The probabililty of 3 rolls of dice and get two 6's

[Philip Wong]

I was asked to help someone to work out the probability above. It was more than a year since I done similar questions and thing do get rusty... So I hope to work it out here and get point out where I did wrong before I show that person the correct answer.

Intuitively there is three ways I came up to solve the problem, which apparently gave different answers. So it's either one or two or all three methods are wrong. Methods I came to mind are:

1) 1/6 * 1/6 * 5/6 = 0.023 -> this is the first idea came to my mind, since there is only 3trials and the probability of getting a 6 is 1/6, so I use the multiply rule to times things together. The results seems rather small, so I doubt this is the correct method.

2) 3C2 * 1/6 * (5/6)^2 = 0.347 -> I got this by using the binomial equation

3) 3C2 * 1/6 * 5/6 = 0.416 -> I got this by using the Bernoulli equation i.e. n*p*q. where n is the number of all possible outcomes, p is the probability of getting a 6, and q is the probability of not getting a 6.

Please check which methods I used was correct, and if neither of them was correct please suggest a way to work this out.

 

[viraltux]

Hi Philip!

The first one is wrong because you just consider the case [6, 6, x], you need to consider also the cases [6, x, 6] and [x, 6, 6], that is 3 \cdot 0.023.. \sim 0.069

The second case is wrong because you raised to the power of two the wrong probability
{3\choose 2} \cdot (1/6)^2 \cdot 5/6 \sim 0.069 which also explains why you also got wrong the Bernoulli formula.

Good luck getting "unrusted"!

 

[jedishrfu]

usually this is solved by considering the converse probability: what's the probability of not getting any 6's in three rolls and then use the 100% - P to get the probability that you want.

 

[chiro]

I was asked to help someone to work out the probability above. It was more than a year since I done similar questions and thing do get rusty... So I hope to work it out here and get point out where I did wrong before I show that person the correct answer.

Intuitively there is three ways I came up to solve the problem, which apparently gave different answers. So it's either one or two or all three methods are wrong. Methods I came to mind are:

1) 1/6 * 1/6 * 5/6 = 0.023 -> this is the first idea came to my mind, since there is only 3trials and the probability of getting a 6 is 1/6, so I use the multiply rule to times things together. The results seems rather small, so I doubt this is the correct method.

2) 3C2 * 1/6 * (5/6)^2 = 0.347 -> I got this by using the binomial equation

3) 3C2 * 1/6 * 5/6 = 0.416 -> I got this by using the Bernoulli equation i.e. n*p*q. where n is the number of all possible outcomes, p is the probability of getting a 6, and q is the probability of not getting a 6.

Please check which methods I used was correct, and if neither of them was correct please suggest a way to work this out.

Hey Philip Wong.

The best way I think to do this is to use a binomial distribution with 3 trials where one state corresponds to rolling a 6 and the other not rolling a 6. Each dice roll is independent (assumed) and has the same probabilities. In this case p = 1/6 and 1-p = 5/6

This means your probability is given 3C2 p^2 x (1-p) = 3C2 x 5/192 = 15/192

You can visualize this by drawing a tree diagram of the events on paper. Use two events: a 6 for one and a non-6 for everything else and then you will get a tree with 8 possibilities with three levels of depth.

 

[Philip Wong]

Hey Philip Wong.

The best way I think to do this is to use a binomial distribution with 3 trials where one state corresponds to rolling a 6 and the other not rolling a 6. Each dice roll is independent (assumed) and has the same probabilities. In this case p = 1/6 and 1-p = 5/6

This means your probability is given 3C2 p^2 x (1-p) = 3C2 x 5/192 = 15/192

You can visualize this by drawing a tree diagram of the events on paper. Use two events: a 6 for one and a non-6 for everything else and then you will get a tree with 8 possibilities with three levels of depth.

Hi Philip!

The first one is wrong because you just consider the case [6, 6, x], you need to consider also the cases [6, x, 6] and [x, 6, 6], that is 3 \cdot 0.023.. \sim 0.069

The second case is wrong because you raised to the power of two the wrong probability
{3\choose 2} \cdot (1/6)^2 \cdot 5/6 \sim 0.069 which also explains why you also got wrong the Bernoulli formula.

Good luck getting "unrusted"!

Hi guys,
Thanks for your help. So what I should do if I were to keep using the Bernoulli formula (n.p.q) what I should really do is:
{3\choose 2} \cdot (1/6)^2 \cdot 5/6

I knew I had either misplaced or missed something in the formula.
So if I used what I've provided before, i.e. {3\choose 2} \cdot (1/6) \cdot (5/6)^2 what it really meant was, the probability of only getting one 6 out of the three dice roll right?

thanks

 

[viraltux]

So if I used what I've provided before, i.e. {3\choose 2} \cdot (1/6) \cdot (5/6)^2 what it really meant was, the probability of only getting one 6 out of the three dice roll right?

thanks

Nope, it would be the probability of getting anything but 6 in both dices.