![]() ![]() ![]() This results in Meredith Stout winning over Gilchrist. Not wiping the malware from the chip and making a deal with Maelstrom will result in a shootout with Maelstrom, and you will have to fight Royce in a mech in the final room. If you take this route, you will find out from Anthony Gilchrist that Meredith Stout is dead and that he sent the Militech soldiers. You will instead have to escape All Foods while being attacked by Militech soldiers. If you wipe the malware from the credchip or warn Royce that the credchip has a virus, you can choose to pay off Malestrom instead of fighting them or stealing the Flathead. In this case Stout will win over Gilchrist. In this scenario you will end up having to fight Royce in a mech in the final room. This option will continue the conversation and allow you to make a deal with the Maelstromers.Ĭhoosing not to tell Jackie to sit down will bring you into a fight with the Maelstromers. Since it is unlikely that you'll have 10,000 eddies with you at this point, the mission will almost always end up as a gun fight against the Maelstromers.Īt the beginning of the meeting with Maelstrom, you will have the option to calm Jackie by telling him to sit down on the couch when the Maelstromers are forcing him to do so. ![]() If you don't meet with Meredith Stout or if you refuse the credchip with malware from her, the only two options left to you will be to pay for the Flathead with your own money, or to have a shootout with the Maelstromers. Removing the malware or warning Royce that the chip has a virus and then striking a deal with the Maelstroms will result in you fighting against Militech instead. This credchip, however, has malware on it which you can choose to remove or not remove. Talking with Meredith Stout will give you the choice of accepting or refusing a credchip with enough eddies to strike a deal with Maelstrom. The Pickup has many approaches you can take, so it's highly recommended that you check this flow chart to ensure you get the outcome you are looking for. ![]()
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