Are you aware of the dangers foxes pose to you and
Sweden
Help, I'm being held prisoner in an overtext typing facility! I don't have much time, they could find out at any m
I'm that weirdo whose number of badges sold prior to yesterday Bail Organa is keeping track of

I was thinking a little about whether there could be any strategy to this game, and it seemed to me I could come up with one rule: never jump just one number.
Every number you call has the same chance of losing, of course, so a move has to be judged on how it sets up the next player. When you split the remaining numbers in two sets, it is less advantageous for the next player to end up in the smaller set, but the chance of this happening is inversely proportional to its size, so it evens out... except in the case of jumping one number, which is guaranteed to leave just one big set.
For a simple proof, consider a twoplayer game where the current number is 40 and the last one was 44. If the call is 42 and not a loss, then the game must end next move on 41 or 43. If instead 41 or 43 is called and not a loss, then the next player still has a 50% chance of winning.
So given that the game is meant to give equal chances of winning to each player including the writer, I propose the writer's "bot move" should be to jump 2 (and of course if only one number remains they have lost anyway).

Fabio Fioravanti
Italy Roma italy

First thanks to your interest on this game :)
Kaffedrake wrote: I was thinking a little about whether there could be any strategy to this game, and it seemed to me I could come up with one rule: never jump just one number. At the begin of play you are right, never jump just one number. This is not always true. Think on a multiplayer game (5, 6 or more players), when the remain range is very low, you have same chance to lost for every number you say, but if you want to lose your neighbor on the right you have to jump one.
Kaffedrake wrote: Every number you call has the same chance of losing, of course, so a move has to be judged on how it sets up the next player. When you split the remaining numbers in two sets, it is less advantageous for the next player to end up in the smaller set, but the chance of this happening is inversely proportional to its size, so it evens out... except in the case of jumping one number, which is guaranteed to leave just one big set.
For a simple proof, consider a twoplayer game where the current number is 40 and the last one was 44. If the call is 42 and not a loss, then the game must end next move on 41 or 43. If instead 41 or 43 is called and not a loss, then the next player still has a 50% chance of winning. In this case, 2 players, the range is 40 to 44 you have to call 42. You have 33% to hit, your opponent will have 100%. If you call 41 your chance is same 33% but your opponent could have 50%
Kaffedrake wrote: So given that the game is meant to give equal chances of winning to each player including the writer, I propose the writer's "bot move" should be to jump 2 (and of course if only one number remains they have lost anyway). The writer must have same chance of other players, if he jump 2 every player have to say 2 numbers. Do you agree?
Anyway using the official link published you don't need of a writer ;) the number is random generated by a code.

Are you aware of the dangers foxes pose to you and
Sweden
Help, I'm being held prisoner in an overtext typing facility! I don't have much time, they could find out at any m
I'm that weirdo whose number of badges sold prior to yesterday Bail Organa is keeping track of

bubbaxis wrote: First thanks to your interest on this game :)
I made use of it once when buying groceries for a game night dinner.
bubbaxis wrote: At the begin of play you are right, never jump just one number. This is not always true. Think on a multiplayer game (5, 6 or more players), when the remain range is very low, you have same chance to lost for every number you say, but if you want to lose your neighbor on the right you have to jump one.
I'm not entirely sure what you mean here. I suppose there's a possibility that the turn could come around to you and at that point you'd wish you hadn't contributed to shrinking the range of numbers previously, but it seems to me it would generally (if not always) be a safer bet to minimize the chance that you will have to take another turn. As a simple and constant writer's rule, at least, it seems to me that stepping two numbers approximates good play for nonwriters much closer than stepping one.
bubbaxis wrote: The writer must have same chance of other players, if he jump 2 every player have to say 2 numbers. Do you agree?
I'm not suggesting a change to the rules beyond the method for determining which number the writer is forced to call; by "jump 2" I meant "add/subtract 2". Then when you have arrived at a number in this way, that is the only number the writer calls, as before. Sorry if that wasn't clear.

Fabio Fioravanti
Italy Roma italy

Kaffedrake wrote: I made use of it once when buying groceries for a game night dinner. :)
I use it every day at work. It's a coffee war with ranking scoreboard ;)
Kaffedrake wrote: I'm not entirely sure what you mean here. I suppose there's a possibility that the turn could come around to you and at that point you'd wish you hadn't contributed to shrinking the range of numbers previously, but it seems to me it would generally (if not always) be a safer bet to minimize the chance that you will have to take another turn. As a simple and constant writer's rule, at least, it seems to me that stepping two numbers approximates good play for nonwriters much closer than stepping one. I think the only strategy is the one you called: "minimize the chance that you will have to take another turn"
Kaffedrake wrote: I'm not suggesting a change to the rules beyond the method for determining which number the writer is forced to call; by "jump 2" I meant "add/subtract 2". Then when you have arrived at a number in this way, that is the only number the writer calls, as before. Sorry if that wasn't clear. If I understood good you say the writer have to add/subtract 2. This meaning the writer call 2 numbers. I think this is an unbalanced rule vs the writer.

Are you aware of the dangers foxes pose to you and
Sweden
Help, I'm being held prisoner in an overtext typing facility! I don't have much time, they could find out at any m
I'm that weirdo whose number of badges sold prior to yesterday Bail Organa is keeping track of

bubbaxis wrote: If I understood good you say the writer have to add/subtract 2. This meaning the writer call 2 numbers. I think this is an unbalanced rule vs the writer.
Well, I'm definitely not trying to suggest a rule unbalanced against the writer! Only one number is picked each turn by everyone, just as before. So again for clarity, if the last number called was 40 and the written number is higher, by the current rules the writer would go one higher (add one, jump one, step one) and call 41. With my suggested amendment the writer would go two higher (add two, etc.) and call 42 (and only 42). The only change is to the rule for forcing the writer's (single) pick, allowing them the same "optimal" play as everyone else.

Fabio Fioravanti
Italy Roma italy

Kaffedrake wrote: bubbaxis wrote: If I understood good you say the writer have to add/subtract 2. This meaning the writer call 2 numbers. I think this is an unbalanced rule vs the writer. Well, I'm definitely not trying to suggest a rule unbalanced against the writer! Only one number is picked each turn by everyone, just as before. So again for clarity, if the last number called was 40 and the written number is higher, by the current rules the writer would go one higher (add one, jump one, step one) and call 41. With my suggested amendment the writer would go two higher (add two, etc.) and call 42 (and only 42). The only change is to the rule for forcing the writer's (single) pick, allowing them the same "optimal" play as everyone else.
In your example (range 40....lower) with your amendment the writer have to call 42. This meaning like call 41 and 42. Meaning the writer called two numbers. If the hidden number is 41 or 42 the writer lost. Is correct?

Are you aware of the dangers foxes pose to you and
Sweden
Help, I'm being held prisoner in an overtext typing facility! I don't have much time, they could find out at any m
I'm that weirdo whose number of badges sold prior to yesterday Bail Organa is keeping track of

bubbaxis wrote: In your example (range 40....lower) with your amendment the writer have to call 42. This meaning like call 41 and 42. Meaning the writer called two numbers. If the hidden number is 41 or 42 the writer lost. Is correct?
Well, no, not unless I have completely misunderstood the rules to the game as it is. You always call just one number, right? So if the last number guessed was 40, and the answer is 41, the writer, who is going next, would call 42 (with my version of the rule for picking the writer's number), then declare that the answer is lower (just as if anyone else had called 42), then the next player will have to pick 41 and lose. At no point does anyone call more than one number. Or am I wrong about how the game works now?

Fabio Fioravanti
Italy Roma italy

Ok Understand. Is my mistake on rules explanation. The writer have to call a number near the last called. Then if the last call was 40 and the answer is 41 the writer have to call 41 and lost the game.
I try to explain in this sentence: "If the turn come back to the writer he have to call a 1 or +1 from the last called number"
Do you think is there a better way to explain this rule?

Are you aware of the dangers foxes pose to you and
Sweden
Help, I'm being held prisoner in an overtext typing facility! I don't have much time, they could find out at any m
I'm that weirdo whose number of badges sold prior to yesterday Bail Organa is keeping track of

To be honest I'm struggling to see where the misunderstanding comes in, because what you're describing now seems to be how I interpreted the rules. They seem perfectly understandable to me, unless I've completely misunderstood how the game is supposed to work. As I've understood it:
* While playing the game, each player calls one number from a dwindling set until someone calls the right answer and loses.
* The possibility should exist for the person who chose the right answer to be an equal participant, but since they obviously can't get to pick a number freely, the game has to provide a simple rule for picking their number each time it's their turn.
* Apart from that, their pick works just the same as everyone else's, i.e. they either hit the answer and lose, or else pass the turn as normally.
My suggested change is only that if the writer is supposed to have equal chances with everyone else, the rule the game uses to pick their number should reflect this. The change concerns only which one number they pick. Everything else works the same.
I don't understand at all what you mean when you say that if someone picks a number two numbers away from the previous pick, then they also pick the number in between. That doesn't seem to follow from the rules in any way.
Do you agree that if the last number is 40, and a player who is not the writer picks 42, they have not also picked 41? If you agree about that, then it seems to me it should also be possible to imagine how the writer picks 42 without also picking 41. And if you disagree, then I don't understand the game any more.

Fabio Fioravanti
Italy Roma italy

Kaffedrake wrote: Do you agree that if the last number is 40, and a player who is not the writer picks 42, they have not also picked 41? If you agree about that, then it seems to me it should also be possible to imagine how the writer picks 42 without also picking 41. And if you disagree, then I don't understand the game any more. :)
Agree, last call was 40. A player (not the writer) call 42. If the hidden number is 41, last call (42) is greater and the next player lost the game. If the hidden number is > 42 then 42 is lower and the game continue.
I'm sorry but the problem is my english speaking. The best to solve is to try to play. Do you want? I will be the writer. I will send you in in your private mailbox the number. Please don't read until game end. You have to call a number from 0 to 999. :)

Are you aware of the dangers foxes pose to you and
Sweden
Help, I'm being held prisoner in an overtext typing facility! I don't have much time, they could find out at any m
I'm that weirdo whose number of badges sold prior to yesterday Bail Organa is keeping track of

Whoops, I opened the geekmail before checking the thread, so I spoiled myself. However, I can save us a lot of time by seeing what would have happened. I would have started with 998. If it's 999 you lose, otherwise you say 997 according to the current rule. Then I say 995, you lose if it's 996 or 994, and so on. For each losing number for me, there are two losing numbers for you, giving me a 67% chance of winning instead of 50%. In this particular case, I would have been the one to say 335, making you lose on 334.
If instead we had been playing with a rule saying you also jump 2, then we would have called 998, 996, 994 and so on, and after exchanging 332 geekmails I would have said 334 and lost.

Fabio Fioravanti
Italy Roma italy

Kaffedrake wrote: Whoops, I opened the geekmail before checking the thread, so I spoiled myself. However, I can save us a lot of time by seeing what would have happened. I would have started with 998. If it's 999 you lose, otherwise you say 997 according to the current rule. Then I say 995, you lose if it's 996 or 994, and so on. For each losing number for me, there are two losing numbers for you, giving me a 67% chance of winning instead of 50%. In this particular case, I would have been the one to say 335, making you lose on 334. In this case, if me and you was around a table, and I knew your math skills I can choice to write 998 to catch you in a trap. ;) But you are right. In two players the nonwriter could have 67% of winning chance instead of 50%.
Kaffedrake wrote: If instead we had been playing with a rule saying you also jump 2, then we would have called 998, 996, 994 and so on, and after exchanging 332 geekmails I would have said 334 and lost. :) Please can you explain who call 998, who call 996 and so on?

Are you aware of the dangers foxes pose to you and
Sweden
Help, I'm being held prisoner in an overtext typing facility! I don't have much time, they could find out at any m
I'm that weirdo whose number of badges sold prior to yesterday Bail Organa is keeping track of

bubbaxis wrote: In this case, if me and you was around a table, and I knew your math skills I can choice to write 998 to catch you in a trap. ;)
I should start with a random number, then, to counter your one free decision in the game.
bubbaxis wrote: Please can you explain who call 998, who call 996 and so on?
I would start with 998. Then the new rule says that you as the writer add or subtract 2 as appropriate (just as the current rule says to add or subtract 1), so here you would subtract 2 and call 996. If the answer was 997 I would then have to call it and lose, but since it's not, you will say the answer is lower, I will then say 994, you will subtract 2 again and get 992, and so on. The actual rules change is very small, but should allow the writer the same chance to win.

Fabio Fioravanti
Italy Roma italy

Kaffedrake wrote: I would start with 998. Then the new rule says that you as the writer add or subtract 2 as appropriate (just as the current rule says to add or subtract 1), so here you would subtract 2 and call 996. If the answer was 997 I would then have to call it and lose, but since it's not, you will say the answer is lower, I will then say 994, you will subtract 2 again and get 992, and so on. The actual rules change is very small, but should allow the writer the same chance to win. You are right and you have convinced me :)


