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I think that Jack has three general strategies: 1) the numbers game, 2) the decoy, 3) the jailbreak.
1) The numbers game: The idea here is to try to preserve as many characters as possible. This is a difficult strategy because of Holmes and the information the Inspector can deduce from
Jack's moves. I think that most players would be pleased if they had 4 characters left by the 7th turn. However, if the Inspector got three Holmes draws, then the odds are that at least one of
the three innocents was drawn is 88%. With three characters left, the odds are 71% that one or more have been drawn. Two characters (Jack and an innocent) is pretty common. 57% of the time,
Holmes won't draw the innocent after three tries. That sounds pretty good but then the Inspector has only two suspects to choose from. The odds of the Inspector not drawing the card and not
guessing correctly are only 28%. The real odds are much worse because the Inspector can deduce information based on Jack's moves. Overall, not a high probability strategy but the most
popular.
2) The decoy: In this strategy, Jack tries to get the Inspector to think that he is someone else. Jack can do this by being particularly protective of one of the innocent characters, or by
making an escape attempt with a decoy character. This works best if Jack already has the witness card for the decoy to ensure that the Inspector won't draw it so Jack will want to take Homes in
the first round. This strategy has some pitfalls, though. For example, if you make an escape attempt with your decoy and the Inspector doesn't see it, then your decoy is useless as the Inspector
will probably figure it out when there are few characters left.
3) The jailbreak: This is the least common strategy but can therefore be the most effective. Escape attempts by only a single character are pretty easy to thwart. The Inspector is on guard
for even turns so the best bet is for odd turns. Try to get several characters at once positioned to escape. The idea is that so many characters can escape that the Inspector must guess as to
which one is Jack. With multiple characters, the odds are improved. The jail break characters need not even include Jack. The best possible outcome is that the Inspector will make an accusation
against an innocent. The problem with this strategy is that you can't be sure which characters will be drawn on an odd numbered turn and some characters can even prevent a jailbreak (such as
Bert, Lestrade or Goodley). Sometimes the jailbreak can continue into the even numbered turn and Jack can escape.
I think that the key (and I haven't mastered it yet) is sensing which of the three strategies will work best given the conditions in the first turn.
Cette contribution a été rédigée par jpganong sur mrjack.hurricangames.com.
Les 2 colonnes A et B représentent indifféremment le nombre de cartes alibi
récoltées et/ou le nombre de suspects innocents (hors Jack donc). La colonne A représente le plus grand des 2 nombres (entre cartes alibi et suspects innocents) et la colonne B le plus petit (ou
égal). La colonne "En commun" indique le nombre d'éléments communs aux cartes alibis et suspects innocents (les suspects disculpés par les alibis). La colonne "Probabilité" indique la probabilité
de se trouver dans cette situation, traduite en % dans la dernière colonne (c'est plus parlant).
Chacun-e en tirera ses propres conclusions, mais tout le monde conviendra que
le poids de Holmes n'est pas négligeable. Dans une situation finale à 2 suspects, 3 cartes alibis donnent plus de 70% de chance de victoire à l'inspecteur. Même face à 4 suspects, un inspecteur
avec 3 cartes alibis aura 40% de chance d'accuser Jack. Cela peut être très frustrant pour un Jack qui se sera démené pour conserver le plus de suspects possibles.
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