The fictional game called \"Sousgammon\" consists of four positions labeled A, B

Question

The fictional game called "Sousgammon" consists of four positions labeled A, B, C, D. Whenever you reach position A or B, you roll some dice whose results determine what position you will move to next. From position A, there is a 1/4 chance you move to position B, a 1/12 chance you move to position C, a 1/6 chance you move to position D, and a 1/12 chance you stay at the position A and re-roll the dice. From position B, there is a 1/3 chance you move to position A, a 1/6 chance you move to position C, a 1/6 chance you move to position D, and a 1/3 chance you stay at position B and re-roll the dice. Whenever you reach position C or D, the game is over and you win some ice cream cake.

Supposing you begin the game at position A, what is the probability that you end the game at position C?

Explanation / Answer

Actually there are following steps in order to calculate the probability of game starting at A and ending at C i) rolling dice at A and moving to position C will have 1/12 chance ii) rolling dice at A and moving to B ,then further rolling dice at B and moving to position C will have (1/4)(1/6) = 1/24 chance iii) rolling dice at A and moving to B ,then further rolling dice at B and moving to position A and then rolling at A moves it to position C will have (1/4)(1/6)(1/4) =1/96 chance iv) rolling dice at A and moves it to A ,so by re-rolling dice ,it moves you to position will have (1/12)(1/12) =1/144 so total probability that you start game at position A and end at position C will be (i) +(ii)+(iii)+(iv) => 1/12 + 1/24 + 1/96 +1/144 =0.08333 +0.041666 +0.0104167 +0.006944 =0.142357

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