Why $N(d_1)$ and $N(d_2)$ are different in Black & Scholes

I'm struggling to understand the meaning of $d_1$ and $d_2$ in Black & Scholes formula and why they're different from each other.

As per the formula, $$C = SN(d_1) - e^{-rT}XN(d_2)$$

which means if the call option gets exercised, one would receive the stock and pay the strike price.

Clearly, the strike price payment is conditional on the option finishing in the money, ie. future value of this payment is $X \cdot \mathbb{P} (S_T > X)$ and thus discounted present value would be $$e^{-rT}X \cdot \mathbb{P} (S_T > X)$$

From this we can see that Black & Scholes $N(d_2)$ is the probability of the option exercise (under risk neutral measure).

But then it should logically follow that the first part of the formula - conditional receipt of the stock - should equally depend on the above probability, in which case its future value would be $S e^{rT} \cdot \mathbb{P} (S_T > X)$, with the present value $S \cdot \mathbb{P} (S_T > X)$, and so Black & Scholes formula "should be" $$C = SN(d_2) - e^{-rT}XN(d_2)$$

But the formula does use $N(d_1)$ and since $d_1 > d_2$ my understanding is that it gives higher probability to receiving the stock than to paying the strike price.

I went through Understanding $N(d_1)$ and $N(d_2)$: Risk-Adjusted Probabilities in the Black & Scholes Model and also found this explanation on Quora helpful, but still don't see what is fundamentally wrong with the train of thought I described above.

The time-$t$ price of a European call on a non-dividend paying stock with spot price $S_t$, when the strike is $K$ and the time to maturity is $\tau = T − t$, is the discounted expected value of the payoff under the risk-neutral measure $Q$ $$C(t,{{S}_{t}},K,T)={{e}^{-r(T-t)}}\mathbb{E}_{t}^{Q}\,[{{({{S}_{T}}-K)}^{+}}]={{e}^{-r\tau }}\mathbb{E}_{t}^{Q}[({{S}_{T}}-K\,)\,{{1}_{\,{{S}_{T}}>K}}]$$ where $1$ is the indicator function,thus we have $$C(t,{{S}_{t}},K,T)=\underbrace{{{e}^{-r\tau }}\,\mathbb{E}_{t}^{Q}[S_T\,1_{S_T>K}]}_{J}-K e^{-r\tau}\,\underbrace{\mathbb{E}_{t}^{Q}[1_{S_T>K}]}_{I}$$ We can therefore write $$I=\mathbb{E}_{t}^{Q}[ 1_{S_T>K}]=Q(S_T>K)$$ Indeed,the expected value $\mathbb{E}_{t}^{Q}[ 1_{S_T>K}]$ is the probability of the call expiring in-the-money under the measure $Q$.
Evaluating $J={{e}^{-r\tau }}\,\mathbb{E}_{t}^{Q}[S_T\,1_{S_T>K}]$ requires changing the original measure $Q$ to another measure $Q^S$.Consider the Radon-Nikodym derivative $$\frac{dQ^S}{dQ}=\frac{{{S}_{T}}/{{S}_{t}}}{{{B}_{T}}/{{B}_{t}}}$$ where $d{{B}_{t}}=r{{B}_{t}}dt$ or $B_t=e^{rt}$. As a result $${{e}^{-r\tau }}\mathbb{E}_{t}^{Q}[{{S}_{T}}{{1}_{{{S}_{T}}>K}}]={{S}_{t}}\mathbb{E}_{t}^{Q}\left[{{e}^{-r\tau }}\frac{{{S}_{T}}}{{{S}_{t}}}{{1}_{{{S}_{T}}>K}}\right]={{S}_{t}}\mathbb{E}_{t}^{Q}\left[\frac{{{S}_{T}}/{{S}_{t}}}{{{B}_{T}}/{{B}_{t}}}{{1}_{{{S}_{T}}>K}}\right]={{S}_{t}}\mathbb{E}_{t}^{{{Q}^{S}}}[{{1}_{{{S}_{T}}>K}}]$$ in othere words $$J={{e}^{-r\tau }}\,\mathbb{E}_{t}^{Q}[{{S}_{T}}{{1}_{{{S}_{T}}>K}}]={{S}_{t}}\mathbb{E}_{t}^{{{Q}^{S}}}[{{1}_{{{S}_{T}}>K}}]=S_t Q^S(S_T>K)$$ This implies that the European call price of Equation be written in terms of both measures as $$C(t,{{S}_{t}},K,T)=S_t Q^S(S_T>K)- {{e}^{-r\tau}}K Q(S_T>K)$$ Under measure $Q$, $S_T$ is distributed as lognormal with mean $m=\ln S_t+(r-\frac{1}{2}\sigma^2)\tau$ and variance $\Sigma=\sigma^2\tau$ then
$$Q(S_T>K)=N\left(\frac{m-\ln K}{\sqrt{\Sigma}}\right)=N\left(\frac{\ln(S_T/K)+(r-\frac{1}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}}\right)=N(d_2)$$
Note that, for the first term, it is \begin{align*} e^{-rT} \mathbb{E}(S_T \mathbb{I}_{S_T >X}), \end{align*} which is not equal to $e^{-rT}S\,\mathbb{P}(S_T >X)$. Here, $\mathbb{P}$ is the risk-neutral measure and $\mathbb{E}$ is the corresponding expectation operator.
Let $\tilde{\mathbb{P}}$ be the probability measure with the stock price process as the numeraire. Then \begin{align*} e^{-rT} \mathbb{E}(S_T \mathbb{I}_{S_T >X}) = S \tilde{\mathbb{P}}(S_T>X). \end{align*} That is, $N(d_1) = \tilde{\mathbb{P}}(S_T>X)$, while $N(d_2) = \mathbb{P}(S_T>X)$. The different probability measures caused the difference between $N(d_1)$ and $N(d_2)$. See also the discussion in Question Understanding $N(d_1)$ and how to use the stock itself as the numeraire?.
• @andreister: For paying the strike, as $X$ is a constant, we can write $\mathbb{E}(X\mathbb{I}_{S_T >X})$ as $\mathbb{E}(X)\mathbb{E}(\mathbb{I}_{S_T >X}) = X \mathbb{E}(\mathbb{I}_{S_T >X})$. However, for receiving the stock, as $S_T$ and $\mathbb{I}_{S_T >X}$ are not independent, you can not write $\mathbb{E}(S_T\mathbb{I}_{S_T >X})$ as $\mathbb{E}(S_T)\mathbb{E}(\mathbb{I}_{S_T >X})$. By a measure change, this can be simplified and valued. That is, for receiving the stock, a different measure is needed. – Gordon Jul 29 '15 at 20:54