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Combinatorics, EXT1 A1 2014 HSC 14b

Two players  `A`  and  `B`  play a game that consists of taking turns until a winner is determined. Each turn consists of spinning the arrow on a spinner once. The spinner has three sectors `P`, `Q` and `R`. The probabilities that the arrow stops in sectors `P`, `Q` and `R` are `p`, `q` and `r` respectively.
 

2014 14b

The rules of the game are as follows:

• If the arrow stops in sector `P`, then the player having the turn wins.

• If the arrow stops in sector `Q`, then the player having the turn loses and the other player wins.

• If the arrow stops in sector `R`, then the other player takes a turn.

Player `A` takes the first turn.

  1. Show that the probability of player `A` winning on the first or the second turn of the game is  `(1-r) (p + r)`.   (2 marks)

  2. Show that the probability that player `A` eventually wins the game is `(p + r)/(1 + r)`.   (3 marks)

Show Answers Only

a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

a.    `text(Show)\ P text{(}A\ text(wins)\ T_1\ text(or)\ T_2 text{)} = (1-r)(p + r)`

`P (A\ text(wins on)\ T_1) = p`

`P (A\ text(wins on)\ T_2)= P text{(} A\ text(lands on)\ R,\ B\ text(lands on)\ Q text{)}= rq`
 

`:.\ P text{(}A\ text(wins)\ T_1\ text(or)\ T_2 text{)}= p + rq`

♦♦ Mean mark 25%.
HINT: Expand out the solution `(1-r)(p+r)=p+r-rp-r^2` to get a better idea of what you need to prove.
 

`text(S)text(ince)\ q = 1-(p + r),`

`P text{(} A\ text(wins)\ T_1\ text(or)\ T_2 text{)}`

`= p + r [1-(p + r)]`

`= p + r-rp-r^2`

`= (1-r)(p + r)\ \ \ text(… as required.)`
 

b.    `text(Show)\ P text{(}A\ text(wins eventually) text{)} = (p + r)/(1 + r)`

`P text{(} A\ text(wins)\ T_1\ text(or)\ T_2 text{)} = (1-r)(p + r)`

`P text{(} text(No result)\ T_1\ text(or)\ T_2 text{)} = r * r = r^2`
 

`:.\ P text{(} A\ text(wins eventually) text{)}`

`= underbrace{(1-r)(p + r) + r^2 (1-r)(p + r) + r^2*r^2 (1-r)(p + r) + …}_{text(GP where)\ \ a = (1-r)(p + r),\ \ r = r^2}`
 

`text(S)text(ince)\ \ 0 < r < 1 \ \ =>\ \  0 < r^2 < 1`

♦♦♦ Mean mark 11%.
HINT: The use of ‘eventually’ in the question should flag the possibility of solving by using `S_oo` in a GP.
`:.\ S_oo` `= a/(1-r)`
  `= ((1-r)(p + r))/(1-r^2)`
  `= ((1-r)(p + r))/((1-r)(1 + r))`
  `= (p + r)/(1 + r)\ \ \ text(… as required)`