[Guest]

I suppose you all know jenga game:

Assuming that each block is exactly at 1 cm. height and there are 18x3=54 blocks;

How tall can be averagely a jenga tower, in which, all the movable blocks are used, and the game is locked = everyone knows that if anybody moves any of the blocks, the tower will fall down.

I mean, every player has played so fine that they all managed to not to drop any block till there is no other block to move since there is no plausible move avoiding a fall down.

[Guest]

I played right, you can have one block per level, so 54 cm high. I'e done that before.

[Guest]

I played right, you can have one block per level, so 54 cm high. I'e done that before.

You posted the answer of that question:

How tall can be a jenga tower at most?

But I asked "how tall averagely a jenga tower will be, unless a player makes an awkward motion?

[Guest]

I think nobody is right on this one.

I'm not sure what you mean by "averagely" tall. But nobody met the criteria you had for averagely, "in which, all the movable blocks are used, and the game is locked = everyone knows that if anybody moves any of the blocks, the tower will fall down."

[Guest]

I think nobody is right on this one.

I'm not sure what you mean by "averagely" tall. But nobody met the criteria you had for averagely, "in which, all the movable blocks are used, and the game is locked = everyone knows that if anybody moves any of the blocks, the tower will fall down."

Let me explain: When you're playing jenga, you may come to a point that all the blocks are so settled that because of physical forces, it is impossible to take out any more block. If players are talent, this is not a seldom case.

I'm saying "averagely tall", because in the case I told, there is not a definite height. According to the setting up the tower, this may be x or y cm. By average I mean if you play this game to given point millions of time what will be x+y+.../million ?.

I don't want to tell more, because if I write some more, this won't be a question anymore, instead a will be a boring blog.

[Guest]

Ok. Ignoring tactics (skilled players would probably construct a tower of maximum height) and assume bricks are picked essentially at random. Then there is a 1/3 chance that a row will have its middle brick removed, and a 2/3 chance that one of the edges will be removed, forcing the second edge to be removed later in the game.

which I think boils down to 4 bricks making three layers of tower on average, so the average height would be 54 x 3/4 = 40.5 cm

[Guest]

Ok. Ignoring tactics (skilled players would probably construct a tower of maximum height) and assume bricks are picked essentially at random. Then there is a 1/3 chance that a row will have its middle brick removed, and a 2/3 chance that one of the edges will be removed, forcing the second edge to be removed later in the game.

which I think boils down to 4 bricks making three layers of tower on average, so the average height would be 54 x 3/4 = 40.5 cm

I appreciate your work but in jenga, bricks are not picked by only chance. The trick point in my question is that. For example by chance, you intend to pick a definite brick, but you can't pick it because the tower is so set up that if you pick that brick, the tower will fall down. Thus you can't pick which you want. To make a taller tower, a skilled player wants to leave the middle brick and pick the two other edge bricks. But this is possible only when the middle brick is highest among these 3 bricks, then the odds for this is 1/3, otherwise the middle brick is picked only and 2 edge bricks are left (2/3 odds). At 2/3 of rows, there will be 2 bricks, at 1/3 of rows there will be only middle brick. Consequently, 5 bricks will make 3 rows. Total= 54*3/5=162/5--> 32+1 = 33 rows= 33 cm.

Of course a super skilled player may suggest that he can pick both edge bricks even the middle brick is not the highest, but I ignore him.

[Guest]

the tallest a tower can be is 54cm, the shortest it can be is 27cm. (54+27)/2=40.5cm

40.5cm is the average of the 2 extremes.

[Guest]

But this is possible only when the middle brick is highest among these 3 bricks,

i thought we could assume that all the bricks were the same? but even if we didn't, then the probability Armcie gave for picking a center brick vs both edge bricks is the same as what you said. If you picked randomly, then the center brick has a 1/3 chance at being picked, and the outsides have 2/3 chance. Without randomizing the picks, it also happens that the center has a 1/3 chance of being a tad taller than the others, leaving the outside bricks a 2/3 chance of being picked.

the tallest a tower can be is 54cm, the shortest it can be is 27cm. (54+27)/2=40.5cm

40.5cm is the average of the 2 extremes.

i think we need to take a hybrid of both your answer and Armcie's

say that the average height of a locked janga puzzle is indeed between 27 cm (every layer having 2 exterior bricks) and 54 (every layer having a center brick only). Since we know that any given layer has a 2/3 chance of having a single center brick remaining, the average height will be 2/3 closer to the center brick scenario (2/3rds closer to 54 than 27). 2/3 * (54-27) + 27 = 45. 45 layers is the average height of a locked jenga puzzle. But then again, Locked jenga puzzles are no fun, it means the games over with no destruction.

Edit: added spoiler

Edited February 17, 2009 by DanCDow

[Guest]

I'm really surprised how all you suggest that you can make a row consist of a single middle brick when even the highest brick is not this middle brick. And, how you can assume that all bricks are same. A little imagination shows that if any edge brick is the highest, you can't make a row of a single middle brick.

[Guest]

I'm really surprised how all you suggest that you can make a row consist of a single middle brick when even the highest brick is not this middle brick. And, how you can assume that all bricks are same. A little imagination shows that if any edge brick is the highest, you can't make a row of a single middle brick.

given that we were assuming perfect players, I'd also assumed perfect jenga bricks. But i'll agree with your logic using sufficiently* uneven bricks.

*in my experience there is an element if choice available when three bricks in a row are of approximately equal height, ad the cog of the tower above a row can also play a factor in which bricks are available to pick.

[Guest]

I'm really surprised how all you suggest that you can make a row consist of a single middle brick when even the highest brick is not this middle brick. And, how you can assume that all bricks are same. A little imagination shows that if any edge brick is the highest, you can't make a row of a single middle brick.

From your OP.

I suppose you all know jenga game:

Assuming that each block is exactly at 1 cm. height and there are 18x3=54 blocks;

How tall can be averagely a jenga tower, in which, all the movable blocks are used, and the game is locked = everyone knows that if anybody moves any of the blocks, the tower will fall down.

I mean, every player has played so fine that they all managed to not to drop any block till there is no other block to move since there is no plausible move avoiding a fall down.

Edited February 18, 2009 by kkehoe5

[Guest]

I suppose you all know jenga game:

Assuming that each block is exactly at 1 cm. height and there are 18x3=54 blocks

From your OP.

And, how you can assume that all bricks are same.

Sorry for my conflict. From that point of view, you're quite right. I should have said "how can you claim that all bricks are at 1,0000000 cm" . In OP, I tried to describe an ordinary jenga game and wanted to inform you that all bricks are made in an intention of being 1 cm. Really I'm not sure, maybe 1/2 inch or whatever. But if you've played jenga, you know that in a starting tower, some bricks among others are more loose, and some are more stuck. This means that they are not exactly at same height in practice . Only they are intended to be so. I may seem to be in conflict with me, but please allow my bad wording.

[Guest]

I'm really surprised how all you suggest that you can make a row consist of a single middle brick when even the highest brick is not this middle brick. And, how you can assume that all bricks are same. A little imagination shows that if any edge brick is the highest, you can't make a row of a single middle brick.

I see what you are saying. I was a little confused in my reasoning. I guess then the answer should be

1/3 * (54-27) + 27 = 36 layers

[Guest]

Taken from the official Hasbro Jenga site

"The highest JENGA tower on record stood 40 complete tiers with two blocks into the 41st "

That brings the upper limit to 41cm, we might need to change our paramaters to find the average.

[Guest]

Taken from the official Hasbro Jenga site

"The highest JENGA tower on record stood 40 complete tiers with two blocks into the 41st "

That brings the upper limit to 41cm, we might need to change our paramaters to find the average.

That would require 122 bricks. You only have 54.

Maybe I don't understand what's being asked, but I don't see how there can possibly be a solution, without knowing the probability of a brick being stuck. I don't see how one can assume it to be 1/2, 1/3, or any other value. And anyway, I used to be pretty good at tapping or flicking those out...

[Guest]

Maybe I don't understand what's being asked, but I don't see how there can possibly be a solution, without knowing the probability of a brick being stuck. I don't see how one can assume it to be 1/2, 1/3, or any other value. And anyway, I used to be pretty good at tapping or flicking those out...

Take a single row, consist of 3 bricks, into account. Only and only if the highest (I mean thickest) brick is the middle one, you can pick the edge bricks and leave the middle brick alone, and so can make a taller tower as a result. If the thickest is at the edge, you have to leave 2 bricks at that row. The odds that the thickest is the middle one, by simple thought, is 1/3. So at the end of a bright game, assuming that non of the players is coward, the tower will consist of rows that have 1 or 2 bricks. Consider 3 rows, two of them will have 2 bricks, one of them will have 1 bricks. So 5 bricks will form a group as asterix

O O

_O

O O

* *

_*

* *

O O

_O

O O

So 50 bricks will form 30 rows. Count of rows formed by remaining bricks varies according to the first row at the base. If it is a one brick row, remaining 3 bricks will make two rows. As a result,1+30+2= 33 rows are formed. If the base row has two bricks, and the row above it has one rows, then remaining 4 bricks will form 2 rows and one brick still remains, I do not know how to handle this. Result: 30+2=32 rows and one remainder. If row above the base 2 bricks is again has two bricks, then 54-50-2= 2 rows will remain and form a single row, total 1+30+1= 32 rows. Average: 33x1/3 + 32x2/3=32.333. Of course this may change if you take the remainder row into account.

This is my original question, thus I may be wrong, but yet I haven't been able to explain the question thoroughly (because of terrible wording, I suppose).

ps: The world record of 41 rows is a record, I ask the average. But not the average of all games, only the games played by talent players, so that the result tower will be formed by 1 or 2 brick towers, no fall down is occured before.

Edited February 19, 2009 by nobody