There are two players, A and B. At the beginning of the game, each starts with 4 coins, and there are 2 coins in the pot. A goes first, then B, then A,. . . . During a particular player's turn, the player tosses a 6-sided die. If the player rolls a:

• 1, then the player does nothing.

• 2: then the player takes all coins in the pot.

• 3: then the player takes half of the coins in the pot (rounded down).

• 4,5,6: then the player puts a coin in the pot.

A player loses (and the game is over) if they are unable to perform the task (i.e., if they have 0 coins and need to place one in the pot). We define a cycle as A and then B completing their turns. The exception is if a player goes out; that is the final cycle (but it still counts as the last cycle). We are trying to determine the expected number (and maybe even the distribution) of cycles the game will last for. I'm guessing that you can use"first-step" analysis to get the expected value. 

Answer:

1) Given

Two players A and B

--> B can accept or reject A's proposal

---> A proposes a split of $10 into integer units

Therefore Proposal of A can be

(A,B) = (0,10),(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1),(10,0)

I.e 11 proposals where $10 can be divided into integers where A and B get an amount

Note : '0' is an integer

Probability of B accepting or rejecting A proposal so there is only two outcomes i.e reject or accept

---Probabiliy of accepting an offer 1/2

Probability of an offer to be the player A offer nothing to B ( and $10 to himself ) i.e (10,0) =1/11

---Probability player B accept an offer of $0 for player A to offer nothing to B ( and $10 to himself) = 1/2 *1/11 = 0.04545

The answer in 3 decimals---0.0454

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2) Given

Two players A and B and already 1 coin in the pot.

Players can put 1, 2 or 3 coins in the pot

The player to add 30th coin in the pot losses.

Let A and be the winner and B be the loser ( for explantion)

For A to be winner 29th coin should be put by A which is possible by

1. Putting 1 coin---Where 28th is put by B

2. Putting 2 coins ---where 27th is put by B

3. Putting 3 coins - where 26th is put by B

And 28th can be put by B only if 25th is put by A because if 27th was put by A, B can put two coins and put 29th coin.

If 26th was put by A , B can put 3 coins and put 29th coin by B

If 25th coin should be put by A

-21st coin should be put by A

-17th coin should be put by A

-13th coin should be put by A

-9th coin should be put by A

-5th coin should be put by A

Therefore who put 5th coin in the pot can win the game

There is already one coin in the pot first player can not put the 5th coin because maximum 3 coin can put by a player

Therefore the no-of coins put by the first mover does not matter since this game features a second mover advantage.