"Help me with some D6 odds..." Topic

I never was overly good with probabilities:

1. What are the odds of me rolling equal or greater than my opponent on 1D6?

2. What are the odds of me rolling greater than my opponent on 1D6?

Thanks, maths-whizzes!

They are both 1 in 6 chances. There is always a 1 in 6 chance any given side will come up on a die roll. It is like flipping a coin. If you are looking at a single instance flip there will always be a 50% chance of getting either side (whether getting the same as your opponent, or the flip side). Just think if you flip 99 heads in a roll you will still have a 50% chance of getting a head on your next flip (well, approximately 50% as there is a very minute chance of getting the edge). So for a single instance there is always a 1 in 6 chance of getting any particular number, whether it is higher or equal to your opponent. The odds, of course, change if you are looking at finding the odds for a given number of rolls.

Forest

equal or greater 7 in 12.
greater 5 in 12.
That is you both throwing the dice together then comparing dice
x42

Yep.
Mght be more insightful to think of it as
21/36 and
15/36

Drawing up a 6x6 table and counting squares always works when only dealing with 2 dice.

To further split the results:

6/36 Chance both players roll equal
15/36 Chance player A rolls higher
15/36 Chance player B rolls higher

Griefbringer

you will increase your odss dramatically if you throw more dice than your opponent.

because someone had to say it.
R.

I just had this asked of me…

What are the odds of six players all rolling a six followed by a five?

This is for the B-17 solitaire game, where in a specific table a 2d6 die roll is really a 1d6X+1d6Y (X=[1d6 x 10] and Y=[1d6 x 1])…

So the question really was what are the odds of six players all rolling a 65?

Would it still the example that Griefbringer gave?

Thanx
Rallypoint

What are the odds of six players all rolling a six followed by a five?

Well, a 6-5 is 1/36, so it would be 1/36 to the 6th power:
1 in 2,176,782,336. This applies if the dice are distinct rolls: I roll once to get a 6, then roll again for a 5.

Thanx, rktsci.

In theory, there are two distinct rolls… in reality you designate two die, one for 10's and one for 1's and then roll. (wow, I feel like I'm back in school!)

Should work the same.

Someone with better skills is welcomed to correct me (it's been a while since my probability and combinatorics course) but the odds are different if you roll two dices at the same time or one dice at a time.

Propability to get a specific result on a D6 is 1/6. Propability to get a specific sequence n times is 1/6 to the power n. While you are asking about six players rolling at the same time, it's really no different than rolling a single dice 12 times in a row, thus propability to get 6 and then 5 six times in a row would be 1/6 to the power of 12.

Things get slightly more complicated if you roll both dices at the same time. To get a six and a five is still 1/6 x 1/6. But since both dices have a six and a five, chance get that result is doubled (the difference is that it doesn't matter any longer which dice will roll a six or five). Thus for two dices to roll 6 and 5 would be 1/18. Chance to roll that result six times in a row (or simultanously) is 1/18 to the power of 6.

Thanks for the help!

<quote>I never was overly good with probabilities:

1. What are the odds of me rolling equal or greater than my opponent on 1D6?

2. What are the odds of me rolling greater than my opponent on 1D6?

Thanks, maths-whizzes!</quote>

it looks as though everybody has answered with the actual lab condition probabilities. but that of course doesn't take chaos theory into consideration.

For that, you need to specify WHO your opponent is.

A prime example is when playing any dice based game against your children: In these instances, the probability of rolling higher when you want lower, and vice vesa are significantly increased.

I played a board game with my son last night. He beat me by a long stretch by rolling a combination of 4,5,6's whilst I was constantly rolling 1 (I kid you not!) 4 times in a row, which must go against normal probability!

the only time I threw a higher number – (3), it landed me on the GO BACK TO START square. He then rolled the finishing move.

I never was overly good with probabilities:

1. What are the odds of me rolling equal or greater than my opponent on 1D6?

2. What are the odds of me rolling greater than my opponent on 1D6?

Thanks, maths-whizzes!