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

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

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

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

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

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

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. $E_n|S$ means the expected value of betting on machine $n$ given that the predictions were the same. Likewise, $E_n|D$ 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.

$E_1|S = 0.5 \times £250-101 = £24$
$E_2|S = 0.5 \times £166-101 = -£18$
$E_1|D = {4\over13} \times £250-101 = -£24.076923076923076923076923076923$
$E_2|D = {9\over13} \times £166-101 = £13.923076923076923076923076923077$

$E_3|S = {12\over54} \times \left\lfloor {£100\over0.4} \right\rfloor-101 = -£45.444444444444444444444444444444$
$E_4|S = {12\over54} \times \left\lfloor {£100\over0.3} \right\rfloor-101 = -£26.925925925925925925925925925926$
$E_3|D = {18\over46} \times \left\lfloor {£100\over0.3} \right\rfloor-101 = £29.304347826086956521739130434783$
$E_4|D = {28\over46} \times \left\lfloor {£100\over0.4} \right\rfloor-101 = £51.173913043478260869565217391304$

$E_1|S = \frac{(1-p_1)(1-p_2)}{p_1p_2+(1-p_1)(1-p_2)}\times\frac{100}{1-p_1}-101$

$E_1|S = \frac{1-p_2}{2p_1p_2-p_1-p_2+1} \times 100-101$

$E_2|S = \frac{(1-p_1)(1-p_2)}{p_1p_2+(1-p_1)(1-p_2)} \times \frac{100}{1-p_2}-101$

$E_2|S = \frac{1-p_1}{2p_1p_2-p_1-p_2+1} \times 100-101$

$E_1|D = \frac{(1-p_1)p_2}{(1-p_1)p_2+p_1(1-p_2)} \times \frac{100}{1-p_1}-101$

$E_1|D = \frac{p_2}{p_1+p_2-2p_1p_2} \times 100-101$

$E_2|D = \frac{p_1(1-p-2)}{(1-p_1)p_2+p_1(1-p_2)} \times \frac{100}{1-p_2}-101$

$E_2|D = \frac{p_1}{p_1+p_2-2p_1p_2} \times 100-101$

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

$P_1 = \left\lfloor \frac{£100}{1-0.6} \right\rfloor = £250$ 

$P_2 = \left\lfloor \frac{£100}{1-0.4} \right\rfloor = £166$

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.

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

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

$E(M_1|S) = \frac{(1-p_1)(1-p_2)}{p_1p_2+(1-p_1)(1-p_2)}\times\frac{100}{1-p_1}-101$

$E(M_1|S) = \frac{1-p_2}{2p_1p_2-p_1-p_2+1} \times 100-101$

$E(M_2|S) = \frac{(1-p_1)(1-p_2)}{p_1p_2+(1-p_1)(1-p_2)} \times \frac{100}{1-p_2}-101$

$E(M_2|S) = \frac{1-p_1}{2p_1p_2-p_1-p_2+1} \times 100-101$

$E(M_1|D) = \frac{(1-p_1)p_2}{(1-p_1)p_2+p_1(1-p_2)} \times \frac{100}{1-p_1}-101$

$E(M_1|D) = \frac{p_2}{p_1+p_2-2p_1p_2} \times 100-101$

$E(M_2|D) = \frac{p_1(1-p-2)}{(1-p_1)p_2+p_1(1-p_2)} \times \frac{100}{1-p_2}-101$

$E(M_2|D) = \frac{p_1}{p_1+p_2-2p_1p_2} \times 100-101$

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

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

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

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

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

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