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

I had the same result for the ABCDEFG puzzle as Remellion. I used a spreadsheet and recalculated every step. How did you do it, Remellion?

I reason that on the last round players A through F must each have 0.64

Therefore on the last round G must pay out 6*(0.64)

G must also end with 1.28

So on the last round G = 6(0.64) + 1.28 = 5.12

What he had on the first round will be that divided by 2^6 which is 0.08

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At the start of the 6th round each player other than G and F must have 0.32

So F will pay out 5(0.32) plus whatever G has on round 6 which I knew because of the above.

After the 6th round F must end with 0.64

So at the beginning of round 6  F= 0.64 + (5*0.32) + (0.08*2^5) = 4.8

So at the 1st round F = 4.8 / 2^5 = 0.15

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etc.

I did this with a calculator and pencil.

Did you use this reasoning also?

ABCDEFG - brute force calculation like what you two did.

ABC - if there is a "king of the hill" (so not the initial state), and A is sitting out, A has a [sum n=1 to infinity (1/2^3n)] = 1/7 chance of winning from that point. The rest is a simple finite probability tree.

Remellion's solution is correct.

solve backward:

5.12= .64*6 +1.28

4.8 = .32*5 +.64 +2.56

Hmm, I had to imagine (well, I drew it to be sure) it as two branches. In one, B is the first winner and the next possible winners progress BAC repeating. In the 2nd branch C is the first winner and it goes CAB. That's why I was adding separate summations.

I wasn't sure if there was a way to calculate it all at once somehow.

The way I understand my own puzzle:

The probability of going from one play again state to another play again state is 1/8 ( play again state: play after sitting ). This is because in a play again state a player loses with 1/2 chance permanently ( because the other player wins again ), wins with 1/4 chance permanently ( because he wins two in a row ), so he has a remaining chance of 1/4  for sitting again, during which he can lose with 1/2 chance permanently again ( because the player ho has beaten him wins again ), so he has an overall 1/8 chance of getting into a new play again state from a previous play again state.

Since the probability of winning a play again state is 1/4, and the first play again state has a probability of 1 ( Adam can play at least once after his first sitting, no matter what ), for Adam you have 1/4 * ( 1 + 1/8 + 1/8^2 + 1/8^3 + ...) = 1/4 * ( 1 / ( 1 - 1/8 ) ) = 1/4 * 8/7 = 2/7. The other two follow from symmetry and that the probabilities should add up to 1.

no #205

Question:

>"Three boys, Adam, Bob and Chris play a game in rounds, In each round one of them sits out, the other two toss a fair coin. The one losing the coin toss sits out in the next round and the winner plays with the one sitting out in the previous round. The boy winning two tosses in a row wins the game. In the first round Adam sits out.

What are the probabilities of each boy winning the game?"

my answer:

A = 28.57...%

B = C= 35.71...%

solution:

in the figure,

(100*2^-1 = 50%,  100*2^-2 = 25%,   100*2^-3 = 12.5% .... )

wins by all are in Green.

B's wins in Left part are in rounds 2, 5, 8, ....

            in Right part are in rounds 4, 7, 10, ...

B’s probablity (in percent)

= 100*(2^-2 +2^-5 + 2^-8 + …) + 100*(2^-4 +2^-7 + …)

= 100* { (2^-2 +2^-5 + 2^-8 + …) + (2^-4 +2^-7 + …) }

Now using the formula below, (infinite geometric series),

r1 & r2 = 2^-3, a1= 2^-2, a2= 2^-4

= 100* { (2/7) + (1/14)}

= 250/7     (=35.71…%)

B and C are symmetric. so C’s probablity = 250/7

A’s probablity = 100 – 2*250/7 = 200/7   (=28.57…%)

can you guys solve this without an engine?

its extremely difficult. it took me several days.

black to move and Win

Here is a puzzle, if you have time, how to cut a pizza into 11 equal slices?

There are as many possible chess games as there are stars in the sky.  I have played all of them and lost every last one of them.

in my calculation the longest game is around 2990 moves long.

It is proven that it is 11797 halfmoves long under the 50 move rule. I have already linked this in this thread:

http://de.wikipedia.org/wiki/50-Z%C3%BCge-Regel#Schachmathematik

The number of legal games depends on the arbitrary 50 move rule and what you consider distinct games. Anyway it is astronomical and of little importance.

What is really important is the number of legal positions. It is well defined and much less than the number of legal games ( no matter how you count legal games ). The only hope to fully solve chess is through generating all legal positions and work backwards from mates ( the tablebase method ). It is hopeless to solve chess by tree search from the starting position ( the chess engine method ), where the number of legal games matters.

Math puzzle

If you have time:

How do you cut a round pizza into 11 equal slices?

cant read that. do you have any english link?

You can translate it with google, but the essence is very simple: there are 30 non king pieces, which means 30 captures, there are a maximum of 6 * 16 = 96 possible pawn moves, this would be 30 + 96 = 126 fifty move cycles. However to use all your pawn moves, 8 pawn moves have to be at the same time captures. This gives you 126 - 8 = 118 fifty move cycles ( each 100 halfmoves ). This is 11800 halfmoves, so this gives you the rough idea why the 11797 number may be correct.

Concentric circles wouldn't be too hard... except for the part where you measure and cut lol. It's interesting (to me) to note this would produce 11 pieces in 10 cuts while the standard slice would need 11 cuts.

In real life I'd use something like string to approximately measure the circumference, divide by 11, mark, and slice.

Really big = infinity, good to know.

There is an ingenious 'real life' solution to this problem. I wonder if anybody can guess it. And it cuts it to normal slices, that you are used to.

Think about it: a real life solution that can generate 2 * Pi / 11 * n angles, for n = 0, 1, ... 10.

I looked it up afterwards, and yes, there is a very nice practical solution I'll remember for a long time :)

11 cuts needed. if you cut through the height then you need 10 cuts.

This is what I mean by concentric circles. 10 cuts would give 11 pieces.