Math People...What are these odds?

Djeta Thernadier

First Post
Just got back from a gaming session during which MojoGM and I rolled the same number SIX TIMES in a row at the exact same time.

First we rolled the same initiative. Then the same attack. Then we did that again. Then we noticed it was weird so we each rolled a random roll to see what would happen. We both got a 10. Then we both rolled the same initiative again (we both got natural ONES that time).

Can someone with some knowledge of statistical odds give me the odds of this? We're both still amazed...

Djeta
 

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these were d20 rolls?

then the odds are 1/20 x 1/20 x 1/20 x 1/20 x 1/20 x 1/20.

0.0000015625% or 1 out of 64,000,000.

quick! buy a lottery ticket! :D
 
Last edited:


You set up a condition with your first die roll that he had to match. He had a 1 in n chance of getting that number (n is the number of sides on the die). So, to find the total, you just do 1/(n<sub>1</sub>*n<sub>2</sub>*n<sub>3</sub>*n<sub>4</sub>*n<sub>5</sub>*n<sub>6</sub>) . Just substitute in the appropriate n's.

If they were all d20's, then it would indeed be 1 / 64 000 000
 

I disagree. The odds in rolling ANY number on a D20 are one in twenty... The odds of two people rolling the same number on 1D20 are 1/20 x 1/20, or one in four hundred. The odds of two people rolling the same number (on 1D20) six times in a row are 400 to the sixth power (or about one in 4,096,000,000,000,000 - a bit over four quadrillion). Assuming I did the math correctly. :p
 

Steverooo said:
I disagree. The odds in rolling ANY number on a D20 are one in twenty... The odds of two people rolling the same number on 1D20 are 1/20 x 1/20, or one in four hundred. The odds of two people rolling the same number (on 1D20) six times in a row are 400 to the sixth power (or about one in 4,096,000,000,000,000 - a bit over four quadrillion). Assuming I did the math correctly. :p

Nope. Those would be the odds if the question was what are the odds of 2 people rolling a 10 and then rolling a 1 and then rolling a 6 and then rolling ....

However I think she is in fact asking what are the odds of the second person always rolling the same as the first person ... so what the first person rolls is irrelevant and it's only the second person's roll that counts and they have a 1 in 20 chance of rolling whatever the first person rolled.
 

Correct.

It's pretty unlikely, but not that unlikely ;)

That's the problem with statistics. Most people could do the math, but they don't know what's in question.

Don't worry, Steverooo, it happens to the pros too (more often than they'd like to admit).
 

Steverooo said:
I disagree. The odds in rolling ANY number on a D20 are one in twenty... The odds of two people rolling the same number on 1D20 are 1/20 x 1/20, or one in four hundred. The odds of two people rolling the same number (on 1D20) six times in a row are 400 to the sixth power (or about one in 4,096,000,000,000,000 - a bit over four quadrillion). Assuming I did the math correctly. :p

The odds in rolling ANY number on a d20 is 1. Each time you roll the dice you're 100% likely to get some number, if we count out any freak accidents like breaking the dice upon landing. So you're wrong.

If you want more proof, the chances for rolling the same dice once is equal to:

P(Djeta throws 1)*P(MojoGM throws 1)+
P(Djeta throws 2)*P(MojoGM throws 2)+
P(Djeta throws 3)*P(MojoGM throws 3)+
.
.
.
+P(Djeta throws 20)*P(MojoGM throws 20) =

20 * 1/20 * 1/20 = 1/20. Doing that six times in a row is equal to

(1/20)^6.


So the first answer was correct.
 


Psionicist said:
It is as impossible to roll 20,20,20,20,20 as it is to roll 10,13,6,5,16 or any other possible serie. :)


So what is upshot of this?

Neither Djeta nor Mojo can expect the win the lottery for A VERY LONG TIME. ;)
 

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