Question regarding two six-sided dice

[mikemcsomething]

Hello everyone! I am having a problem here in a discussion. I will be writing a paper for a small online publication attempting to clear up the various math errors that occur during this process:

I have a question regarding probability - in a game where you roll one six-sided die, and if you roll less than a 3 (so a 2 or 1) you get one more chance to roll 3 or higher again by rolling a second six-sided die, are your chances of rolling a 3 or above 88% or 75%?

If I make a chart showing all of the possible rolls I get 12/16 or 75%, however 66% + 1/3 of 66% (22%) is 88%

If you would like some background, it's a tabletop wargame, and this is represented as the roll "To Hit" your target. You declare your shot, and roll to hit. If you roll a certain number or above (in this case, 3) you "hit". If you do not "hit", certain models on the board have the option of rolling one more time to try to hit again.

Thanks for your time everyone.

 

[soroban]

Hello, mikemcsomething!

[sie=120]You roll one six-sided die.
If you roll less than a 3 (a 2 or a 1), you get one more chance to roll 3 or higher again by rolling a second six-sided die.
Are your chances of rolling a 3 or above 88% or 75%? [/size]

As I understand the rules, we have two rolls to get 3-or-higher.


\(\displaystyle \text{For two rolls, there are }6^2 = 36\text{ outcomes.}\)

\(\displaystyle \text{The only outcomes in which we do }not\text{ get 3-or-higher are: }\1,1),\1,2),\2,1),\2,2).\)

\(\displaystyle \text{Hence, we get 3-or-higher on the other 32 outcomes.}\)

\(\displaystyle \text{Therefore: }\;P(\text{3-or-higher}) \;=\;\frac{32}{36} \:=\:\frac{8}{9} \;\approx\;89\%\)

 

[mikemcsomething]

It's actually one roll, and if you fail that initial shot you try again, so you would not be rolling the second six-sided die if you 'hit' the first time (you would not follow a success with a success). Would this adjust the odds of hitting? Since your total possible results are:

6 hit
5 hit
4 hit
3 hit
2 miss - reroll (3+ hits) so you can get 2,1 2,2 2,3 2,4 2,5 2,6
1 miss - reroll (3+ hits) so you can get 1,1 1,2 1,3 1,4 1,5 1,6
I show that as 16 total results with 12 hitting - would this be the correct way to do it as the second roll is dependent upon the first?

 

[Denis]

1st roll: 2/3 ; so 1/3 of time, 2nd roll required:
2nd roll: 1/3 * (2/3) = 2/9

Total: 2/3 + 2/9 = 6/9 + 2/9 = 8/9 = ~89%