Probability of trade's exit orders being triggered in random-walk market

Question

When placing a trade with Stop Loss and Take Profit orders in a hypothetical random market (i.e. 0.5 probability of up tick and 0.5 probability of down tick), assuming:

x is the distance in ticks of the SL order from the entry price. y is the distance in ticks of the TP order from the entry price.

How do we calculate the probability of x being triggered first?

Answer 1

The answer can be found here under 1.3) Random Walk Hitting Probabilities (when events have equal probability of $\frac{1}{2}$ each).

\begin{equation} p(a) = \frac{b}{a+b} \end{equation}

$p(a)$ would be the probability of take-profit hit first. To look at probability of stop-loss being hit first, just take 1 minus the above, resulting with $a$ on the top (where $a$ is take profit and $b$ is hit stop loss level, respectively).

You can run this script I wrote in R, to verify:

tw<-0; d<-function() sample(c(-1,1),size=1)

#x=sl; y=tp
walk<-function(x,y){
for(i in 1:1000){
tw<- sum(tw+d())
if(tw== x || tw==y) break()}
return(tw)}

sl<- -10; tp<- 20
res<-replicate(1000,walk(sl,tp))
resx<-length(res[res==sl])
resy<-length(res[res==tp])
resx/(resx+resy)

Result for hitting x (stop loss first, where x= -10) is

resx/(resx+resy) = 0.673716

while, tp/(tp+stop) = 20/(20+10) gives 0.6666667, in agreement.