Ok so I know most of the reasoning but I just need it explained a bit better because thinking about it is hurtng my brain.
So you have 2 chances of 20% rolls to hit your goal at least once, 2 D5 dice, win condition need one of them to land on 5 would be an example, the chances of either one hitting isn’t 40%, it’s actually more like 36%. I THINK this is because you are hitting 2 out of 2 in some realities so that’s wasted potential, therefore your odds of hitting at least one are lower than 40%… But I really don’t get that last bit at all.
In: Mathematics
20% of the time you’ll hit on the first die, 20% on the other.
However, 4% of the time you’ll hit on both dice, so simply the probabilities together counts that case twice. So you have to remove it, 20% + 20% – 4% = 36%.
Here is another way to look at it. Instead of rolling two dice, you roll on die twice. If you hit on the first roll, you stop, no need to roll again, you already won.
20% of the time you’ll win on the first roll, 80% you roll again. Of that 80 %, you win 20% of the time as well. 80% * 20% = 16%, 20% + 16% = 36%.
Consider the extreme case where you have 5 D5 dice. If you add 20% 5 times, you would get 100%. Obviously, it is not guaranteed that rolling 5 D5 dice will always result in a 5 being rolled, so the chance must be less than 100%. Therefore you can’t simply add the probabilities.
Let’s say your 20% chance is rolling a 2 dice with 5 sides. You win if it either comes up a 5.
All the possible rolls are:
(1,1)(2,1)(3,1)(4,1)(5,1)
(1,2)(2,2)(3,2)(4,2)(5,2)
(1,3)(2,3)(3,3)(4,3)(5,3)
(1,4)(2,4)(3,4)(4,4)(5,4)
(1,5)(2,5)(3,5)(4,5)(5,5)
There are 9 results with at least one 5, a win, of the 25 outcomes.
9/25 = .36 so a 36% win chance
Hopefully one of the explanations given is sufficient; assuming so, I won’t add another. But I did want to add that my favorite (layman) book on the subject is *The Drunkard’s Walk* by Leonard Mlodinow. And if someone would like to add their fave, that’d be great too.
The important part is the “at least one of two”. 20% of the time when you get one hit, you also get another hit, because the two are independent of each other. But that second hit doesn’t count, so to speak, because you already have one hit, which is all you need.
So there’s a 20% chance of a hit on the first, and then a 20% chance on the second, but also a 20% chance *of that chance* that you already have a hit. So it’s 20% plus (20% times 80%), for a total of 36%.
Plenty of correct answers above, but if you extend this out you can also see a practical example of why just adding the 20% together doesn’t work. If it did think of what would happen of you had 5 tries. 20%+20%+20+20%+20% =100% so if that system worked it would mean after 5 attempts you would be garunteed a success. In your example of rolling a D5 you can see how this doesn’t neccesarly work, assuming rolling a 5 is the success and 1-4 failure you can probably we see how even rolling 5 times would not automatically give you a successful roll. It further would get complicated if you had 6 possible attempts to roll that as 20% times 6 would mean a 120% chance of success, which is of course impossible.
With two rolls, you can go through the whole probability tree:
Roll 1 fail & roll 2 fail: 80% * 80% = 64%
Roll 1 success & roll 2 fail: 20% * 80% = 16%
Roll 1 fail & roll 2 success: 80% * 20% = 16%
Roll 1 success & roll 2 success: 20% * 20% = 4%
If you sum up all the cases where you have at least one success, it is 36%.
You can generalize the case when you only want at least once success with the equation 1 – (1-p)^n where p is the chance to succeed and n is the number of rolls. Basically, it rephrases the question to “What are the chances I’ll fail every roll? Then subtract that chance from 100%”
You have a 4/5 chance to miss, and then a 4/5 chance to miss again because one roll doesn’t inform the other. You can fail the same way again so your odds don’t jut add together – otherwise you could guarantee a heads in two coin flips.
This means 4/5 x 4/5, which is 16/25 – a 64% chance of failure.
It would be 4/5 x 3/4 (60% failure chance) only if roll one removed that possibility from roll two so you can’t have the same roll again.
Another way to look at it is that the chance of hitting at least one of two is 100% minus the chance of not hitting either. The chance of that is 80% (chance of NOT hitting A) times 80% (chance of NOT hitting B), or 0.8 * 0.8 = 0.64, or 64%. (You multiply because you need both events to happen — not hit A AND not hit B). So the chance of not hitting either is 64%, which means the opposite — the chance of hitting one OR the other OR both — is 100% – 64%, or 36%.
There is a scenario where you hit the goal the first time (20%)
There is a scenario where you hit the goal the second time (20%)
If you just add them up, you get 40%.
The problem is the scenario where you hit *both* goals. This happens in both the first scenario *and* the second scenario. So if you add up the odds, you are double counting that specific occurrence. So you need to subtract one instance of it.
The odds of hitting both goals is 20%*20% = 4%.
And 40% – 4% = 36%
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