3 Dice totals - Same result three rolls

If I understand correctly the odds of a specific triple thrown three times in a row would be 1/36*1/36*1/36 = 1 in 46,656?
If I already have the first outcome, then 1/36*1/36 = 1 in 1296?

Assuming I get the basics, which I am not sure I do, can some one help with my actual question as this is too confusing for me?

If I have three dice and the outcome of the first roll is known, what is the probability of repeating the same combination a further two times for a total of thee?

Isn't the odds different depending on the initial 3 dice combination? For example an initial throw of 6-6-6 would be more difficult to repeat a further two times than 1-2-3 due to the various combinations of the latter? (for my purposes 1-2-3 is the same regardless of which dice show which number).

I would be very grateful for any help with this.

I think you're off on the triples. the odds of getting a 6 on each die is 1/6. so the odds of getting 666 is 1/6^3, which is 1:216. The odds of doing it 3 times in a row is (I think) 1/216^3, which is about 1:10,000,000. the odds of doing it twice in a row (or 3 times after the first time) is about 1:46,650.


123 is easier, because the first die can be any one of the numbers (3/6), the second can be either of the other two (2/6), and only the third needs to be a specific number (1/6). So the odds of getting 123 is (3/6)*(2/6)*(1/6)=1/36.

I agree with the rdw4potus post above.

Quote: rdw4potus

I think you're off on the triples. the odds of getting a 6 on each die is 1/6. so the odds of getting 666 is 1/6^3, which is 1:216. The odds of doing it 3 times in a row is (I think) 1/216^3, which is about 1:10,000,000. the odds of doing it twice in a row (or 3 times after the first time) is about 1:46,650.


123 is easier, because the first die can be any one of the numbers (3/6), the second can be either of the other two (2/6), and only the third needs to be a specific number (1/6). So the odds of getting 123 is (3/6)*(2/6)*(1/6)=1/36.

Hi rdw, thanks for the quick reply.

I think I wasn't very clear in my first sentence. I was thinking any triple from 1-1-1 through to 6-6-6. So there are six possibilities out of 216 total combinations. 6/216*6/216*6/216* or 1/36*1/36*1/36??? Am I still wrong :(

If I'm still not getting it can I just confirm I understand that if the current roll is known, i.e. 6-6-6 the probability the second and third rolls repeat are 1:46,650.

Thanks in advance for your help

If your first roll is 666 its 1 in 46656 that the next 2 rolls are 666
If your first roll is 554 its 1 in 5184 that the next 2 rolls are 554
If your first roll is 321 its 1 in 1296 that the next 2 rolls are 321

Is that what you wanted or something else?

Yes, exactly what I was trying to find, thank you.

I was trying to work out what is the lowest probability of two repetitions and the highest. 1-1-1 through to 6-6-6 would be the highest probability at 1 in 46,656?

Would you be able to tell me what combination is the lowest? Would your 123 example be that? That is, would any combination that is three different numbers, i.e. 123, 234, 456 have a probability of 1 in 1296?

Likewise in your second example, a double number and a single number, 554, or maybe 112, 113, 223 etc all have odds of 1 in 5184?

Thanks again for your help!

If what you are asking is: if you roll the dice three times, what is the probability of getting tripples? i was looking for the same answer and know understand it. i was tutored in this subject. it is 6/216 or 1/36 chances.( 6*6*6= 216 and there are six possible numbers.) 1/36 is you answer or in percentage terms: 2.8%