Let’s say that on a roll of a 10-sided die a 7 or higher is considered a “success.” That means on each die roll there’s a 40% chance of getting a success (4/10).
Let’s say now that you’re rolling a batch of three 10-sided dice. How would you calculate the chances of getting either exactly no successes, exactly one success, exactly two successes, or exactly three successes?
Hit the spoiler for my failed attempt to do it:
[spoiler]Chances of getting no successes:
6/10 * 6/10 * 6/10 = .216 or ~21%
Chances of getting one success:
4/10 * 6/10 * 6/10 = .144 or ~14%
Chances of getting two successes:
4/10 * 4/10 * 6/10 = .096 or ~9%
Chances of getting three successes:
4/10 * 4/10 * 4/10 = .064 or ~6%
These should be the only four outcomes (no successes, one success, two successes, three successes) but their combined percent chances don’t add up to anywhere near 100%.[/spoiler]
Help would be appreciated. And no, it’s not for homework.
Nearly right. But you can get 1 and 2 each in 3 different ways, so the chances are 3 times as great as those you calculated. They all should add to 1.000 because you have to get either 0,1,2 or 3 right.
You’re on the right track, but you have to triple the probabilities for the “one success” and “two successes” options, since there are three ways of getting them. In other words:
(4/10 * 6/10 * 6/10) + (6/10 * 4/10 * 6/10) + (6/10 * 6/10 * 4/10) = 43.2%
(6/10 * 4/10 * 4/10) + (4/10 * 6/10 * 4/10) + (4/10 * 4/10 * 6/10) = 28.8%
These, added to the 21% and 6% you already got, sum to 100%.
Chances of getting no successes: ~21%
Chances of getting exactly one success: ~42%
Chances of getting exactly two successes: ~27%
Chances of getting exactly three successes: ~6%
Total: ~96% (didn’t hit 100 because of how I rounded)
Chances of getting at least one success: ~79%
Chances of getting at least two successes: ~33%
I just wanted to pipe in to say that it’s good that you’re doing the sanity check. Too many students would declare themselves done once they got numbers, and not care that they don’t add up to 100%.
If this is for tabletop roleplaying games like I expect it is (and if not that’s a crazy coincidence), you’re going to want to use formulas for larger numbers of dice. In this case, the binomial distribution is helpful: Binomial distribution - Wikipedia
It’s a little complicated but you can plug it into excel pretty easily. I made these charts for Shadowrun (where you roll 6-sided dice, and are looking for 5s and 6s for successes) using the formula:
As others have responded, this is an example of a Binomial Distribution:
You have a fixed number n of trials (the 3 die rolls)
You have a fixed and independent probability of “success” on each trial (the 40% chance of getting a 7 or better)
You want to know the probability of getting a specific number of successes (out of the n trials).
You can calculate this by using an appropriate formula, by looking it up on a table, by using a built-in calculator function, by using an online probability calculator, etc. I’m sure the links given above detail how to do at least some of these things.
To get the probability of no success, you have to multiply by the failure rate. The chance of each die yielding 1-6 is 60%. The chance of all three failing to top 6 would be .6x.6x.6, or 21.6%, leaving a 78.4% chance of hitting at least one.
Consider a forecast calling for a 40% chance or rain today, 10% chance tonight, and 30% chance tomorrow. What are the chances that it will rain during that time. Well, the chances of no rain are 60%, 90% and 70% in each day part. Multiplied out (.6x.9x.7), there is a 38% chance that all three of those periods will be rainless, and a 62% chance that it will rain on your newly-washed car at least once.
