| JackWhite | 25 Oct 2008 9:13 a.m. PST |
While playing a board game in which a five and six was a hit on a six-sider, one of the players claimed that rolling two of them gave him a 67% chance of success. I'm no math wizard, but even I know that all he has is two chances at 33%. What I don't know is how much his chances do improve. At least I would hope that rolling a hundred six siders would give a higher percentage than rolling just one. How much do the odds improve per die rolled? JW |
Formerly Regiment Games  | 25 Oct 2008 9:24 a.m. PST |
It's 2/3 that he won't hit times 2/3 that he won't hit with the other. So it's 4/9 he won't hit, leaving 5/9, about 55%, that he'll hit with at least one of them. Notice that 1/3 of a chance times 1/3 of a chance gives 1/9, and is not the formula! It's one (100%) minus the odds that you WON'T hit. |
| x42brown | 25 Oct 2008 9:32 a.m. PST |
First dice success 1/3
Second dice succeeds when first fails 2/3 x 1/3 = 2/9
chance of at least one success = 1/3 + 2/9 = 5/9 = 56% approx a good bit less than he expects. x42 |
| Daffy Doug | 25 Oct 2008 11:03 a.m. PST |
I never worry about such questions: no matter what the odds, I usually roll what I don't need…. |
 Editor in Chief Bill  | 25 Oct 2008 11:33 a.m. PST |
Maybe his dice are special? |
 John the OFM  | 25 Oct 2008 12:30 p.m. PST |
Loaded dice skew the odds. |
| JackWhite | 25 Oct 2008 3:06 p.m. PST |
That's still a lot higher than I would have expected. I wouldn't have thought it would have been more than a few percentage points per extra die. Players laugh at me when I play a game in which I should roll only one die and roll two, designating the result die. My Arkham Horror Group always tries to desuade me from using two clue tokens for rerolls, not wanting me to "waste" them. I always thought it was because the dice knocked together, exchanging pips, causing one to roll high and the other low. JW |
| JackWhite | 25 Oct 2008 3:17 p.m. PST |
Regiment Games Let me see if I have the math right. So with three dice, there is an 8/27 chance of failure, which gives a 19/27 chance of success? That's a tad better than 2/3 the other way. Those are pretty good odds, and it is now surprising that there are so many "misses", especially when rolling higher multiples of dice, say five or more. Now that I know that, I'll never lose another die roll again. (Heh-Heh) (Just kidding, dice gods.) :-) (See? I'm grinning and everything.) Of course, needing two or more hits now complicates the math that much more. JW |
| JackWhite | 25 Oct 2008 3:21 p.m. PST |
I never noticed the "edit" function before now. I really like that feature. Kudos, Bill. JW |
| pphalen | 25 Oct 2008 6:50 p.m. PST |
"Never tell me the odds!" |
| pphalen | 25 Oct 2008 6:50 p.m. PST |
Besides, what Doug said…
I could probably find a way to fail a die roll with a 100% cahnce of success! |
| Whatisitgood4atwork | 25 Oct 2008 9:06 p.m. PST |
Or the same calculation arrived at a slightly different way: There are 36 possible outcomes: 1–1, 1–2, 2–1,….6–6. 20 of those 36 outcomes have at least one 5 or 6 in them. Or 55.55'%. |
| Whatisitgood4atwork | 25 Oct 2008 9:14 p.m. PST |
With 3 dice, there are 216 possible outcomes, and the chances rise to just over 70% of rolling at least one 5 or 6. So the odds your friend was counting on are closer to the odds when using 3 dice. And of course, no matter how many dice you use, the odds of getting a 5 or 6 are never quite 100%. |
