Edited appa_whitepaper.adoc with Atlas code editor

appa_whitepaper.adoc

@@ -101,10 +101,10 @@ As an additional firewall, a new key pair should be used for each transaction to
 ==== Calculations
 We consider((("calculations", "in Bitcoin whitepaper", secondary-sortas="Bitcoin whitepaper", id="calculate-whitepaper"))) the scenario of an attacker trying to generate an alternate chain faster than the honest chain. Even if this is accomplished, it does not throw the system open to arbitrary changes, such as creating value out of thin air or taking money that never belonged to the attacker. Nodes are not going to accept an invalid transaction as payment, and honest nodes will never accept a block containing them. An attacker can only try to change one of his own transactions to take back money he recently spent.
 
-The race between the honest chain and an attacker chain can be characterized as a Binomial Random Walk. The success event is the honest chain being extended by one block, increasing its lead by +1, and the failure event is the attacker's chain being extended by one block, reducing the gap by -1.
+The race between the honest chain and an attacker chain can be characterized as a ((("Binomial Random Walk")))Binomial Random Walk. The success event is the honest chain being extended by one block, increasing its lead by +1, and the failure event is the attacker's chain being extended by one block, reducing the gap by -1.
 
 ++++
-<p>The probability of an attacker catching up from a given deficit is analogous to a Gambler's Ruin problem. Suppose a gambler with unlimited credit starts at a deficit and plays potentially an infinite number of trials to try to reach breakeven. We can calculate the probability he ever reaches breakeven, or that an attacker ever catches up with the honest chain, as follows <a href="#ref_eight">[8]</a>:</p>
+<p>The probability of an attacker catching up from a given deficit is analogous to a ((("Gambler&#x27;s Ruin problem")))Gambler's Ruin problem. Suppose a gambler with unlimited credit starts at a deficit and plays potentially an infinite number of trials to try to reach breakeven. We can calculate the probability he ever reaches breakeven, or that an attacker ever catches up with the honest chain, as follows <a href="#ref_eight">[8]</a>:</p>
 ++++
 
 p = probability an honest node finds the next block