Dice Rolling: Stuck in the middle of question that instantly come into my mind

S

Selim47

New member
Joined
Dec 7, 2019
Messages
6
There are 2 pieces of 6 sided dice we have. We roll them 5 times one after another. What is the probability that the numbers on the front side are same (same numbers) at least one times of 5 turns?

What I did so far is simplfy the question like what if we roll once:

1st dice could be any number(6/6)
2nd dice have to be same number with 1st dice(1/6)

Then result for the problem is (6/6).(1/6)=1/6

What if I roll them 5 times and I get same numbers at least one times? I guess each turn are independent from each other and I can't go for this way.

For this time, I thought that I have to look from different angle like what are the chances NOT to get same results and then substarct from 1. Could you help me about it?
 
So there is a 1/36 chance that you are going to get a match of any number (1/6)x(1/6). The goal is to figure out the chances of this happening not five times in a row, but at least once out of five rolls.

Can you tell me which course you are taking so that I can see how to help?
 
Selim47 said:
For this time, I thought that I have to look from different angle like what are the chances NOT to get same results and then substarct from 1. Could you help me about it?
Click to expand...

This is exactly how I would approach this problem, using the complement rule as it will greatly simplify matters. Can you compute the probability for one roll of the two dice where you do not get a double?
 
Selim47 said:
There are 2 pieces of 6 sided dice we have. We roll them 5 times one after another. What is the probability that the numbers on the front side are same (same numbers) at least one times of 5 turns? What I did so far is simplfy the question like what if we roll once:
Click to expand...
Can you explain what in the world that means? In particular the pieces part.
 
I suspect that it means "two 6-sided dice"; that is, two pieces of a set of dice. It's understandable, though not standard English.

Both native English speakers and others have trouble with the word "dice", often not fully grasping that it is the plural of "die", and sometimes (as here?) thinking as if "dice" were a mass noun, like "dirt" (you can't talk about "a dirt", but about "a piece of dirt").
 
Dr.Peterson said:
I suspect that it means "two 6-sided dice"; that is, two pieces of a set of dice. It's understandable, though not standard English.
Both native English speakers and others have trouble with the word "dice", often not fully grasping that it is the plural of "die", and sometimes (as here?) thinking as if "dice" were a mass noun, like "dirt" (you can't talk about "a dirt", but about "a piece of dirt").
Click to expand...
Thanks for your complete and historical answer. I have never before understood the confusion. I am a seventh generation member of one of America's deepest south states, but the confusion of "die & dice" never was an issue. Even after eight years of collegiate study, I cannot remember a single instance that that came.
 
firemath said:
So there is a 1/36 chance that you are going to get a match of any number (1/6)x(1/6). The goal is to figure out the chances of this happening not five times in a row, but at least once out of five rolls.

Can you tell me which course you are taking so that I can see how to help?
Click to expand...
Can you please look at the entire sample space (36 entries) and see if there is only one match? 1/36 seems a bit low.
 
MarkFL said:
Yes, and so by extension, what is the probability of 5 successive rolls of the two dice not having a double?
Click to expand...
(5/6).(5/6).(5/6)(5/6).(5/6) = 55/65

So this is the quantity of total samples that we are NOT looking for.

Then answer should be; 1 - (55/65) = 0.598
 
Selim47 said:
(5/6).(5/6).(5/6)(5/6).(5/6) = 55/65

So this is the quantity of total samples that we are NOT looking for.

Then answer should be; 1 - (55/65) = 0.598
Click to expand...

Yes, the probability that we will not get doubles for 5 successive rolls is:

[MATH]\left(\frac{5}{6}\right)^5[/MATH]
Let's call this event \(X\). Let event \(Y\) be that we get at least one double in our 5 rolls (which is what we're actually trying to find). It is certain we will either get no doubles OR we will get at least one double. Thus:

[MATH]P(X)+P(Y)=1[/MATH]
Or:

[MATH]P(Y)=1-P(X)=1-\left(\frac{5}{6}\right)^5=\frac{4651}{7776}\approx0.598\quad\checkmark[/MATH]
 
The probability of getting the same numbers on both dice when you roll a pair of dice is NOT (1/6)(1/6)= 1/36. That is the probability of getting a specific number on both dice. Since there are 6 possible "specific" numbers, the probability of getting the same number on a pair of dice is 6(1/36)= 1/6. You could also argue that you could get anything on the first die, which, or course, have probability 1, then roll the same number, probability 1/6, on the second die. Again that gives (1)(1/6)= 1/6.
 
Jomo said:
Can you please look at the entire sample space (36 entries) and see if there is only one match? 1/36 seems a bit low.
Click to expand...

I'm trying to give a reference point here.

For rolling two times *in a row*, the probability is (1/36).
I wanted OP to go from there.
 
Last edited:
firemath said:
I'm trying to give a reference point here.

For rolling two times *in a row*, the probability is (1/36).
I wanted OP to go from there.
Click to expand...

If you wanted "any" double - 1,1 or 2,2 or 3,3 or 4,4, or 5,5 or 6,6 - the probability is 1/6

If you wanted a particular double (say 5,5) then the probability is 1/36
 
firemath said:
So there is a 1/36 chance that you are going to get a match of any number (1/6)x(1/6). The goal is to figure out the chances of this happening not five times in a row, but at least once out of five rolls.
Click to expand...
firemath said:
I'm trying to give a reference point here.

For rolling two times *in a row*, the probability is (1/36).
I wanted OP to go from there.
Click to expand...
You may have meant the right thing, but even now it isn't clear. Events have to be very clearly stated.

In the original, "match any number" could mean either "match any one specific number" (in which case you are right, but it's irrelevant to the problem), or "have the same number, whatever it is" (in which case you are wrong, but it is what the problem is about).

What you said just now is even less clear, and doesn't address the issue of definition. What are you rolling two times in a row?

The problem is about rolling the same number on both dice, for which the probability is 1/6 (as the OP correctly said). Your 1/36 is not a good place to start.
 
Subhotosh Khan said:
If you wanted "any" double - 1,1 or 2,2 or 3,3 or 4,4, or 5,5 or 6,6 - the probability is 1/6

If you wanted a particular double (say 5,5) then the probability is 1/36
Click to expand...
Now I see. So we aren't trying for a particular number?
 
OP said that s/he was looking for a "match" that was "the same"- which I took to mean that they were looking for a particular number.
 
You must log in or register to reply here.
Top