Yes, it is possible for you and your friend to beat the system.  The key to this is sharing with each other what predictions the machines are making, and changing your bets based on whether the predictions are the same or are different.

For the first two machines, we can construct the following probability table:

$\begin{matrix} & 1W & 1L \\ 2W & 0.24 & 0.16 \\ 2L & 0.36 & 0.24 \end{matrix}$

These are the four possible outcomes, and their probabilities were calculated by multiplying each machine's probability.  For example, the 1L 2W cell, which holds the probability that machine 1 loses and machine 2 wins, is the probability that machine 1 loses times the probability that machine 2 wins.

$0.4(1-0.6) = 0.16$

At this point, we need to combine the knowledge of the two machines' predictions to form two groups of results.  Either the machines agree, or they disagree.  If both machines predict the same team, then either both must win, or both must lose.  This puts us in the upper-left or lower-right quadrants of the table.  Conversely, if the machines predict different teams, then we are in either the upper-right or lower-left quadrants.  Once we have restricted the possible positions, we can recalculate the probabilities.

The probability of both machines winning was equal to the probability of both machines losing, so if the machines chose the same result, each now has a

>!$50\%$ chance of winning.  $\frac{0.24}{0.24+0.24}$

If the machines have different predictions, the probability that the first machine will win (and the second will lose) is

>!$\frac{0.36}{0.36+0.16} = 0.69230769230769230769230769230769 = {9\over13}$

Similiarily, the probability that the second machine will win given that the machines have different predictions is

>!$\frac{0.16}{0.36+0.16} = 0.30769230769230769230769230769231 = {4\over13}$

Each machine's payout is rounded down to the nearest pound, so our results will be slightly less than one might initially calculate.

>!$P_1 = \left\lfloor \frac{£100}{1-0.6} \right\rfloor = £250$<br><br>
>!$P_2 = \left\lfloor \frac{£100}{1-0.4} \right\rfloor = £166$

Now we can multiply each outcome's probability with its payout to find its expected value.  Any bets placed on a machine that wins pays $£0$, so we don't need to bother including them.  $E(M_n|S)$ means the expected value of betting on machine $n$ given that the predictions were the same.  Likewise, $E(M_n|D)$ means the expected value of betting on machine $n$ given that the predictions were different.  Each expected value is lowered by $£101$ to account for the initial bet and the betting fee.

>!$E(M_1|S) = 0.5 \times £250-101  =  £24$  
>!$E(M_2|S) = 0.5 \times £166-101  = -£18$  
>!$E(M_1|D) = {4\over13} \times £250-101 = -£24.076923076923076923076923076923$  
>!$E(M_2|D) = {9\over13} \times £166-101 =  £13.923076923076923076923076923077$

From these results, we can see that 

>!Bets should be placed on machine 1 when the machines predict the same team, and on machine 2 when the machines predict different teams.

The estimated return following this strategy is the sum of (the probability of each action taken times the expected value of that action).  The probability of taking the first action is

>!$0.24 + 0.24 = 0.48$

The probability of taking the second action is

>!$0.36 + 0.16 = 0.52$

So the strategy's expected return is

>!$(0.48 \times £24) + (0.52 \times £13.923076923076923076923076923077) = £18.76$

Splitting the betting profits with your friend, each person can expect to make

>!$\frac{£18.76}{2} = £9.38$ per week.


The third and fourth machines can be solved in the same manner.

$\begin{matrix} & 3W & 3L \\ 4W & 0.42 & 0.28 \\ 4L & 0.18 & 0.12 \end{matrix}$

>!$E(M_3|S) = {12\over54} \times \left\lfloor {£100\over0.4} \right\rfloor-101 = -£45.444444444444444444444444444444$  
>!$E(M_4|S) = {12\over54} \times \left\lfloor {£100\over0.3} \right\rfloor-101 = -£26.925925925925925925925925925926$  
>!$E(M_3|D) = {18\over46} \times \left\lfloor {£100\over0.3} \right\rfloor-101 =  £29.304347826086956521739130434783$  
>!$E(M_4|D) = {28\over46} \times \left\lfloor {£100\over0.4} \right\rfloor-101 =  £51.173913043478260869565217391304$

The strategy we should use here is to 

>!Bet on both machines when they differ, and neither when they agree.

The expected return for just using machines 3 and 4 is

>!$(0.18 \times £29.304347826086956521739130434783) + (0.28 \times £51.173913043478260869565217391304) = £19.603478260869565217391304347826$

or

>! $£9.801739130434782608695652173913$ per person per week

If we consider all four machines, then we should look at the actions we have so far.  We have four actions that all gain money, but not all can be taken at the same time.

>!If machines 1 and 2 agree, and machines 3 and 4 also agree, then our strategy tells us to bet on machines 2, 3, and 4, but the friend (who has access to machines 2 and 4) only has enough money to bet on one machine.  In this case, the friend should take the action that has the higher value, which is to bet on machine 4.  This is the only situation in this scenario that has a conflict.

This scenario has only three different sets of actions to take.

>!If machines 3 and 4 differ, bet on both of them.  Otherwise, follow the original strategy for machines 1 and 2, which itself has two possible actions.

The expected return for following our strategy using all four machines is

>!The expected value of machines 3 and 4 plus (the expected value of machines 1 and 2 times the probability of machines 3 and 4 being the same (causing us to consider 1 and 2))

That is,

>!$£19.603478260869565217391304347826 + (0.54 \times £18.76) = £29.733878260869565217391304347826$

or

>!$£14.866939130434782608695652173913$ per person per week


In the general case, two machines having probabilities $p_1$ and $p_2$, respectively, have the following probability table:

$\begin{matrix} & 1W & 1L \\ 2W & p_1p_2 & (1-p_1)p_2 \\ 2L & p_1(1-p_2) & (1-p_1)(1-p_2) \end{matrix}$

These machines then have the following expected values:

>!$E(M_1|S) = \frac{(1-p_1)(1-p_2)}{p_1p_2+(1-p_1)(1-p_2)}\times\frac{100}{1-p_1}-101$<br><br>
>!$E(M_1|S) = \frac{1-p_2}{2p_1p_2-p_1-p_2+1} \times 100-101$<br><br>
>!$E(M_2|S) = \frac{(1-p_1)(1-p_2)}{p_1p_2+(1-p_1)(1-p_2)} \times \frac{100}{1-p_2}-101$<br><br>
>!$E(M_2|S) = \frac{1-p_1}{2p_1p_2-p_1-p_2+1} \times 100-101$<br><br>
>!$E(M_1|D) = \frac{(1-p_1)p_2}{(1-p_1)p_2+p_1(1-p_2)} \times \frac{100}{1-p_1}-101$<br><br>
>!$E(M_1|D) = \frac{p_2}{p_1+p_2-2p_1p_2} \times 100-101$<br><br>
>!$E(M_2|D) = \frac{p_1(1-p-2)}{(1-p_1)p_2+p_1(1-p_2)} \times \frac{100}{1-p_2}-101$<br><br>
>!$E(M_2|D) = \frac{p_1}{p_1+p_2-2p_1p_2} \times 100-101$

With two machines, 

>!Simply bet on whichever outcomes ($S_1$, $S_2$, $D_1$, $D_2$) that occurred are positive.

With multiple sets of machines, 

>!Bet on the highest (positive) outcomes that occurred.  For example, if you have the following outcomes:<br><br>
>!$S_1=5$  
>!$S_2=-5$  
>!$D_1=10$  
>!$D_2=5$  
>!$S_3=20$  
>!$S_4=-20$  
>!$D_3=-10$  
>!$D_4=15$<br><br>
>!And machines 1 and 2 are the same, and 3 and 4 are different, then bet on machines 1 and 4.  This is because the valid cases are:<br><br>
>!$S_1=5$  
>!$S_2=-5$  
>!$D_3=-10$  
>!$D_4=15$<br><br>
>!And of these, $S_1$ is greater than $D_3$ and $D_4$ is greater than $S_2$.

With more than two (say, $n$) machines for a single pair of teams, 

>!Build a bigger ($n \times n$) probability table, calculate the expected values the same way, and bet on the outcomes with the highest (positive) expected values.

These strategies wouldn't work in real life, because 

>!They operate under the assumption that the machines generate predictions independently.  In reality, the machines probably use a lot of the same input data, which makes their predictions partially dependent.

This is my first post on Stack Exchange (long time lurker), so forgive the poor formatting.  I'm trying to figure out how it works, but would appreciate any tips.