One of the rules we enforce on Jailbreak is no favouritism, meaning that all prisoners should be treated equally, in the case that you may let one prisoner off for doing something but kill another for doing the same thing. One of the games I've found this collides with is Jeopardy. Most Jeopardy/Quiz minigames involve a podium for the warden or host, and a certain amount of podiums for the prisoners, which is mostly 3. Sometimes the warden will decide to do Jeopardy with 4, 6, 9 (etc.) prisoners which causes some problems. The main thing a warden will do here is pick out 3 prisoners and tell the remaining prisoners to take the podiums when the prisoner previously on that podium has died. What this can lead to is prisoners sitting off doing nothing while 3 prisoners risk their lives. It's almost like letting them skip out of a game. The prisoner last picked to go on the podium doesn't have to risk his life nearly as much as the prisoners who were picked first, which makes it much easier for them to win the game. So what I want to propose is a system that goes along the lines of this:
The warden can only pick Jeopardy if the amount of reds is a multiple of the number of podiums. For example, if the map has 3 podiums, there must be 3, 6, 9, 12, etc. reds participating. The warden will pick 3 and ask a question. The warden then has a choice; ask two questions and kill one prisoner per question, leaving one alive, or ask one question and only let the person who gets the answer right live. The survivor sits off to the side and another 3 reds are chosen. This goes on until all the reds that have survived are off to the side. Now we encounter a problem. Let's say we had 12 reds. 1 person wins each round, leaving us with 4 reds once we are finished. 4 would not be a valid number to do more Jeopardy with. So I propose two solutions; A) Play a small game at this time to get the amount of reds to a multiple of 3 or B) Instead of the warden being able to pick Jeopardy when the amount of reds is a multiple of the number of podiums, the amount of reds must instead be a power of the number of podiums. For example, if there are 3 podiums, there must be 3 reds, 9 reds, or 27 reds. If there are 4 podiums, there must be 4 reds or 16 reds. This means that after all reds are off to the side, you will still have a multiple of 3. To demonstrate:
27 reds. 3 move to the podium. 1 lives, 2 die. Repeat this 9 times (because 9*3=27) and you end up with 9 prisoners. Ask 3 to move up to the podiums and the other 6 to sit off. 1 person lives, 2 die. Repeat this 3 times and you end up with 3 prisoners. Ask those 3 to move to the podium. 1 person lives, 2 die. The last person wins.
I understand that this explanation may be confusing and that, if you are able to understand what I'm getting at, the system would also be rather confusing to players. But I feel like we need to address the issue of reds being able to sit off while other reds risk their lives. Any other ideas that are less confusing or more effective than mine are very appreciated.
The warden can only pick Jeopardy if the amount of reds is a multiple of the number of podiums. For example, if the map has 3 podiums, there must be 3, 6, 9, 12, etc. reds participating. The warden will pick 3 and ask a question. The warden then has a choice; ask two questions and kill one prisoner per question, leaving one alive, or ask one question and only let the person who gets the answer right live. The survivor sits off to the side and another 3 reds are chosen. This goes on until all the reds that have survived are off to the side. Now we encounter a problem. Let's say we had 12 reds. 1 person wins each round, leaving us with 4 reds once we are finished. 4 would not be a valid number to do more Jeopardy with. So I propose two solutions; A) Play a small game at this time to get the amount of reds to a multiple of 3 or B) Instead of the warden being able to pick Jeopardy when the amount of reds is a multiple of the number of podiums, the amount of reds must instead be a power of the number of podiums. For example, if there are 3 podiums, there must be 3 reds, 9 reds, or 27 reds. If there are 4 podiums, there must be 4 reds or 16 reds. This means that after all reds are off to the side, you will still have a multiple of 3. To demonstrate:
27 reds. 3 move to the podium. 1 lives, 2 die. Repeat this 9 times (because 9*3=27) and you end up with 9 prisoners. Ask 3 to move up to the podiums and the other 6 to sit off. 1 person lives, 2 die. Repeat this 3 times and you end up with 3 prisoners. Ask those 3 to move to the podium. 1 person lives, 2 die. The last person wins.
I understand that this explanation may be confusing and that, if you are able to understand what I'm getting at, the system would also be rather confusing to players. But I feel like we need to address the issue of reds being able to sit off while other reds risk their lives. Any other ideas that are less confusing or more effective than mine are very appreciated.
