Probability of getting caught on the chase (quiz show)

Question

There is a quiz show in which a contestant must correctly answer 5 multiple choice questions to reach the 'bank'. A 'chaser' simultaneously answers questions, but starts 3 answers behind. How do I write a formula to calculate the contestant's chance of reaching the bank as variables are changed, for example:

A = the contestant's chance of correctly answering any given question
B = the chaser's chance of correctly answering any given question
C = the contestant's headstart
D = number of correct answers to reach the bank

Each of these can change from show to show, and contestants are offered more or less money depending on the headstart they accept.

I ran an extremely rudimentary simulation in excel, the result was if
A = 0.7333
B = 0.5333
C = 3
D = 5

The contestant has a ~65% chance of reaching the bank vs a ~35% chance of being caught.

Answer 1

This is a random walk, with the probabilities of forward, stay still, backwards, given by:

• $A(1-B)$
• $AB+(1-A)(1-B)$
• $(1-A)B$

and is asking for the expected chance of reaching point D-C.

Answer 2

You can use a recurrence $$P(A,B,C,D) = (1-A)(1-B)\,P(A,B,C,D) + (1-A)B\, P(A,B,C-1,D) \\ + A(1-B) \,P(A,B,C+1,D-1) + AB\, P(A,B,C,D-1)$$ which reduces to $$P(A,B,C,D) = \tfrac{(1-A)B \, P(A,B,C-1,D) + A(1-B)\, P(A,B,C+1,D-1) + AB\, P(A,B,C,D-1)}{A+B-AB}$$

where $P(A,B,C,D)=1$ when $D=0$, and $P(A,B,C,D)=0$ when $C=0$ and $D>0$