Discussion
Came across this logic puzzle recently. I don’t have the verified answer. Interested in the thoughts of others:
There is a company board of directors with ten members. They all agree the board is too large and the number of members should be reduced. They plan to hold a series of votes. If the number voting “no” to a reduction equals or exceeds the number voting “yes” then they stop, otherwise the last member to join the board has to resign and there is a new vote.
All members agree the board is too large and vote accordingly but their overriding priority is to remain on the board.
How many will there be on the board when the voting stops?
There is a company board of directors with ten members. They all agree the board is too large and the number of members should be reduced. They plan to hold a series of votes. If the number voting “no” to a reduction equals or exceeds the number voting “yes” then they stop, otherwise the last member to join the board has to resign and there is a new vote.
All members agree the board is too large and vote accordingly but their overriding priority is to remain on the board.
How many will there be on the board when the voting stops?
I'm assuming they all know how long they've all been on the board? If so, and we name them A to J in order of time served, A being the longest serving Then in the 1st vote A-I all vote yes and J votes no as he wants to stay, so it's 9-1 and he goes. And so on until only A B & C are left, when they vote 2-1 and C goes. Then A & B vote, it's 1-1 and they both stay.
It seems so straightforward that I must be completely wrong.
It seems so straightforward that I must be completely wrong.
TwigtheWonderkid said:
I'm assuming they all know how long they've all been on the board? If so, and we name them A to J in order of time served, A being the longest serving Then in the 1st vote A-I all vote yes and J votes no as he wants to stay, so it's 9-1 and he goes. And so on until only A B & C are left, when they vote 2-1 and C goes. Then A & B vote, it's 1-1 and they both stay.
It seems so straightforward that I must be completely wrong.
That was my first thought but eventually settled on a different answer.It seems so straightforward that I must be completely wrong.
NickCQ said:
^ yes, but all the board members can figure that out so all except A and B vote against any reduction as they know they are gone (eventually) otherwise. So no reduction in the size of the board as the very first vote fails.
That was my second thought but I think there is another answer (although I’m not sure it is correct).TwigtheWonderkid said:
I'm assuming they all know how long they've all been on the board? If so, and we name them A to J in order of time served, A being the longest serving Then in the 1st vote A-I all vote yes and J votes no as he wants to stay, so it's 9-1 and he goes. And so on until only A B & C are left, when they vote 2-1 and C goes. Then A & B vote, it's 1-1 and they both stay.
By round 5 (with A, B, C, D & E still in), C & D will be able to see that if they vote Y then they will be forced out in a subsequent round, so C & D will side with E and the vote will be 2 v 3 and all 5 will stay in.Given that a split vote will result in the voting being stopped, in round 1 F - J could realise their only hope is voting against anyone leaving giving a 5 v 5 split and the vote being stopped.
My thinking was:
“A” can’t be voted off so will always vote “yes” to reduce.
“B” is also safe because if it came to the last two they can vote No and save themselves
“C” can’t allow it to get down to the last three as they would be voted out so will always vote yes until it gets down to the last four as they can count on D voting No at that point (which saves them both).
As A to D will always vote Yes until it gets down to just four left, then E, F, G and H will all get eliminated if the board drops below 8 members, so at the vote for 8 they will all vote No, stopping the vote. I and J will get eliminated because A to H will vote them out.
“A” can’t be voted off so will always vote “yes” to reduce.
“B” is also safe because if it came to the last two they can vote No and save themselves
“C” can’t allow it to get down to the last three as they would be voted out so will always vote yes until it gets down to the last four as they can count on D voting No at that point (which saves them both).
As A to D will always vote Yes until it gets down to just four left, then E, F, G and H will all get eliminated if the board drops below 8 members, so at the vote for 8 they will all vote No, stopping the vote. I and J will get eliminated because A to H will vote them out.
Esceptico said:
That was my second thought but I think there is another answer (although I’m not sure it is correct).
These puzzles are generally unsatisfying as you have to make some assumption about how many steps ahead in the game the players anticipate and whether they perfectly anticipate the reactions of other players to those moves. It gets unsolvably recursive at some point.Gassing Station | Science! | Top of Page | What's New | My Stuff