[Guest]

I don't believe this one has been posted.

You and your friend decide to go to the local pub. When you get there, your friend challenges you to a friendly contest, which is as follows:

You are given an unlimited amount of empty beer bottles and a perfectly circular table. One at a time, your friend and you must take turns placing a bottle on the table. Whoever runs out of space to place a bottle first is the loser, and must buy the other a drink. You may not move preexisting bottles on the table.

Your friend kindly offers to let you have the choice of going either first or second.

Questions:

1. Should you choose to go first or second to have a better chance of winning? If it depends, explain.

2. Is there a way to place the beer bottles to totally insure your victory? Again, if it depends, explain.

3. Can you do so whether you choose to go first or second?

The size of the table and bottles is irrelevant.

[Guest]

I'd think the size of the table did matter, after all if it were only big enough for one bottle whoever went first would obviously win and were it big enough for 2, whoever went 2nd would win.

[Guest]

I'd think the size of the table did matter, after all if it were only big enough for one bottle whoever went first would obviously win and were it big enough for 2, whoever went 2nd would win.

Mmm... not quite. Try imagining the situation in your head, or better yet, make a model.

[Guest]

I see no'where in the rules that it says you can't stack bottles, if that is the case, i suppose it would be a matter of who went first and how tall the ceiling was.

If that isn't the case, then I would think it'd be a matter of whoever went first, if the tables were both the same size, then depending on the size, you could lose depending on your choice to go first or second.

[Guest]

You want to always go first and always place the first bottle exactly in the middle of the table. Then every time it is your turn, place your bottle the same distance from the center on the exact opposite side of where your friend put their bottle.

[Guest]

Do the bottles have to be touching each other?

If you can find the area of the table surface, and the area of the bottom of the bottle surface(assuming the bottom is the widest part, like a typical beer bottle)There's a formula to calculate the maximum number of smaller circles(bottles) that will fit inside the larger one(table)...from that if the number is odd, go first, if it is even go second..of course it all depends on placing them correctly to ensure that the maximum number of bottles will fit(If I remember right I think a hexagonal pattern is best, been a while)..other than that I don't see an optimal placement strategy to be victorious, especially since your buddy could ruin your strategy at any point during the game..also without knowing the areas of the table and bottle bottoms, I don't see how you could decide to go first or second, without it being a geuss, and having luck on your side..but who knows..

Edited July 14, 2009 by James8421

[Guest]

and set the bottle down in the center of the table. If there is room for a second, there is room for another layer more. You set the third bottle down directly across from your opponent, he sets the fourth and you the fifth, taking up the available positions. Each succeeding incremental size of the table will allow the addition of only an even number of bottles, making for a total of an odd number. You started, you win.

Edited July 14, 2009 by bonanova
Spoiler added.

[bonanova]

If you play first, put your bottle in the exact center.

On successive turns, mirror your opponent's move: place your bottle the same distance from the center, 180^o away.

So long as your opponent has a move, you do as well.

Opponent will run out of space first.

If you play 2nd, mirror your opponent's move and hope he does not cover the center point with any of his moves.

If he does, you cannot mirror his move, and you cannot force a win.

The size of the table does not matter so long as it is at least 2 bottle diameters in diameter.

[Guest]

let him go first as long as the table is at least big enough for 20 beers (assuming its a guy or a girl that can hold her beer). When the other player is placing his beer and not looking pour most of your beer out. Then chug the rest of your beer very quickly and make a big deal about how fast your chugging your beers. Your opponent will then try to show you how fast he can chug beers and I guarantee by the 8ish beer your opponent will either be a drunkered or so full that a conveniently placed nudge could end the game in a victory for you. (anyone that has ever done a case rase, minikeg race, or box of wine challenge will agree)

[Guest]

i already thought about the whole placing it opposite your opponents thing but it is after all plausible for a circle to be made out of an odd number of things, i.e the circumfrence of a certain layer is only big enough to hold an odd amount of bottles

[Guest]

maybe it doesn't matter if you go first or not, if the table is small enough, simple smash the beer bottle into dozzens of ieces and spread them out to take up as much as the table as necessary.

[Guest]

The size of the table does not matter so long as it is at least 2 bottle diameters in diameter.

The table must be at least 1 (one) bd in diameter for the first bottle to fit. Thereafter, given the first bottle is centered, no bottle can be added unless the table is at least 3 bd's in diameter.

(sorry about the spoiler miscue)

[bonanova]

Brighter: Right on!

[Guest]

I would think that you would always win if you went SECOND.

Edited July 15, 2009 by yragcom1

[Guest]

I would think that you would always win if you went SECOND, because you would always be the last to have to fill 51% of the table. If there's a table that fit 100 bottles for example, player #1 would place all the odd numbered bottles (1-3-5-7....), and the second player would place all the odd numbered bottles (2-4-6-8...). The 101st bottle would have to be placed by player #1. In other words, Player #1 would have to place the bottle that tried to occupy the space past 51% of the total surface. Thus, Player #2 would win.

[Guest]

I would think that you would always win if you went SECOND, because you would always be the last to have to fill 51% of the table. If there's a table that fit 100 bottles for example, player #1 would place all the odd numbered bottles (1-3-5-7....), and the second player would place all the odd numbered bottles (2-4-6-8...). The 101st bottle would have to be placed by player #1. In other words, Player #1 would have to place the bottle that tried to occupy the space past 51% of the total surface. Thus, Player #2 would win.

If player one centers bottle one on the table, and if there is room for more bottles, and if player one then always places his (odd numbered) bottle diametrically opposed to and equal distance from the center as his opponent's (even numbered) bottles, then the game will end with an odd number of bottles every time. Player one wins------always.

[Guest]

If you play first, put your bottle in the exact center.

On successive turns, mirror your opponent's move: place your bottle the same distance from the center, 180^o away.

So long as your opponent has a move, you do as well.

Opponent will run out of space first.

If you play 2nd, mirror your opponent's move and hope he does not cover the center point with any of his moves.

If he does, you cannot mirror his move, and you cannot force a win.

The size of the table does not matter so long as it is at least 2 bottle diameters in diameter.

Don't we also have to assume that the table is symmetric in this strategy?

Edited July 16, 2009 by Terminator

[Guest]

yah terminators right what if its a "perfectly circular" asymmetrical table.

[Guest]

If player one centers bottle one on the table, and if there is room for more bottles, and if player one then always places his (odd numbered) bottle diametrically opposed to and equal distance from the center as his opponent's (even numbered) bottles, then the game will end with an odd number of bottles every time. Player one wins------always.

You proved my point. The game ends when the first person over 50% of the area tries to place his bottle. That would be bottle 101, or Player #1. There wouldn't be space for the 101st bottle, so player #2 wouldn't get his turn.

Edited July 16, 2009 by yragcom1

[Guest]

You proved my point. The game ends when the first person over 50% of the area tries to place his bottle. That would be bottle 101, or Player #1. There wouldn't be space for the 101st bottle, so player #2 wouldn't get his turn.

You have managed to either bypass or ignore the concept in its entirety.

[Guest]

I would think that you would always win if you went SECOND, because you would always be the last to have to fill 51% of the table. If there's a table that fit 100 bottles for example, player #1 would place all the odd numbered bottles (1-3-5-7....), and the second player would place all the odd numbered bottles (2-4-6-8...). The 101st bottle would have to be placed by player #1. In other words, Player #1 would have to place the bottle that tried to occupy the space past 51% of the total surface. Thus, Player #2 would win.

The goal isn't to see how many the two of you can fit on the table the goal is to place your bottles so he can't place one witout moving another one.

If the table is two bottles in diameter and you take the center space then he can't go. If he can go then you can place your bottle directly opposite his. so that's the almost perfect strategy. He could foil it in two ways: if, by natural human variation, the bottles were placed so as to leave one hole he could play in, or if he used a different size bottle than you then the one with smaller bottles would have a large advantage. different sized bottles or smashing a bottle on the table would probably be considered cheating unless that's what he has in mind.

[Guest]

This problem is similar to the "Losing Noughts and Crosses" game. The same rules apply as in normal Noughts and Crosses, but each player tries to force his opponent to form three symbols in a row. The first player can escape defeat by placing his "X" in the centre and then mirroring his opponent's move.