Probability: getting two of the same when rolling 2 or more dices

[Nikitin]

Hate over probability!

I just can't figure this out: what is the probability of rolling a pair (like, 6,6 or 3,3) when throwing n amount of dices? I mean, let's say I throw 3 dices total, shouldn't the amount of outcomes be something like this: 6*1*6 ? Or Maybe 6*1*6 should be multiplied by 3!/2! ? Just a (most likely wrong, I think) hunch I'm having

 

[HallsofIvy]

If you roll three dice You can get "66N", "6N6", or "N66" where "N" stands for "anything other than a 6" The probability of "6" is 1/6, the probability of "N" is 5/6 so the probability of anyone of "66N", "6N6", or "N66" is (5/6)(1/6)(1/6). The probability of 2 "6"s with 3 dice is 3(5/6)(1/6)(1/6).

Generally, the probability of rolling exactly 2 "6"s in n rolls is (1/6)^2(5/6)^{n-2} times the number of ways of ordering 2 "6"s and n-2 "N"s which is, of course, N!/2!(n-2)!, the binomial coefficient _nC_2.
That is, the probability or rolling exactly 2 "6"s in n rolls of the die is
\frac{n!}{2!(n-2)!}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{n-2}


By the way, "dices" is not a word. "Dice" is already plural- it is the plural of "die".

 

[chiro]

Hate over probability!

I just can't figure this out: what is the probability of rolling a pair (like, 6,6 or 3,3) when throwing n amount of dices? I mean, let's say I throw 3 dices total, shouldn't the amount of outcomes be something like this: 6*1*6 ? Or Maybe 6*1*6 should be multiplied by 3!/2! ? Just a (most likely wrong, I think) hunch I'm having

Just a tip for you since i had problems when I did probability in high school, but since then have understood it a lot better.

When it comes to starting probability or dealing with situations that aren't in standardized models (like binomial, geometric, poisson distribution etc) draw a tree diagram.

The tree diagram will help you visualize probabilities and help you calculate them easily through graphical means. You can then use this to understand the models that you will commonly use (like binomial for example), and all of those 'urn' problems won't seem so bad.

The tree diagram works by having successive events branch off of the associated prior event. Each branch leads to a node with a specific probability. To get the probability of a final event (which is a leaf of your tree), you just multiply the probabilities from the leaf all the way to the root to get your final probability.

Hope that helps!

 

[Nikitin]

First of all, sorry for my english

thanks for the advice, Chiro. I'l make sure to do it the next time! though hopefully I don't have to deal with this probability getting harder when I start studying engineering at the university in 2 years...

Ivy, due to my terrible OP (my fault 100%), you misunderstood me. I'l try to improve my writing.

I was asking for the probability of throwing at least two same numbers in 3 rolls of the die. While your model explains the probability of exactly getting 2 specific numbers in 3 rolls of the die... I tried using your way of thinking and create a proper formula for my problem, but I can't seem to do it. Can anyone pls help?

edit: oh and thanks for explaining the entire dice-die thing

 

[arildno]

By the way, "dices" is not a word. "Dice" is already plural- it is the plural of "die".

That is not entirely correct, Halls.
Once the number of sides exceeds 8, it is called "a dice", with plural "douse".

 

[Nikitin]

aaaah thank god I just came back to think on this and I think I finally solved it, and the solution was so simple all along! :-/

Basically, you can't make one single formula for my problem, but you need to make several binomial-probability calculations and add them up. Yes, sorry, my math-english is pretty bad. I'l make an example instead:

What is the probability to get at least two same numbers in 4 rolls of the die? Answer:

[(6/6)*(1/6)*(5/6)^2]*(4C2) + [(6/6)*(5/6)(1/6)^2]*(4C3) + [(6/6)*(1/6)^3]*(4C4) = the probability to get at least two same numbers in 4 rolls of the die

Argh it was just a simple binomial distribution.. damn

 

[JohnMorriss]

Any time I see words like "at least", or "any pair", I immediately think about reversing the question!

In this case, it is esier to find the probability of ALL the rolls being different, and then subtract this from 1. You avoid the danger of missing counting one of the ways of accomplishing the double result.

What is the probability of getting at least one number at least twice when you roll 4 eight-sided dice?

Switch to: what is the prob that all the rolls will be different?

First roll has 8/8 chance of being different (there's nothing for it to match!)
Second is 7/8; eight possible values and one's already used
Third is 6/8, and fourth is 5/8

Multiplying, we get 0.410156...as the probability of four different results, so 0.58984...as the prob of at least one pair.

Note that this method will give a value of exactly zero for the probability of all different if we roll 9 eight-sided dice...

 

[uart]

aaaah thank god I just came back to think on this and I think I finally solved it, and the solution was so simple all along! :-/

Basically, you can't make one single formula for my problem, but you need to make several binomial-probability calculations and add them up. Yes, sorry, my math-english is pretty bad. I'l make an example instead:

What is the probability to get at least two same numbers in 4 rolls of the die? Answer:

[(6/6)*(1/6)*(5/6)^2]*(4C2) + [(6/6)*(5/6)(1/6)^2]*(4C3) + [(6/6)*(1/6)^3]*(4C4) = the probability to get at least two same numbers in 4 rolls of the die

Argh it was just a simple binomial distribution.. damn

I hadn't been following this thread so I only noticed this mistake when John (above) bumped it. OP I'm not sure if you're still following this but you've made a fundamental mistake (and a very common mistake) in how you've added probabilities there!

The addition of the probabilities for the "2 same" "3 same" "4 same" cases is fine, however where you multiplied these probabilities by 6 (which is effectively adding the probability for the repeated ones, repeated twos, repeated threes ... cases) is INCORRECT.

Can you see why? Here's a hint, it would have been ok to do that for the case of fewer than 4 dice but it is incorrect (gives too higher probability) for the case of 4 or more dice.

 

[HallsofIvy]

That is not entirely correct, Halls.
Once the number of sides exceeds 8, it is called "a dice", with plural "douse".

One difficulty with the internet is that one cannot administer what is often the appropriate response- a two by four about the head and shoulders!

 

[arildno]

One difficulty with the internet is that one cannot administer what is often the appropriate response- a two by four about the head and shoulders!