Which is the probability formula for tossing a dice until the rth six appears?

Question

Find a formula for the next cases:

a. toss a coin until the 2nd head occours (α = number of tosses)
b. toss a dice until the 2nd six occours (α = number of tosses)
c. toss a dice until the 3rd six occours (α = number of tosses)
d. toss a dice until the rth six occours (α=number of tosses)

Answer 1

for the coin, you have a 1in 2 chacen of hitting heads.
For the die, you have a 1 in 6 chance of hitting a six.

For either case, each role is independent of the other, meaning just because you flipped the coin and got tails doen't mean the next will be definitely heads. Still 50/50 on every flip. The same independence for the dice.

So for a) (1/2)(1/2) = (1/2)^2 = 1/4 or 25%
b) (1/6)(1/6) = (1/6)^2 = 1/36 or 2.78%
c) (1/6)(1/6)(1/6) = (1/6)^3 = 1/216 = 0.463%
so for d)
(1/6)^r = 1 / (6^r)

Answer 2

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a.
Probability of tossing non-1 on first roll: 5/6
Probablity of tossing 1 on second roll: 1/6
5/6 * 1/6 = 5/36 (13.89%)

b.
Probability of tossing non-1 on first roll: 5/6
Probability of tossing non-1 on second roll: 5/6
Probability of tossing 1 on thrid roll: 1/6
5/6 * 5/6 * 1/6 = 25 / 216 (11.57%)

c.There is a probablity of 5/6 for every non-1 roll, and a probability of 1/6 for the last roll (for the 1). Therefore, the probility that the first 1 appears on the kth toss is
(1/6) * (5/6)^(k-1)

Answer 3

prob is 1 / (6 exponent r)