A Quant's Gamble

Solution

Answer: -$117

Let \(W\) be the number of winning hands out of \(50\). If each hand is \(\$100\), then to have a \(3\%\) house edge, the expected profit per hand for the quant must be \[ 100\bigl(p - (1-p)\bigr) = 100(2p-1) = -3. \] \[ 2p - 1 = -0.03 \] \[ p = 0.485. \] So we model \[ W \sim \text{Binomial}(50,\,0.485). \] Now the quant’s blackjack profit after \(50\) hands is \[ X = 100(W - (50-W)) = 100(2W-50). \] So:

- he makes money when \(X>0\), which means \[ 2W-50>0 \quad \Longleftrightarrow \quad W>25; \] - he loses money when \(X<0\), which means \[ W<25; \] - he breaks even when \[ W=25. \] The blackjack EV itself is easy: \[ E[X] = 50(-3) = -\$150. \] Now let \(Y\) be the side-bet payoff with the friend:

- \(Y=+200\) if the quant loses money overall,
- \(Y=-200\) if the quant makes money overall,
- \(Y=0\) if he breaks even.

So \[ E[Y] = 200P(W\le 24) - 200P(W\ge 26). \] We can use a normal approximation. For the binomial distribution, \[ \mu = np = 50(0.485)=24.25 \] and \[ \sigma = \sqrt{np(1-p)} = \sqrt{50(0.485)(0.515)} \approx 3.53. \] The probability of finishing ahead is \[ P(W\ge 26) \approx P\left(Z \ge \frac{25.5-24.25}{3.53}\right). \] This gives \[ P(W\ge 26) \approx P(Z\ge 0.354). \] Using the normal table, \[ P(Z\ge 0.354) \approx 0.362. \] The probability of finishing behind is \[ P(W\le 24) \approx P\left(Z \le \frac{24.5-24.25}{3.53}\right). \] So \[ P(W\le 24) \approx P(Z\le 0.071). \] Using the normal table, \[ P(W\le 24) \approx 0.528. \] So the side bet has expected value \[ E[Y] \approx 200(0.528) - 200(0.362). \] \[ E[Y] \approx 200(0.166) = \$33.2. \] Therefore the quant’s total expected value is \[ E[X+Y] = E[X] + E[Y]. \] \[ E[X+Y] \approx -150 + 33.2 = -\$116.8. \] So the side bet helps, but it does not fully offset the casino disadvantage. The final answer is approximately \[ -\$117. \]