Maximising probability for financial advice I have the following problem:

A financial advisor tries to impress his clients if immediately
  following a week in which the ftse index moves by more than
  $5\%$ in some direction he correctly predicts that this is the last
  week during the calendar year that it moves more than $5\%$ in
  that direction.
Suppose that in each week the change in the index is independently
  up by at least $5\%$, down by at least $5\%$ or neither of
  these, with probabilities $p$, $p$ and $1 - 2p$ respectively ($p\le1/2$).
  He makes at most one prediction this year. With what strategy
  does he maximize the probability of impressing his clients? 

Source
I think the goal is to use backwards induction. I start to formalize everything. Let $X_i$ be a random variable which represents how the market change in week $i$, therefore $P(X_i>0.05)=p=P(X_i<-0.05), P(X_i\in(-5/100, 1/100))=1-2p$ We have $52$ weeks per year so $i\in\{1,...,52\}$ Now lets assume we have $k$ weeks per year where index does not change more that 5%, and $i-k$ where it does. 
My guess is that he has to wait a specific number a week and than makes the guess, but how can I prove that?
 A: As the article recommends (and as you suggested), using backward induction seems like a good idea. As you put this idea forward, I'll assume that you are familiar with backward induction and the reasons why it should lead us to the optimal solution, so I will not comment on that. Below is my attempt to start solving the problem. From what I have seen, the answer will depend on the value of $p$ and determining the function associating a best strategy to any value of $p$ might be tedious for 52 periods. See if you like the way I begin and if you find a way to finish the argument. 
I will refer to the beginning of a week (say week 13) as the moment at which the financial adviser has observed the move of the index in the past week  (here week 12) but does not yet knows about the move in the index for the present week (here week 13), and can therefore make her prediction trying to impress her clients.


*

*Beginning of week 52. Suppose the adviser has not made any prediction up to now. There are three possible states of the world
1) The market went up more than 5% in week 51, Probability = $p$


*

*Two possible actions:

*

*Predict that last week was the last week of the year that the market went up by more than 5%. Probability to impress = $1-p$

*Stay quiet. Probability to impress = 0 (not optimal).



2) The market went down by more than 5% in week 51, Probability = $p$


*

*Two possible actions:

*

*Predict that last week was the last week of the year that the market went up down by more than 5%. Probability to impress = $1-p$

*Stay quiet. Probability to impress : 0 (not optimal).



3) The market did not move by more than 5%. Probability = $1-2p$


*

*Probability to impress = 0



Probability to impress by waiting until last week and playing optimal in last week = $p(1-p) + p(1-p) + (1-2p)0 = 2p(1-p)$.


*

*Beginning of week 51. Suppose the adviser has not made any prediction up to now. There are three possible states of the world
1) The market went up more than 5% in week 50, Probability = $p$


*

*Two possible actions:

*

*Predict that last week was the last week of the year that the market went up by more than 5%. Probability to impress = $(1-p)*(1-p)$

*Stay quiet. Probability to impress = Probability to impress by waiting until last week and playing optimal in last week = $p*(1-p) + p*(1-p) = 2p(1-p)$.



2) The market went down by more than 5% in week 50, Probability = $p$


*

*Two possible actions:

*

*Predict that last week was the last week of the year that the market went up down by more than 5%. Probability to impress = $(1-p)*(1-p)$

*Stay quiet. Probability to impress = Probability to impress by waiting until last week and playing optimal in last week = $p*(1-p) + p*(1-p) = 2p(1-p)$.



3) The market did not move by more than 5%. Probability = $1-2p$


*

*Only available strategy : stay quiet. Probability to impress = Probability to impress by waiting until last week and playing optimal in last week = $p*(1-p) + p*(1-p) = 2p(1-p)$.



Notice that which strategy is optimal in the two first states of the world depends on the value of $p$. So your answer will also depend on the value of $p$.
Probability to impress by waiting until week 51 and playing optimal in week 51 
\begin{align} = \begin{cases} 2*p(1-p)(1-p) + (1-2p)2p(1-p),& \text{   if  } (1-p)(1-p) \geq 2p(1-p) \\ 2*2p^2(1-p) + (1-2p) 2p(1-p), & \text{   otherwise  }  \end{cases}\end{align}
I hope I did not make any algebra mistake. At least, I checked that these probabilities are between 0 and 1, and they are (under the stated conditions), which is reinsuring.
From here, I would identify the cut-off value of $p$ ( such that $(1-p)(1-p) = 2p(1-p)$). Then see under which conditions on $p$ it is better, once in week 51, to adopt the speaking or the quiet strategy. In other words, try to identify the value of $p$ which makes it optimal to speak in period 51. Then move on to the next iteration.  
If you are lucky, after not too many iterations, you can see some pattern emerging in the value resulting from each strategies, and in the cut-off values of $p$. For instance, you might realize after a few iterations that, no matter the value of $p$, the probability to impress decreases as you go go back (i.e. as the adviser waits less). Then you may conclude that she should wait until the end. Or you can associate to each period a value for $p$ which makes it optimal to speak once arrived at this period. 
If you are not lucky, the pattern is hard to find or in-existent, and the number of cut-off values explodes as you go further back in time.
Good luck anyways. If you end up finding a solution, think about coming back and answering your own answer. Hope this helps you get there, or that it gives you an idea about an easier way to solve the problem.
