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WIFOM (Game): Difference between revisions

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(have s'more tables)
(and how about some more theory too)
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If we ignore all combinations where one player is guaranteed to win and earn 0 points, there are 13 combinations. Of those 13, four of them are effectively duplicates (such as NAA and NPP), leaving us with a total number of 9 unique combinations, listed below:
If we ignore all combinations where one player is guaranteed to win and earn 0 points (and therefore have no reason to be chosen), there are 13 combinations. Of those 13, four of them are effectively duplicates (such as NAA and NPP), leaving us with a total number of 9 unique combinations, listed below:


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Four of those nine combinations are won by not drinking anything (those of which the drinker is not poisoned will always be won by not drinking). Four of those nine can be won by drinking Glass A. Four of those nine can be won by drinking Glass B. Three can be won by drinking both glasses.
Four of those nine combinations are won by not drinking anything (those of which the drinker is not poisoned will always be won by not drinking). Four of those nine can be won by drinking Glass A. Four of those nine can be won by drinking Glass B. Three can be won by drinking both glasses.
Note that three of the combinations carry less reward to the pourer and more to the drinker by having only one correct choice, whereas the other six have two correct choices. Statistically both have the same expected value assuming random choice by the drinker, though this only matters after a reasonable number of rounds.

Revision as of 21:42, 1 May 2013

WIFOM (Game) is a game developed by Who and Zaicon via the MafiaScum Site Chat.

The standard version of this game is 2-player, although a 3-player version of this game is playable.

The Setup

One player will pour the wine; the other player will be the drinker. Turns alternate after every round.

The pourer must first choose whether to poison the drinker or not. Then, the pourer can place either an antidote, some poison, or nothing in each glass (all combinations are allowed, one per glass).

After that, the game can begin. If there is a mod, the pourer should notify the mod of his/her choices at this point.

The Play

Typically, the drinker will ask questions to help determine what to drink (if anything). The drinker can choose to drink from neither glass, one glass (the drinker must specify whose glass to drink from), or both glasses.

The drinker wins if he/she is still alive. One dose of antidote will cancel out one dose of poison (being poisoned to start with counts as one dose of poison). However, the antidote by itself is also harmful and will kill the drinker if there is no poison to cancel it out.

Scoring

If the drinker lives, the drinker gets points equal to the number of wrong choices there were. If the drinker dies, the pourer gets points equal to the number of right choices there were.

Right/wrong choices refer to the different options available to the pourer (drink from no glasses, drink from glass A, drink from glass B, and drink from both glasses). A "right" choice is a choice that allows the drinker to live.

Example

Let's say Player A is the pourer and Player B is the drinker.

Privately, Player A decides to poison Player B and put a dose of antidote into each glass. Once the decision has been made, Player B can start asking questions of the pourer. Eventually, Player B decides to drink from both glasses.

Since Player B has a dose of poison and two doses of antidote in his/her system, he/she dies (one of the antidotes does not cancel out). Player A gets a total of two points because there were two options available to Player B that would have allowed him/her to live (by drinking either glass, but not both).

Theory

There are a total of 18 possible combinations for the poison and antidotes (including whether or not the drinker starts poisoned). They are listed below.

Letter Code Drinker Glass A Glass B Points for Drinker Points for Pourer How to win
NAA Not poisoned Antidote Antidote 3 1 Do not drink
NAN Not poisoned Antidote Nothing 2 2 Do not drink / Drink B
NAP Not poisoned Antidote Poison 2 2 Do not drink / Drink both
NNA Not poisoned Nothing Antidote 2 2 Do not drink / Drink A
NNN Not poisoned Nothing Nothing 0 - Any action
NNP Not poisoned Nothing Poison 2 2 Do not drink / Drink A
NPA Not poisoned Poison Antidote 2 2 Do not drink / Drink both
NPN Not poisoned Poison Nothing 2 2 Do not drink / Drink B
NPP Not poisoned Poison Poison 3 1 Do not drink
PAA Poisoned Antidote Antidote 2 2 Drink A / Drink B
PAN Poisoned Antidote Nothing 2 2 Drink A / Drink both
PAP Poisoned Antidote Poison 3 1 Drink A
PNA Poisoned Nothing Antidote 2 2 Drink B / Drink both
PNN Poisoned Nothing Nothing - 0 none
PNP Poisoned Nothing Poison - 0 none
PPA Poisoned Poison Antidote 3 1 Drink B
PPN Poisoned Poison Nothing - 0 none
PPP Poisoned Poison Poison - 0 none

If we ignore all combinations where one player is guaranteed to win and earn 0 points (and therefore have no reason to be chosen), there are 13 combinations. Of those 13, four of them are effectively duplicates (such as NAA and NPP), leaving us with a total number of 9 unique combinations, listed below:

Letter Code Drinker Glass A Glass B Points for Drinker Points for Pourer Do not drink Drink A Drink B Drink both
NAA
NPP
Not poisoned Antidote
Poison
Antidote
Poison
3 1 Win Loss Loss Loss
NNA
NNP
Not poisoned Nothing Antidote
Poison
2 2 Win Win Loss Loss
NAN
NPN
Not poisoned Antidote
Poison
Nothing 2 2 Win Loss Win Loss
NAP
NPA
Not poisoned Antidote
Poison
Poison
Antidote
2 2 Win Loss Loss Win
PAP Poisoned Antidote Poison 3 1 Loss Win Loss Loss
PPA Poisoned Poison Antidote 3 1 Loss Loss Win Loss
PAA Poisoned Antidote Antidote 2 2 Loss Win Win Loss
PAN Poisoned Antidote Nothing 2 2 Loss Win Loss Win
PNA Poisoned Nothing Antidote 2 2 Loss Loss Win Win

Four of those nine combinations are won by not drinking anything (those of which the drinker is not poisoned will always be won by not drinking). Four of those nine can be won by drinking Glass A. Four of those nine can be won by drinking Glass B. Three can be won by drinking both glasses.

Note that three of the combinations carry less reward to the pourer and more to the drinker by having only one correct choice, whereas the other six have two correct choices. Statistically both have the same expected value assuming random choice by the drinker, though this only matters after a reasonable number of rounds.