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

  • Thread starter Philip Wong
  • Start date
  • Tags
    Dice
In summary: So it would be the probability of getting 1 out of 3 rolls correct and 2 out of 3 incorrect. To get the probability of getting 1 out of 3 rolls correct, you would have to multiply this by 3, since there are 3 possible ways to get just one 6. So it would be 3 * {3\choose 2} \cdot (1/6) \cdot (5/6)^2 = 45/192 = 15/64.
  • #1
Philip Wong
95
0
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.
 
Physics news on Phys.org
  • #2
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 [itex] 3 \cdot 0.023.. \sim 0.069 [/itex]

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

Good luck getting "unrusted"! :wink:
 
  • #3
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.
 
  • #4
Philip Wong said:
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.
 
  • #5
chiro said:
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.

viraltux said:
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 [itex] 3 \cdot 0.023.. \sim 0.069 [/itex]

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

Good luck getting "unrusted"! :wink:

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:
[itex]{3\choose 2} \cdot (1/6)^2 \cdot 5/6 [/itex]

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

thanks
 
  • #6
Philip Wong said:
So if I used what I've provided before, i.e. [itex]{3\choose 2} \cdot (1/6) \cdot (5/6)^2 [/itex] 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.
 

1. What is the probability of rolling two 6's in three rolls of a single die?

The probability of rolling two 6's in three rolls of a single die is 1/36. This is because the probability of rolling a 6 on a single roll is 1/6, and the probability of rolling two 6's in three rolls is (1/6)^2 = 1/36.

2. How does the number of dice rolled affect the probability of getting two 6's?

The more dice that are rolled, the higher the probability of getting two 6's. For example, the probability of rolling two 6's in three rolls of two dice is 1/6, while the probability of rolling two 6's in three rolls of three dice is 1/6 * 1/6 = 1/36. This is because the chances of rolling a 6 on any single die is 1/6, and the more dice rolled, the more chances there are of rolling a 6.

3. Is the probability of getting two 6's in three rolls the same for a pair of dice and a single die?

No, the probability is not the same. The probability of getting two 6's in three rolls of a single die is 1/36, while the probability of getting two 6's in three rolls of two dice is 1/6. This is because when rolling two dice, there are more possible combinations that can result in two 6's (such as a 6 on one die and a 6 on the other) compared to rolling a single die.

4. Is it possible to calculate the exact probability of getting two 6's in three rolls?

Yes, it is possible to calculate the exact probability using probability theory and the concept of independent events. The probability of getting two 6's in three rolls can be calculated as the product of the probability of rolling a 6 on a single roll, raised to the power of two (since we want two 6's), and the probability of not rolling a 6 (which is 5/6) raised to the power of the remaining rolls (in this case, one roll).

5. Can the probability of getting two 6's in three rolls be influenced by external factors?

No, the probability of getting two 6's in three rolls is solely dependent on the number of dice rolled and the probability of rolling a 6 on a single die. It cannot be influenced by any external factors such as lucky charms or previous rolls.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
994
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
  • Set Theory, Logic, Probability, Statistics
2
Replies
41
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
16
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
32
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
1K
  • Set Theory, Logic, Probability, Statistics
2
Replies
42
Views
4K
  • Set Theory, Logic, Probability, Statistics
Replies
11
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
15
Views
1K
Back
Top