23125/1166886 1.2%
8: sffss 6845/4667544 0.15%
9: sffsf 34225/1166886 2.9%
10: sfff 34225/222264 15%
11: fsss 50/27783 0.18%
12: fssfs 250/583443 0.043%
13: fssff 5000/583443 0.86%
14: fsfss 370/583443 0.063%
15: fsfsf 7400/583443 1.3%
16: fsff 1850/27783 6.7%
17: ffsss 370/583443 0.063%
18: ffssf 7400/583443 1.3%
19: ffsf 1850/27783 6.7%
20: fff 5476/9261 59%
Now lets rearrange these to show which ten the resistance win and which ten the spies win.
+ Show Spoiler +
Resistance win
1: sss 125/37044 0.34%
2: ssfs 925/222264 0.42%
3: ssffs 4625/4667544 0.099%
5: sfss 925/222264 0.42%
6: sfsfs 4625/4667544 0.099%
8: sffss 6845/4667544 0.15%
11: fsss 50/27783 0.18%
12: fssfs 250/583443 0.043%
14: fsfss 370/583443 0.063%
17: ffsss 370/583443 0.063%
Total chance to win: 29005/1555848 1.9%
Spies win
4: ssfff 23125/1166886 1.2%
9: sffsf 34225/1166886 2.9%
10: sfff 34225/222264 15%
7: sfsff 23125/1166886 1.2%
13: fssff 5000/583443 0.86%
15: fsfsf 7400/583443 1.3%
16: fsff 1850/27783 6.7%
18: ffssf 7400/583443 1.3%
19: ffsf 1850/27783 6.7%
20: fff 5476/9261 59%
Total chance to win: 1526843/1555848 98%
1: sss 125/37044 0.34%
2: ssfs 925/222264 0.42%
3: ssffs 4625/4667544 0.099%
5: sfss 925/222264 0.42%
6: sfsfs 4625/4667544 0.099%
8: sffss 6845/4667544 0.15%
11: fsss 50/27783 0.18%
12: fssfs 250/583443 0.043%
14: fsfss 370/583443 0.063%
17: ffsss 370/583443 0.063%
Total chance to win: 29005/1555848 1.9%
Spies win
4: ssfff 23125/1166886 1.2%
9: sffsf 34225/1166886 2.9%
10: sfff 34225/222264 15%
7: sfsff 23125/1166886 1.2%
13: fssff 5000/583443 0.86%
15: fsfsf 7400/583443 1.3%
16: fsff 1850/27783 6.7%
18: ffssf 7400/583443 1.3%
19: ffsf 1850/27783 6.7%
20: fff 5476/9261 59%
Total chance to win: 1526843/1555848 98%
These numbers hint that using this setup gives the spies an overwhelming advantage, but this is an extreme oversimplification of the rules, lets simulate that the resistance are a little smarter than simply picking random teams, this time teams are picked such as to maximize the chance that the mission is successful. The spies always sabotage if doing so causes a mission to fail.
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We estimate the chance of someone being spy by using what we know from previously sent teams. Initially everyone are regarded to have a 3 in 9 (or 1/3) chance or being spy.
a 1/3
b 1/3
c 1/3
d 1/3
e 1/3
f 1/3
g 1/3
h 1/3
i 1/3
We have no information to base our initial individual estimates so the first team will be random, lets pick a, b, c. (I am well aware that this is not true in real games, regardless, we simplify the problem this way in this simulation),
Team sent: a, b, c
Team 1 succeeds 5/21 23% of the time.
+ Show Spoiler +
Team 1 fails 16/21 77% of the time.
+ Show Spoiler +
a 1/3
b 1/3
c 1/3
d 1/3
e 1/3
f 1/3
g 1/3
h 1/3
i 1/3
We have no information to base our initial individual estimates so the first team will be random, lets pick a, b, c. (I am well aware that this is not true in real games, regardless, we simplify the problem this way in this simulation),
Team sent: a, b, c
Team 1 succeeds 5/21 23% of the time.
+ Show Spoiler +
a, b and c are confirmed resistance and will be sent on all missions henceforth.
a 0
b 0
c 0
d 2/3
e 2/3
f 2/3
g 2/3
h 2/3
i 2/3
Lets also send player d
Team sent: a, b, c, d
Team 2 succeeds 1/3 33% of the time. (Total chance to reach here: 5/63 7.9%).
+ Show Spoiler +
Team 2 fails 2/3 67% of the time. (Total chance to reach here: 10/63 16%).
+ Show Spoiler +
a 0
b 0
c 0
d 2/3
e 2/3
f 2/3
g 2/3
h 2/3
i 2/3
Lets also send player d
Team sent: a, b, c, d
Team 2 succeeds 1/3 33% of the time. (Total chance to reach here: 5/63 7.9%).
+ Show Spoiler +
Excellent, we have 4 confirmed resistance, we will send the same team again and win.
a 0
b 0
c 0
d 0
e 3/5
f 3/5
g 3/5
h 3/5
i 3/5
Team sent: a, b, c, d
Team 3 succeeds 1/1 100% of the time. (Total chance to reach here: 5/63 7.9%).
+ Show Spoiler +
a 0
b 0
c 0
d 0
e 3/5
f 3/5
g 3/5
h 3/5
i 3/5
Team sent: a, b, c, d
Team 3 succeeds 1/1 100% of the time. (Total chance to reach here: 5/63 7.9%).
+ Show Spoiler +
Resistance win
Team 2 fails 2/3 67% of the time. (Total chance to reach here: 10/63 16%).
+ Show Spoiler +
d is confirmed spy, we send another potential resistance member.
a 0
b 0
c 0
d 1
e 2/5
f 2/5
g 2/5
h 2/5
i 2/5
Team sent: a, b, c, e
Team 3 succeeds 3/5 60% of the time. (Total chance to reach here: 2/21 9.5%).
+ Show Spoiler +
Team 3 fails 2/5 40% of the time. (chance to reach here: 4/63 6.3%).
+ Show Spoiler +
a 0
b 0
c 0
d 1
e 2/5
f 2/5
g 2/5
h 2/5
i 2/5
Team sent: a, b, c, e
Team 3 succeeds 3/5 60% of the time. (Total chance to reach here: 2/21 9.5%).
+ Show Spoiler +
Excellent, we have found a 4th resistance member, this is great because now it doesn't matter who we send on mission 4 as the 5th member, because the spies need 2 sabotages to make the mission fail.
a 0
b 0
c 0
d 1
e 0
f 1/2
g 1/2
h 1/2
i 1/2
Team sent: a, b, c, e, anyone
Team 4 succeeds 1/1 100% of the time. (Total chance to reach here: 2/21 9.5%).
+ Show Spoiler +
a 0
b 0
c 0
d 1
e 0
f 1/2
g 1/2
h 1/2
i 1/2
Team sent: a, b, c, e, anyone
Team 4 succeeds 1/1 100% of the time. (Total chance to reach here: 2/21 9.5%).
+ Show Spoiler +
Resistance win
Team 3 fails 2/5 40% of the time. (chance to reach here: 4/63 6.3%).
+ Show Spoiler +
e is confirmed spy
a 0
b 0
c 0
d 1
e 1
f 1/4
g 1/4
h 1/4
i 1/4
For the fourth team we need to send 5 members, and the spies need two sabotages to win, of those that have not been sent yet only 1 can possibly be a spy, therefore, if we send the confirmed resistance and 2 new, we are guaranteed to succeed with the mission, lets send f and g
Team sent: a, b, c, f, g
Team 4 succeeds 1/1 100% of the time. (Total chance to reach here: 4/63 6.3%).
+ Show Spoiler +
a 0
b 0
c 0
d 1
e 1
f 1/4
g 1/4
h 1/4
i 1/4
For the fourth team we need to send 5 members, and the spies need two sabotages to win, of those that have not been sent yet only 1 can possibly be a spy, therefore, if we send the confirmed resistance and 2 new, we are guaranteed to succeed with the mission, lets send f and g
Team sent: a, b, c, f, g
Team 4 succeeds 1/1 100% of the time. (Total chance to reach here: 4/63 6.3%).
+ Show Spoiler +
edit:I did this one wrong, I said this was a resistance win, but its actually not, its really the following:
We have 3 confirmed resistance and 2 succesful missions and 2 failed missions, its down to the final mission to find out if resistance wins or not.
The last mission did not give us any additional information, so we still only know that we have 3 confirmed resistance, 2 confirmed spies and a group of 4 players where one is a spy, out of which we need to select 2 players to go on the final mission and we lose if we pick the spy.
Lets do it.
Team sent: a, b, c, f, g
Team 5 succeeds 1/2 50% of the time. (Total chance to reach here: 2/63 3.2%).
+ Show Spoiler +
Team 5 fails 1/2 50% of the time. (Total chance to reach here: 2/63 3.2%).
+ Show Spoiler +
We have 3 confirmed resistance and 2 succesful missions and 2 failed missions, its down to the final mission to find out if resistance wins or not.
The last mission did not give us any additional information, so we still only know that we have 3 confirmed resistance, 2 confirmed spies and a group of 4 players where one is a spy, out of which we need to select 2 players to go on the final mission and we lose if we pick the spy.
Lets do it.
Team sent: a, b, c, f, g
Team 5 succeeds 1/2 50% of the time. (Total chance to reach here: 2/63 3.2%).
+ Show Spoiler +
Resistance win
Team 5 fails 1/2 50% of the time. (Total chance to reach here: 2/63 3.2%).
+ Show Spoiler +
Spies win
Team 1 fails 16/21 77% of the time.
+ Show Spoiler +
At least one of a, b and c are spies, but there could be more, there is a possibility that two or even all three of the crew are spies, the estimates that a, b and c are spies will be adjusted to reflect this fact, as a reaction, the estimates of the rest of the players will be adjusted to reflect that they have a lower chance of being spies.
There are 9!/(3!6!) = 84 ways to choose 3 members from 9 players, out of these there are:
6!/(3!3!) = 20 ways to choose teams with no spies
6!/(2!4!)*3 = 45 ways to choose a team with exactly 1 spy
6!/(1!5!)*3 = 18 ways to choose a team with exactly 2 spies
6!/(0!6!) = 1 way to choose a team made entirely of spies
Therefore, if we assume that at least 1 spy has been sent, then on average there will be ((45*1)+(18*2)+(1*3))/(45+18+1) = 21/16 = 1.3125 spies sent, therefore we will adjust our estimates to say that each of a, b and c have an estimated spyvalue so to speak, of (21/16)/3 = 7/16, the remaining 6 players will have estimated spyvalues of (3-(21/16))/6 = 9/32.
a 7/16
b 7/16
c 7/16
d 9/32
e 9/32
f 9/32
g 9/32
h 9/32
i 9/32
j 9/32
So these numbers represent our distrust in these players, lets go on to the choosing of the next team. we need to pick 4 players, and we have 2 different groups of players, one group of 3 players where 1 is a spy, and another group of 6 where 2 are spies, lets do some quick math to find out what combination of players from these groups maximizes the chance that our team succeeds.
If we pick all 4 from the group of 6 then we need to pick all resistance, there is therefore only 1 succesful team possible from that selection, and there are 6!/(4!2!) = 15 possible teams, there is therefore a 1 in 15 chance of succeeding.
If we pick 3 from the group of 3 and 1 from the group of 4 we are guaranteed to fail, so lets not do that.
If we pick 3 from the group of 6 and 1 from the group of 3 we have a 2*(4!/(3!1!))/(6!/(3!3!))/3 = 2/15 chance of sending a succesful team.
If we pick 2 from the group of 6 and 2 from the group of 3 we have a (4!/(2!2!))/(6!/(3!3!))/3 = 1/10 chance of sending a succesful team.
The best odds are if we send 1 from the group of 3 and 3 from the group of 6, giving a 2/15 chance.
Team sent: a, d, e, f
Team 2 succeeds 2/15 13% of the time. (Total chance to reach here: 32/315 10%).
+ Show Spoiler +
Team 2 fails 13/15 87% of the time. (Total chance to reach here 208/315 66%).
+ Show Spoiler +
There are 9!/(3!6!) = 84 ways to choose 3 members from 9 players, out of these there are:
6!/(3!3!) = 20 ways to choose teams with no spies
6!/(2!4!)*3 = 45 ways to choose a team with exactly 1 spy
6!/(1!5!)*3 = 18 ways to choose a team with exactly 2 spies
6!/(0!6!) = 1 way to choose a team made entirely of spies
Therefore, if we assume that at least 1 spy has been sent, then on average there will be ((45*1)+(18*2)+(1*3))/(45+18+1) = 21/16 = 1.3125 spies sent, therefore we will adjust our estimates to say that each of a, b and c have an estimated spyvalue so to speak, of (21/16)/3 = 7/16, the remaining 6 players will have estimated spyvalues of (3-(21/16))/6 = 9/32.
a 7/16
b 7/16
c 7/16
d 9/32
e 9/32
f 9/32
g 9/32
h 9/32
i 9/32
j 9/32
So these numbers represent our distrust in these players, lets go on to the choosing of the next team. we need to pick 4 players, and we have 2 different groups of players, one group of 3 players where 1 is a spy, and another group of 6 where 2 are spies, lets do some quick math to find out what combination of players from these groups maximizes the chance that our team succeeds.
If we pick all 4 from the group of 6 then we need to pick all resistance, there is therefore only 1 succesful team possible from that selection, and there are 6!/(4!2!) = 15 possible teams, there is therefore a 1 in 15 chance of succeeding.
If we pick 3 from the group of 3 and 1 from the group of 4 we are guaranteed to fail, so lets not do that.
If we pick 3 from the group of 6 and 1 from the group of 3 we have a 2*(4!/(3!1!))/(6!/(3!3!))/3 = 2/15 chance of sending a succesful team.
If we pick 2 from the group of 6 and 2 from the group of 3 we have a (4!/(2!2!))/(6!/(3!3!))/3 = 1/10 chance of sending a succesful team.
The best odds are if we send 1 from the group of 3 and 3 from the group of 6, giving a 2/15 chance.
Team sent: a, d, e, f
Team 2 succeeds 2/15 13% of the time. (Total chance to reach here: 32/315 10%).
+ Show Spoiler +
Excellent, we have 4 confirmed resistance for mission 3, lets send the same team again.
a 0
b 1/2
c 1/2
d 0
e 0
f 0
g 2/3
h 2/3
i 2/3
Team sent: a, d, e, f
Team 3 succeeds 1/1 100% of the time. (Total chance to reach here 32/315 10%).
+ Show Spoiler +
a 0
b 1/2
c 1/2
d 0
e 0
f 0
g 2/3
h 2/3
i 2/3
Team sent: a, d, e, f
Team 3 succeeds 1/1 100% of the time. (Total chance to reach here 32/315 10%).
+ Show Spoiler +
Now we send anyone together with the previous team of 4, the spies cannot sabotage because they need 2 spies for this mission.
a 0
b 1/2
c 1/2
d 0
e 0
f 0
g 2/3
h 2/3
i 2/3
Team sent: a, d, e, f, anyone
Team 4 succeeds 1/1 100% of the time. (Total chance to reach here 32/315 10%).
+ Show Spoiler +
a 0
b 1/2
c 1/2
d 0
e 0
f 0
g 2/3
h 2/3
i 2/3
Team sent: a, d, e, f, anyone
Team 4 succeeds 1/1 100% of the time. (Total chance to reach here 32/315 10%).
+ Show Spoiler +
Resistance win
Team 2 fails 13/15 87% of the time. (Total chance to reach here 208/315 66%).
+ Show Spoiler +
The first and second mission has failed, and sending an unsuccesful team at any point in time will now mean a spy victory.
We need to send 4 players, but first we should update our distrust metric. it gets a little complicated trying to figure out the chances for each person of being spy, but lets do our best.
We know that among a, b and c there is at least 1 spy.
We know that among a, d, e and f there is at least 1 spy.
After working on it I get the chances to be as such:
a 28/55
b 2/5
c 2/5
d 18/55
e 18/55
f 18/55
g 13/55
h 13/55
i 13/55
The best team to send turns out to be d, g, h, i, it succeeds in 2/11 times, thereby barely beating a number of other combinations that all have 9/55, such as aghi, bdgh, bghi and degh.
Team sent: d, g, h, i
Team 3 succeeds 2/11 18% of the time. (Total chance to reach here 416/3465 12%).
+ Show Spoiler +
Team 3 fails 9/11 82% of the time. (Total chance to reach here 208/385 54%).
+ Show Spoiler +
We need to send 4 players, but first we should update our distrust metric. it gets a little complicated trying to figure out the chances for each person of being spy, but lets do our best.
We know that among a, b and c there is at least 1 spy.
We know that among a, d, e and f there is at least 1 spy.
After working on it I get the chances to be as such:
a 28/55
b 2/5
c 2/5
d 18/55
e 18/55
f 18/55
g 13/55
h 13/55
i 13/55
The best team to send turns out to be d, g, h, i, it succeeds in 2/11 times, thereby barely beating a number of other combinations that all have 9/55, such as aghi, bdgh, bghi and degh.
Team sent: d, g, h, i
Team 3 succeeds 2/11 18% of the time. (Total chance to reach here 416/3465 12%).
+ Show Spoiler +
A succesful mission! That means we have 4 confirmed resistance and we can send these 4 on the next mission together with one more, the spies cannot possibly sabotage this mission because they need 2 spies on this mission but we are sending 4 confirmed resistance.
Lets assume we do not find out whether anyone sabotaged the mission, this is realistic since in this scenario any sent spy would never actually sabotage the mission since that would simply unveil that player as a spy.
Team sent: d, g, h, i, anyone
Team 4 succeeds 1/1 100% of the time. (Total chance to reach here 416/3465 12%).
+ Show Spoiler +
Lets assume we do not find out whether anyone sabotaged the mission, this is realistic since in this scenario any sent spy would never actually sabotage the mission since that would simply unveil that player as a spy.
Team sent: d, g, h, i, anyone
Team 4 succeeds 1/1 100% of the time. (Total chance to reach here 416/3465 12%).
+ Show Spoiler +
It is now time to choose the final team. We have 4 confirmed resistance but need to find 1 last resistance player to win.
There are only 10 possible configurations of spyteams that can give the results that the previous missions have had, these 10 possible spyteams are:
a, b, c
a, b, e
a, b, f
a, c, e
a, c, f
a, e, f
b, c, e
b, c, f
b, e, f
c, e, f
Which interestingly gives every player that is not confirmed resistance an equal chance of being spy, 6/10 or equivalently 3/5.
a 3/5
b 3/5
c 3/5
d 0
e 3/5
f 3/5
g 0
h 0
i 0
So lets send the team a, d, g, h, i and hope for the best.
Team sent: a, d, g, h, i
Team 5 succeeds 2/5 40% of the time. (Total chance to reach here 832/17325 4.8%).
+ Show Spoiler +
Team 5 fails 3/5 60% of the time. (Total chance to reach here 416/5775 7.2%).
+ Show Spoiler +
There are only 10 possible configurations of spyteams that can give the results that the previous missions have had, these 10 possible spyteams are:
a, b, c
a, b, e
a, b, f
a, c, e
a, c, f
a, e, f
b, c, e
b, c, f
b, e, f
c, e, f
Which interestingly gives every player that is not confirmed resistance an equal chance of being spy, 6/10 or equivalently 3/5.
a 3/5
b 3/5
c 3/5
d 0
e 3/5
f 3/5
g 0
h 0
i 0
So lets send the team a, d, g, h, i and hope for the best.
Team sent: a, d, g, h, i
Team 5 succeeds 2/5 40% of the time. (Total chance to reach here 832/17325 4.8%).
+ Show Spoiler +
Resistance win
Team 5 fails 3/5 60% of the time. (Total chance to reach here 416/5775 7.2%).
+ Show Spoiler +
Spies win
Team 3 fails 9/11 82% of the time. (Total chance to reach here 208/385 54%).
+ Show Spoiler +
Spies win
So what does this tell us?
Well first of all it tells us that if all players play as if they were resistance (i.e. it is impossible to tell a resistance player from a spy through psychology) and spies always cause missions to fail if they can do so, then spies win 64% of the time and resistance win 36% of the time.
resistance win 87% of the time if mission 1 succeeds.
Further, under the same restrictions we can say that if resistance succeeds in mission 2 then they are guaranteed to win with proper play.
And finally, resistance is guaranteed to succeed mission 4 given those restrictions.
however, more than half of the games ends in 3 consecutive failed missions.
But of course, this is an extreme oversimplification. A very potent weapon among spies are to not sabotage missions they are sent on to gain the resistance trust and throw off the reasoning behind who is and is not confirmed resistance. This is something that the above calculations do not take into consideration, but for now I can't think of a very good mathematical model to represent such behaviour, so Ill simply leave it as an unsolved mystery as to what is the proper play at what time when considering enemy wifom such as spies not sabotaging.
Anyway, I would love to have some comments regarding if I have done something wrong or if there is something you believe is worth adding. how would you play in situation X? how would you draw out a tricky spy? how would you beat the odds?
