Question: Sum the Odds to One and Stop

I have a problem understanding how the below reasoning works
for Bruss article “Sum the Odds to One and Stop,”

i) I would like to understand the die problem and more specificaly
why the probability is k(1/6)^1*(5/6)^(k-1). I have a hard time understanding this

ii) How can I apply this reasonong to a coin toss example?

I have managed to plot the maximum at 6 for the die example but as I said
I dont understand the equation ie why you multiply by k etc

z := (1/6)*k*(5/6)^(k-1);
plot(z, k = 1 .. 20);

The maximum for a coin toss example seem to be at 1.5

z := (1/2)*k*(1/2)^(k-1);
[seq(evalf(z, 4), k = 1 .. 5, .5)];
plot(z, k = 1 .. 5);

iii) can anyone explain that?

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