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If any of you have seen the quiz show"The Chase", then you will understand the background to what I am about to describe. If not, here are wiki links for the original UK version and the similar US version.

My question is inspired by the "head-to-head" or "individual chase" round. In brief, your goal, as the contestant, is to reach "home" (or the "bank") before the "Chaser" catches up with you. The Chaser starts at step 8 and you can choose to start at step 4, 5 or 6. You are both simultaneously asked a question and if you answer correctly, you move down one step closer to home (step 0), otherwise you stay where you are.

For example, should you choose to start at step 6, and you get the first question wrong but the Chaser gets it right, then you will stay at step 6, but the Chaser will move down from step 8 to step 7, only one step behind you now. If the same thing were to happen on the second question then again you will stay at step 6 and the Chaser will move down from step 7 to step 6 and thus will have caught you and the game is over.

My question is what is the probability of being caught by the Chaser if you choose to start at step 4. Similarly for step 5, and for step 6?

Obviously we need more information. We need to know the respective probabilities of you and the Chaser answering a question correctly. For simplicity's sake, let's rate the intelligence of both you and the Chaser by giving you each a number from 0 to 1, which equates to your probability of answering any given question correctly.

My attempt

Notation:

Let [itex]P_{m,n}[/itex] be the probability of at some point being caught if you are currently at step [itex]n[/itex] and the Chaser is at step [itex]m[/itex], [itex]m>n>0[/itex].

Let [itex]p[/itex] be the probability of you answering any given question correctly.

Let [itex]q[/itex] be the probability of the Chaser answering any given question correctly.

So my question, in more explicit terminology, is what are the values of [itex]P_{8,4}[/itex], [itex]P_{8,5}[/itex] and [itex]P_{8,6}[/itex], in terms of [itex]p[/itex] and [itex]q[/itex]?

If, in addition to the definition of [itex]P_{m,n}[/itex] above, we also define the following:

[tex]P_{r,r}=1\text{, for }r>0[/tex][tex]P_{r,0}=0\text{, for }r=0[/tex]then, without being rigorous (although I could be if you want me to), I think it's pretty obvious to see that:

[tex]P_{m,n}=aP_{m,n}+bP_{m-1,n}+cP_{m,n-1}+dP_{m-1,n-1}\text{ for }m>n>0[/tex]where:

\begin{cases}

a=(1-p)(1-q) &\text{(both you and the Chaser answer the next question incorrectly)}\\

b=(1-p)q &\text{(you are incorrect and the Chaser is correct)}\\

c=p(1-q) &\text{(you are correct and the Chaser is incorrect)}\\

d=pq &\text{(both you and the Chaser are correct)}

\end{cases}or

[tex]P_{m,n}=

\begin{cases}

\frac{bP_{m-1,n}+cP_{m,n-1}+dP_{m-1,n-1}}{1-a} &\text{ for }a\neq 1\\

1 &\text{ for }a=1

\end{cases}[/tex]If [itex]a=0[/itex], then [itex]p=q=0[/itex] and both you and the Chaser would ALWAYS give incorrect answers and the game would continue forever with no outcome - so let's discard this scenario!

[table][tr][td]

[/td][td]m\n

[/td][td]0

[/td][td]1

[/td][td]2

[/td][td]3

[/td][td]4

[/td][td]5

[/td][/tr][tr][td]6

[/td][td]1

0[/td][td]

1[/td][td]

-[/td][td]

-[/td][td]

-[/td][td]

-[/td][td]

-[/td][/tr][tr][td]

[/td][td]2

0[/td][td]

[itex]P_{2,1}[/itex][/td][td]

1[/td][td]

-[/td][td]

-[/td][td]

-[/td][td]

-[/td][/tr][tr][td]

[/td][td]3

0[/td][td]

[itex]P_{3,1}[/itex][/td][td]

[itex]P_{3,2}[/itex][/td][td]

1[/td][td]

-[/td][td]

-[/td][td]

-[/td][/tr][tr][td]

[/td][td]4

0[/td][td]

[itex]P_{4,1}[/itex][/td][td]

[itex]P_{4,2}[/itex][/td][td]

[itex]P_{4,3}[/itex][/td][td]

1[/td][td]

-[/td][td]

-[/td][/tr][tr][td]

[/td][td]5

0[/td][td]

[itex]P_{5,1}[/itex][/td][td]

[itex]P_{5,2}[/itex][/td][td]

[itex]P_{5,3}[/itex][/td][td]

[itex]P_{5,4}[/itex][/td][td]

1[/td][td]

-[/td][/tr][tr][td]

[/td][td]6

0[/td][td]

[itex]P_{6,1}[/itex][/td][td]

[itex]P_{6,2}[/itex][/td][td]

[itex]P_{6,3}[/itex][/td][td]

[itex]P_{6,4}[/itex][/td][td]

[itex]P_{6,5}[/itex][/td][td]

1[/td][/tr][tr][td]

[/td][td]7

0[/td][td]

[itex]P_{7,1}[/itex][/td][td]

[itex]P_{7,2}[/itex][/td][td]

[itex]P_{7,3}[/itex][/td][td]

[itex]P_{7,4}[/itex][/td][td]

[itex]P_{7,5}[/itex][/td][td]

[itex]P_{7,6}[/itex][/td][/tr][tr][td]

[/td][td]8

0[/td][td]

[itex]P_{8,1}[/itex][/td][td]

[itex]P_{8,2}[/itex][/td][td]

[itex]P_{8,3}[/itex][/td][td]

[itex]P_{8,4}[/itex][/td][td]

[itex]P_{8,5}[/itex][/td][td]

[itex]P_{8,6}[/itex][/td][/tr][/table]

In order to find the values of [itex]P_{8,4}[/itex], [itex]P_{8,5}[/itex] and [itex]P_{8,5}[/itex], I find the table above quite useful to visualize what is needed. The equation for [itex]P_{m,n}[/itex] I wrote above shows that you can calculate the value for any cell in this table, provided you already have the values for 3 particular cells around it, eg [itex]P_{6,3}[/itex] can be calculated from the values [itex]P_{5,3}[/itex], [itex]P_{6,2}[/itex] and [itex]P_{5,2}[/itex].

So, if I start at the top-left, I can populate the table until I eventually reach the 3 values I need. The values in column 1 are easy:

[tex]P_{m,1}=\frac{bP_{m-1,1}+cP_{m,0}+dP_{m-1,0}}{1 - a}=\frac{b}{1-a}P_{m-1,1}[/tex]

We know [itex]P_{1,1}=1[/itex]

so [itex]P_{m,1}=\left( \frac{b}{1-a} \right) ^{m-1}[/itex]

However, the rest of the grid I'm finding a lot harder to fill in. For example:

[tex]P_{m,2}=\frac{bP_{m-1,2}+c\left( \frac{b}{1-a}\right) ^{m-1}+d\left( \frac{b}{1-a}\right) ^{m-2}}{1-a}[/tex]and I can't see a simple way of calculating these values other than plugging away. The rest of the columns get more and more convoluted as [itex]m[/itex] increases.

Sorry for the long post, but I can't help feeling that there must be a simpler way of doing this. Can anyone simplify my method, or even provide a more direct method of solving the equation:

[tex]P_{m,n}=aP_{m,n}+bP_{m-1,n}+cP_{m,n-1}+dP_{m-1,n-1}[/tex]?

Thanks for reading! :)

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# "The Chase" Quiz Show -- probability of being caught by the Chaser

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