What has the higher probability?

Question

The problem goes as following : Which of the following has a higher probability of happening :
a) person rolls 4 dice at once, one of them must must fall at 6
b) person rolls 2 dices 24 times, at least one of those 24 times the combo 1-1 must fall?

Any help will be greatly appreciated.

Answer 1

If you work by complement the solutions are much easier

a) $p_a = 1 - p(\text{ no dice falls on 6 }) = 1 - \left(\frac{5}{6}\right)^4 \approx 0.517$

b) $p_b = 1 - p(\text{ no 1-1 combination }) = 1 - \left(\frac{35}{36}\right)^{24} \approx 0.491$

Answer 2

Case 1-

$p(6) = \frac16$

Then,

$q(Other) = 1 - \frac16 = \frac56$

Now according to bionamial distribution

$P(X=r) = \binom nr × (q)^{n-r} × (p)^r$

Here X = 1 and n = 4.

$P(X=1) = \binom 41 × (\frac 56)^3 × (\frac 16)$

Case 2-

At least 1 pair of (1,1) = 1 - None cases with (1,1)

$= 1 - (\frac{35}{36})^{24}$

Answer 3

Case 1 - Two dices each with a probability of rolling 6 at 1 / 6th. Since you need just one of them to be 6. Total probablility is

$$\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{2}{3}$$ $$66.66\%$$ There is a 66.66% probability that one of your two dices will roll a six.

Case 2 -

2 dices rolled 24 times,
number of combinations / roll is given by $\frac{1}{6^{2}}$ since we have 6 possibilities and each with a combination of six more possibilities on die two. Do this 24 times and total probability becomes.

$$=24 * \frac{1}{6^{2}}$$ $$=\frac{3}{4}$$ $$=75\%$$ There is a 75% probability that one of your 24 rolls will result in a 1,1 roll.

Case 2 has a higher probability.

Edit: If you feel my answer is wrong I'd like to know why?