How many distinct chess games are possible, and which is the longest?

Denying the 4 castlings one at a time. I see.

You have ambition trying to compose those stipulations. Composing 50 move retro problems is one of the hardest tasks possible. Nikita Plaksin is probably the master of those, and searching online chess problem databases should turn a few up. At the moment I think I'll share one by Gyula Breyer, though.

Who wins? (OK, you already know the answer, but the proof is long.)

To my understanding of the rules, neither the 50-move rule or the threefold-repetition rule are forced draws. They both require one of the players to claim the draw. Therefore the number of possible games should be infinite, as you can just keep moving pieces back and forth forever and not claim the draw.

5-fold repetition and 75 non-capture/pawn moves are now forced draws though. By FIDE at least.

Castling has nothing to do with the 50 move rule though.

Interesting. That's a new one to me.

Can anyone calculate total distinct games from this position?(condition: 1. no pawn can be moved/captured 2. Game ends at 3fold repetation or at 50th move)

* black to move

hint: only 5 pieces can move.

for a gross assumption, suppose each move there is 4 possible moves, this way total games =

4^(50+50)= 4^100 = 1.606938044… X 10^60  

Actually there will more than this: when the knights will be in middle of the board more than 4 moves are possible

Kilonewton, first off, a total of 8 pieces can move without any further pawn moves, all four knights and all four rooks.

Secondly, in the original position, there are only four possible moves, but in almost every other position obtainable in this game there are more. 

In fact, I think there is only one position in which there is less than four possible moves, and only a handful with exactly four possible moves. 

Here's a look at the maximum number of moves:

and most positions are closer to the 14 than to the 4, so considering there is only one position with only 3 moves, and most positions will have much more than 3 moves, and the number of draws by 3-fold repition is incredably small compared to the total number of distinct games from this position, I would say your estimate of 1.606(really close to phi, weird)x10^60 is probably a good lower bound, but the actual number of games from this position should be much higher.

So, one way we could get an upper bound for this position, unless I'm mistaken, black can have up to 14 potential moves per position, and white can have up to 15, take into account that both black and white has only four from the get go, and an upper bound for the total distinct games in this position is: 4*4*(14^49)*(15^49)=3.36x10^101=10^101.5, just over a googol.

So I think we can say with a decent bit of confidence that the number of distinct games from this position is somewhere between 1.6x10^60 and 1 googol.

I made that question to understand only for single pawn move.

a very rough assumption can be (4^100 +15^100)/2 for single pawn move.

now, assume there are only 16 pawns on the board, try to find total number of combinations. then multiply it with previous number.

exact number isn't calculateable by present technology, even with super computers.

i wonder if it will be in next 1000 years...  or ever!

There, is, no, limit, on, distinct, chess, games.

Assuming that, neither player claims draw on 50-mov-rule.

Benzodiaepine, if you were to read the original prompt of this forum, you would see that we are talking about chess with a few small restrictions that does limit the number of distinct chess games. Specifically, a draw is automatically declared if 50 moves have happened without a pawn move or capture, which makes the number of distinct chess games finite.

TL;DR

My assumption is that neither player would want to claim the 50-mov-rule.

I realized this, without even reading, your post.

My attention span is that of a gold fish.

Also, you forgot a Z there, professor.

Kilonewton, I think there is a flaw in your counting method, basing other pawn moves off this one will give you a large error since many pawn moves free up the bishops, queen, king, and give the rooks much more mobility, which would increase the number of possible moves in each position drastically, so estimating using a position where only knights have freedom, and rooks have extremely limited movement wouldn't give you very good estimates.

this thread isnt about players claimed or not. its about position where 1-0, 0-1 or 1/2-1/2 can be called theoretically.

ok. lets try something calculateable

what is the minimum number of moves to reach this position from starting position? (you play both sides & all legal moves)

I believe that kings and queens are swapped, right?

I always put black king on black square and white king on white square.

No, the queens go on their own colour, not the kings. Short castle is also to the right, from white's perspective. 1680 live?!

Pulling your leg.

Remellion, I can't figure the position out completely, but here's what I have thus far:

This is my first retroanalysis so excuse me if I don't use correct termonology. I'll try to explain as clearly and concisely as I can.

First off, lets talk about where the pawns came from:

each side made 3 captures, the c3 pawn must have gotten there from dxc3, and the c4 pawn must have gotten there from axb3 followed by bxc4 since if it came from e2, you would need a fourth capture to move the f-file pawn onto the e-file to be the e6 pawn. As a consequence of this, the pawn on e6 was originally the e pawn. then the f, g, and h pawns are all missing. since white is only missing pawns, the two captures by black to lead to his doubled pawns had to have been either captures of promoted pawns or captures of pieces dupilcated by promoted pawns. the h pawn couldn't have promoted since h7 is still in its original spot, and white doesn't have any extra captures to get his h pawn around h7, so the f and g pawns both promoted while the h pawn was captured as a pawn. 

Blacks pawns: there are two doubled pawns and three captures, so two of them had to have been used to create the doubled pawns, thus there is only a single unaccounted capture, so the a2 pawn had to have come from the a file, c5 and c6, one of them is the c pawn, the other comes from the d file(which one is which we will discuss later), which leaves e5, this had to have come from f6 where it was the f-file pawn. The g pawn is missing, and it had to have promoted since it(or something that it replaced by promotion) was captured on either c3 c4 or b3.

Okay, now let me talk about a few moves that will come up countless times in my analysis. bxc4 had to have happened before whites bishop got to b1. We know this because white's bishop had to have gome from b3 to a2 to get to b1, and white's pawn on c4 has only had 3 homes, a2, b3 and c4, so to have both a2 and b3 free, the pawn had to already have been on c4, so unless if white's bishop is not on b2, I will not consider bxc4, since it could not have happened.

now lets think about blacks move fxe5... white used all their captures to make that backwards L shape on the left, so the f and g pawns that promoted must have had a clear path to do so, so fxe5 had to have happened before white's second promotion, so while I'm working backwards, I won't consider fxe5 unless I bring white's f pawn back first.

Okay, now onto the analysis... Black just moved. We can see this because white's king, queen, rooks, bishops, and bottom left knight all have nowhere they could have moved from, white's other knight couldn't have been the last move, since black would have ended in check by white's dark bishop. the only other possible move white could have made was dxc3, but this couldn't have happened since that would imply that white's dark bishop was taken on c1(and the one on c8 is from a promoted pawn), but out of blacks three captures, one of them was fxe5, another was either dxc6 or dxc5 and the last one had to have been the g pawn capturing(must have been white's h pawn) in order to allow black's g-pawn to promote on the h-file and white's g-pawn to promote on the g-file, so dxc3 was not the last move.

Thus, black made the last move. fxe5 has already been discussed, so that couldn't have been the last move. the only other options for the last move are dxc6, Rb3(from b4), Ba3(from b4) or Qb5(from b4). Kd6 isn't possible since neither the e6 pawn or the c4 pawn could have moved the turn prior to give the king check at either d5 or d7.

We can exclude Rb3 almost immediately since white has no moves that could have preceded it(bxc4 had to have preceded Bb1, and dxc3 couldn't have happened for the same reasons discussed above). Similarly we can exclude Ba3 since white has no moves that could have preceded it.

dxc6 couldn't have happened. to see this, consider what could have been on c6. it couldn't have been a pawn since white has nowhere near enough captures to make quadrupled pawns, if it was a rook or queen, black would have been in check, so the rook or queen would have just gotten to c6, but that is impossible since either c6 was empty(so blacks king is in check from the b6 rook and its white's move) or c6 had a black piece on it, but white has no non-pawn captures so that couldn't have happened. 
 if it was a bishop, that would imply that white's g pawn(the only pawn that could have turned into a light-square bishop) turned into a light-square bishop, but the g-pawn had to have promoted and then been captured by fxe5 since fxe5 precedes white's f-pawn promotion, so that is a contradiction. If white had a knight on c6, then white has no legal move preceding dxc6(dxc3 still doesn't make sense, and neither of the knights can move without having black in check on white's turn, and no other piece has anywhere to go)...

SO! after exhaustive analysis, the last move was Qb5

From here, white has two potential moves... Rb6 or Ra5(dxc3 still doesn't make sense... in fact, unless if there is a bishop on c1, dxc3 couldn't have happened yet, so exclude it unless if white has a bishop on c1) 

Ra5 is a lot easier to analyse so lets start there. then black either has Qb4 or Ra6. if Ra6, white's last move was either Rb6 or Qa7(from a6), and black has no preceding move in either of those cases(dxc6 still doesn't work for all the reasons described above), so Ra6 is out, so black had to have done Qb4 this forces Rb5. If Qa5 preceded Rb5, white has no predecing move, so Ba4 precedes Rb5(to recap, in this line 1. ... Ba4 2. Rb5 Qb4 3.Ra5 Qb5 to get to the current position.) then white had to have done Rb4. From here, we can exclude blacks Rb3 since that forces Na1, and black has no preceding moves. Black could have also played Qa5 or Ba3.

The problem here(and with the Rb6 as whites last move) is that both of these lines allow for the possibility white's rook on b6 to leave the 6th rank, which makes dxc6 a possibility for black, which I cannot find a way to refute) 

So, I am going to stop here because once that rook can leave the 6th rank, the possibilities seem to explode, so I cant analyse further...

Any hints on how to get around this problem? or did I do something completely wrong in my analysis? 

[UPDATE]

Remellion, I figured out the problem with my earlier analysis, but it just seems to be getting incredably complicated. There are two main lines I've been following:

-7.Ra5 Bb5 -6. Ra6 Qa5 -5. Ra4 Ba3 -4.Rb4 Ba4 -3.Rb5 Qb4 -2.Ra5 Ra6 -1.Rb6 Qb5 
(-5. ...  Qa5 needs to be explored)

and

-7. ... Ba4 -6.Ra6 Qa5 -5.Rb6 Bb5 -4.Ra4 Ba3 -3.Rb4 Ba4 -2.Rb5 Qb4 -1.Ra5 Qb5 
(-4. ... Qa5 still needing to be explored)

Other than the two places indicated needing further exploration, I think all those choices are forced. I'm getting all sorts of lost at this point, and every step seems to open up 10 other possabilities, so I'm going to take a break on this one for a while.