Snow Fall Chess Game

Introduction:

In the game of SnowChess, your goal is to use the received ("gift") and existing pieces on the board to achieve victory, where "gifts" introduce an element of randomness into each player's strategy. SnowFall determines the probability of receiving a "gift", ranging from 10% to 24%, where 10% is the minimum probability and 24% is the maximum.

In Snow chess Version 2.0 it works this way: If probability of receiving gift is about 17% You can imagine it as If You have 2 cubes with 6 sides numbered (1,2,3,4,5,6) and You throw this 2 cubes after every player's move, the most often 7 points together is sum, the points throwed by the cubes are collected, when is reached 35 points, then generator randomly select some pieces coming from set (like "PPPPPPPPLLLLLLLLRRQC" see below) to players and is 35 subtracted from points container. So in case of gift probability about 17% some pieces coming about after every 2nd or 3rd move (or 5. half-move). If the gift probability will be much higher, you can imagine it as having dice with more sides, so 35 points will be reached much sooner, and vice versa if smaller. Before every new game points container is filled randomly like among 0 and 35 points.

These SnowFall probabilities can be viewed in the chat. For each individual SnowFallChess game, a probability is randomly selected within the range of 10 to 24. But more often 17 or arround than other values is randomly selected.

Gift Generation. Strategy:

In general, one should understand that expecting a specific piece on the next move is quite a hopeless endeavor...

The generator considers a set where P- pawn, L- bishop or knight (light piece), R- rook, Q- queen, C- combo: PPPPPPPPLLLLLLLLRRQC.

In other words, we have a total of 20 cases: 8P, 8L, 2R, 1Q, 1C. The probability that both players will receive a pawn on the move is calculated as 5%+5%+5%+5%+5%+5%+5%+5%=40%...

If the player's gift is 'L', then:

Both players receive 'N' (knight): 25%.
Both players receive 'B' (bishop): 25%.
The current player receives 'N' (knight), and the opponent receives 'B' (bishop): 25%.
The current player receives 'B' (bishop), and the opponent receives 'N' (knight): 25%.
Each of these outcomes has an equal probability of 25%.

Additionally, the total number of each type of piece a player can have on the board and in hand is regulated, maximum is total 2 boards of pieces, so 16P, 2Q, 4N etc.

Combo Generation in general:

As mentioned above, Combo has a probability of 1/20 (5%) when some "gift" is generated. These 5% of Combo, in turn, are divided into:

10%: after my move, I'll receive a bishop, and the opponent receives two pawns;
10%: after my move, I'll receive a knight, and the opponent receives two pawns;
10%: after my move, I'll receive a rook and a pawn, and the opponent receives a bishop and a knight;
10%: after my move, I'll receive a queen, and the opponent receives a rook and a pawn;
10%: after my move, I'll receive a queen;, and the opponent receives a bishop and a knight
10%: after my move, I'll receive a queen, and the opponent receives two bishops;
10%: after my move, I'll receive a queen, and the opponent receives two knights;
10%: after my move, I'll receive a queen, and the opponent receives a knight and a pawn;
10%: after my move, I'll receive a queen, and the opponent receives a bishop and a pawn;
10%: after my move, I'll receive two bishops, and the opponent receives two knights.

But. Here is 1/3 alias 33% that gifts will be switched,
this means that there is a one-third chance that the "gifts" will be exchanged between players.
For example:
chance (alias probability) that player will get Queen just before his move if gift will 100% come (in other words, probability to find Queen in gift, which was guarantered), calculated as:
1/20+(6*((1/20)/10))*2/3 = 7%,
chance, that player will get Rook is
2/20+(1*(1/20)/10))*2/3+(1*(1/20)/10)*1/3= 11%

If it's not entirely clear how this is calculated, 1/20 for Queen or 2/20 for Rook represents the probability of getting a queen or a rook from the set PPPPPPPPLLLLLLLLRRQC. 6*((1/20)/10) means that the probability of Combo (1/20) is multiplied by 6 cases when we have 'I'll receive a queen' in the Combo list, you see? The Combo list contains 10 equally probable possibilities. 6 of them satisfy us, the rest don't. In other words, the expression
6*((1/20)/10)
can be rewritten as 6/10*1/20 if that's more convenient for you.

Then (6*((1/20)/10)) needs to be multiplied by 2/3. There is a probability of 1/3 that the opponent will get the queen we need. This is because with a probability of 1/3, the gifts are exchanged between players in the case of Combo. And the necessary 2/3 remains.

For rook:

2/20 is a R from PPPPPPPPLLLLLLLLRRQC set. 1/10 is "I'll receive a rook and a pawn, and the opponent receives a bishop and a knight;" case. This case we multiple on 2/3. Also, "I'll receive a queen, and the opponent receives a rook and a pawn;" case. This case we multiple on 1/3 (gifts switched).

So in this position Description of the in checkmate M1 image

if last move was of white's Kg1 then was chance about 70% that black will get diagonal pieces to checkmate in one by D@f2#, if last move was black's Ne4 then was about 16% chance ie one from about six tries that black will be checkmated with heavy by H@h8#. But the chances counts only if was gift generated ie based on snowfall number. So if SnowFall is lets say 17 ie 17% that will gift be generated, then total probability of white being checkmated in example above with black's following move using gift received is about 12% (0,17*0,7=0,12), chance that black will be checkmated by following white's move using gift received is under 3%.

Playing examples

In the first situation, what is the algorithm for gifts after 1.e4 for SnowFall:10?

1st example Description of the first image

Answers 1st example:

1st example. Due to SnowFall: 10, we have a 10% probability of receiving any gift...

Gifts are drawn from the set PPPPPPPPLLLLLLLLRRQC. In this position, a record number of pawns for both white and black has been reached.

We have 8 pawns on the board and in reserve for both white and black. In other words, neither white nor black can receive a pawn as a gift.

So, how do we calculate in this case? If we have the case P, no one simply gets a pawn as a gift. This was the simplest case.

Now let's consider when C is drawn. After the correction, we will get

after our move, white receives a bishop, and black receives nothing;
after our move, white receives a knight, and black receives nothing;
after our move, white receives a rook and black receives a bishop and a knight;
after our move, white receives a queen, and black receives a rook;
after our move, white receives a queena , and black receives bishop and a knight;
after our move, white receives a queen, and black receives two bishops;
after our move, white receives a queen, and black receives two knights;
after our move, white receives a queen, and black receives a knight;
after our move, white receives a queen, and black receives a bishop;
after our move, white receives two bishops, and black receives two knights.

But here is 1/3 chance that gifts will be switched and in that case:

after our move, white receives nothing, and black receives a bishop;
after our move, white receives nothing, and black receives a knight;
after our move, white receives a bishop and a knight, and black receives a rook;
after our move, white receives a rook, and black receives a queen;
after our move, white receives a bishop and a knight, and black receives a queen;
after our move, white receives two bishops, and black receives a queen;
after our move, white receives two knights, and black receives a queen;
after our move, white receives a knight, and black receives a queen;
after our move, white receives a bishop, and black receives a queen;
after our move, white receives two knights, and black receives two bishops.

In the second situation, what is the algorithm for gifts after 1.e4 for SnowFall:20 and after 1.e4 e5 for SnowFall:20?

2nd example Description of the second image

Answers 2nd example:

2nd example. Due to SnowFall: 20, we have a 20% probability of receiving any gift.

Gifts are drawn from the set PPPPPPPPLLLLLLLLRRQC. In this position, a record number of pawns for white and bishops for black has been reached.

We have 8 pawns on the board and in reserve for white and 2 bishops on the board and in reserve for black.

In other words, white can't receive a pawn as a gift and black can't receive a bishop as a gift.

So, how do we calculate in this case? If we have the case P, black will get P as a gift, not white.

If we have the case L:

25% both sides will get N;
25% white will get B and black will nothing;
25% white will get B and black will get N;
25% white will get N and black will nothing;

Now let's consider when C is drawn. After the correction, after 1.e4 we will get:

after white's move, white receives a bishop, and black receives two pawns;
after white's move, white receives a knight, and black receives two pawns;
after white's move, white receives a rook, and black receives a knight;
after white's move, white receives a queen, and black receives a rook and a pawn;
after white's move, white receives a queen, and black receives a knight;
after white's move, white receives a queen, and black receives nothing;
after white's move, white receives a queen, and black receives two knights;
after white's move, white receives a queen, and black receives a knight and a pawn;
after white's move, white receives a queen, and black receives a pawn;
after white's move, white receives two bishops, and black receives two knights.

But here is 1/3 chance that gifts will be switched and in that case:

after white's move, white receives nothing, and black receives nothing;
after white's move, white receives nothing, and black receives a knight;
after white's move, white receives a bishop and a knight, and black receives a rook and a pawn;
after white's move, white receives a rook, and black receives a queen;
after white's move, white receives a bishop and a knight, and black receives a queen;
after white's move, white receives two bishops, and black receives a queen;
after white's move, white receives two knights, and black receives a queen;
after white's move, white receives a knight, and black receives a queen;
after white's move, white receives a bishop, and black receives a queen;
after white's move, white receives two knights, and black receives nothing.

After the correction, after 1.e4 e5 we will get:

after black's move, white receives nothing, and black receives nothing;
after black's move, white receives nothing, and black receives a knight;
after black's move, white receives a bishop and a knight, and black receives a rook and pawn;
after black's move, white receives a rook, and black a queen;
after black's move, white receives a bishop and a knight, and black receives a queen;
after black's move, white receives two bishops, and black receives a queen;
after black's move, white receives two knights, and black receives a queen;
after black's move, white receives a knight, and black receives a queen;
after black's move, white receives a bishop, and black receives a queen;
after black's move, white receives two knights, and black receives nothing.

But here is 1/3 chance that gifts will be switched and in that case:

after black's move, white receives a bishop, and black receives two pawns;
after black's move, white receives a knight, and black receives two pawns;
after black's move, white receives a rook, and black receives a knight;
after black's move, white receives a queen, and black a rook and a pawn;
after black's move, white receives a queen, and black receives a knight;
after black's move, white receives a queen, and black receives nothing;
after black's move, white receives a queen, and black receives two knights;
after black's move, white receives a queen, and black receives a knight and a pawn;
after black's move, white receives a queen, and black receives a pawn;
after black's move, white receives two bishops, and black receives two knight.

Follows modification B, C of first and second situation, with same result. Only difference is that sum of pieces on board + hand together is 16 P , 4 B.

1st B example

2nd B example

1st C example

2nd C example

BPGN/PGN

In description of snow fall chess variant game we use tag {H:} for info what pieces hold each player after move in hand. For example holding {H:PPPPPPPPbb} means situation on 2nd example when white holds 8 pawns and black 2 bishops.

Game Example

You can watch video with snow fall chess variant (v.1004) example games on Onlinea Youtube channel.