Lifting four at eight with found teams

This is the second of the two situations in which the author of "50 Shades of Red" considers lifting four justified. There were misses, and the town has found teams at eight players: at the table sits a known black trio on one version and a pair on the other. By forcing the town into a red move (a tie vote) we get a mathematical win, or at least provoke the mafia into breaking the vote, which helps to choose the version and also brings the win. The strategy was first proposed and described by Artyom Puraev (game nickname Shakhist) in 2014. It is built on the principle "four are removed now, the fifth in the guessing round".

First read the lesson "Lifting four at eight: a pair of blacks": it explains the general rules of lifts (do not remove the sheriff and the false sheriff before the critical round, explain the scheme to the table and duplicate it with gestures) and the universal tie vote.

The condition

There were misses, the teams are known. Say, nos. 9 and 10 are absent from the table. On one version the mafia is no. 4, no. 5, no. 6, on the other the pair no. 7 and no. 8 is at the table. You remove four players now and the fifth in the guessing round. According to the author this is a mathematical win.

There are only two variations of splitting the table, depending on who is nominated under the first hands.

Variant 1: under the first hands is a player of the critical team

Under the first hands stands a player of the team on whose version the round is critical now, the team that could lose three against three. The author's example for four nominated (first hands no. 7):

for removing no. 7 vote no. 4 and no. 5;
for removing no. 8 vote no. 1 and no. 6;
for removing no. 4 vote no. 7 and no. 8;
for removing no. 5 vote no. 2 and no. 3.

If for removing the candidate under the first hands more than two votes fly in, the mafia from the team that broke the tie vote is removed by all hands.

The rule the author calls natural and logical:

For the candidate under the first hands (from the version with two mafia players) votes the black pair of the other version.
For the candidate under the second hands votes the black pair of the first version.
For the candidate under the third hands vote the remaining black and one red. This candidate is insured against a break: the mafia may not give the vote.
For the last candidate by default go the votes of the remaining reds.

The second and third points can be swapped; then the order of nomination is different and no insurance is needed. The key principle: try to dump all black votes into the two candidates nominated at the start of the vote. Ideally nominate first two candidates from the team with only two players, then two from the trio.

Variant 2: under the first hands is a player of the non-critical team

Under the first hands stands a player of the team on whose version the round is not critical now: they can win three against three. In this case the tie vote of the first variant cannot be used, because the mafia team has three hands, quite enough for a win if two civilians end up on the vote. So, as in the tie vote at nine, the player nominated under the first hands must vote for himself.

The author's example (the same players, nos. 1 to 8; mafia on one version no. 4, no. 5, no. 6, on the other the pair no. 7 and no. 8 at the table); nominated no. 6, no. 5, no. 7, no. 8:

for removing no. 6 vote no. 6 and no. 5;
for removing no. 5 vote no. 7 and no. 8;
for removing no. 7 vote no. 4 and no. 1 (the break is insured by no. 2 or no. 3);
for removing no. 8 vote no. 2 and no. 3.

As a rule:

For the candidate under the first hands votes the candidate himself and the player who is potentially mafia together with him.
For the candidate under the second hands votes the black pair from the other side.
For the candidate under the third hands votes a black from the opposite version and a red.
For the last candidate by default go the votes of the known civilians.

The author notes: the second tie vote is much harder and demands more coordination from the town. So it makes sense to resaw the vote so that under the first hands stands someone from the pair and not from the black trio.

Why it works (explanation of the Academy)

The idea of both variants is that each black vote goes where it cannot harm a civilian, and each candidate gets exactly two hands. If all mafia votes are dumped into the first two candidates, the mafia can break nothing and no insurance is needed. If insurance is needed, a red votes together with a black, and a break is visible at once. In any case the one who breaks gives himself away: the town has enough votes to finish him off.

The author's workshop

The condition of every task: two versions (a trio and a pair), the players absent from the table, and four nominated candidates. Saw the table.

Mafia nos. 2, 3, 5 or nos. 4, 7. Nos. 8 and 10 absent. Nominated nos. 5, 3, 4, 7.
Mafia nos. 1, 7, 10 or nos. 5, 6. Nos. 2 and 9 absent. Nominated nos. 1, 5, 7, 6.
Mafia nos. 2, 3, 4 or nos. 9, 10. Nos. 1 and 5 absent. Nominated nos. 4, 9, 10, 2.
Mafia nos. 6, 8, 9 or nos. 1, 7. Nos. 3 and 10 absent. Nominated nos. 9, 7, 8, 1.
Mafia nos. 8, 9, 10 or nos. 2, 5. Nos. 1 and 4 absent. Nominated nos. 2, 5, 8, 9.
Mafia nos. 1, 4, 6 or nos. 5, 10. Nos. 7 and 9 absent. Nominated nos. 5, 10, 4, 6.
Mafia nos. 4, 7, 9 or nos. 2, 3. Nos. 1 and 6 absent. Nominated nos. 9, 4, 2, 3.
Mafia nos. 1, 2, 3 or nos. 9, 10. Nos. 4 and 5 absent. Nominated nos. 1, 2, 3, 9.

Solution of the first task (Academy)

At the table remain nos. 1, 2, 3, 4, 5, 6, 7, 9. Trio A: nos. 2, 3, 5; pair B: nos. 4, 7. The order of nomination is no. 5, no. 3, no. 4, no. 7; first stands a player of the trio, so this is variant 2.

For no. 5 vote no. 5 and no. 3 (the candidate and his possible partner from the trio).
For no. 3 vote no. 4 and no. 7 (the pair from the other side).
For no. 4 vote no. 2 (the remaining black of the trio) and the red no. 1.
For no. 7 vote the remaining known reds, no. 6 and no. 9.

Each got two hands, eight votes in total, and nobody leaves automatically. Solve the other tasks with the same template and compare with each other.

For the black

This is one of the few places where the reds get a mathematical gain. If the mafia sees the reds preparing such a lift, arguing against the scheme is useless: a break gives away the offender. It is better to work in advance: not let the reds find both teams, doubt the versions and play so that misses do not pile up.

Example (compiled for the Academy)

Eight players. Version A: blacks no. 4, no. 5, no. 6. Version B: blacks no. 7, no. 8. The reds nominate no. 7, no. 8, no. 4, no. 5 and distribute hands by variant 1. The mafia no. 6 "forgets" to put the hand where it should go: everybody sees it and finishes off no. 6.

Self-check questions

What is the difference between variants 1 and 2?
Why in variant 2 does the first candidate vote for himself?
Which principle lets us do without insurance?
What will you do if the mafia broke the scheme?

Practice

Solve the author's eight tasks, then invent three of your own and swap with a partner. Check the scheme with a simple rule: each candidate has two hands, all black votes are dumped into the first candidates, and the insured candidate has a red hand.

Typical mistakes

Mixing up the order of nomination.
Not determining which version is critical.
Not finishing off the offender.
Forgetting insurance where a black vote remained in the third candidate.

Source of the idea. The section "Lifting four at eight players. The version with found teams" from Dmitry Novik's book "50 Shades of Red" (2021). The idea belongs to Artyom Puraev (2014) as the author states. The solution of the first task, the example and the questions were compiled for the Academy; the task conditions belong to the author.