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Just a question about probability.

If I roll one d10 there is a 10% chance of rolling a 1. Assuming that receiving any 1s among the results of a pool of dice is a failure, what are the chances of rolling any 1s as the number of d10s increases?

My instinct tells me that the chance of there being any 1s among the results increases as the number of dice in the pool goes up. However, I am not certain this is correct nor how to determine the probability for pools with greater numbers of d10s.

Thanks.

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    \$\begingroup\$ Good on you, by the way, for not trusting your intuition when it comes to probability and wanting to understand how the math works for yourself! This is a healthy skepticism that can serve you well in many situations. \$\endgroup\$ – Kevin - Reinstate Monica Jun 25 '16 at 7:43
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    \$\begingroup\$ Statistics for RPG mechanics is on topic here. Also @T.E.D., comments are expected to be nice. \$\endgroup\$ – SevenSidedDie Jun 26 '16 at 1:11
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Yes it does.

Your instinct is right. The more dice, the more likely you are to roll some 1s.

If I'm reading you right you're just interested in whether any 1s appear in your rolled pool. It may not seem obvious but the easiest way to think of this is to model the probability of rolling all 2s-through-10s. That is

$$P(\text{not }1)=\frac 9 {10} = 0.9$$

Then for n dice, we have the probability of n "not 1"s is equal to the probability of each independent event multiplied:

\begin{align*} P(n \text{ not } 1) &= \underbrace{P(\text{not }1) \times P(\text{not }1) \times \dots\times P(\text{not }1)}_{n \text{ times}}\\ &= \left[P(\text{not }1)\right]^n \\ &= 0.9^n \end{align*}

The probability, then, of some die/dice being a 1 is the complement of all n d10 not being a 1:

$$P(\text{some } 1 \text{ in } n \text{d}10) = 1-P(n\text{ not }1) = 1-0.9^n$$


Here's the visual:

enter image description here

And the tabulation:

\begin{array}{cc|cc} n & P(\text{some }1\text{ in }n\text{d}10) & n & P(\text{some }1\text{ in }n\text{d}10)\\ \hline 0&00.00\%&10&65.13\%\\ 1&10.00\%&11&68.62\%\\ 2&19.00\%&12&71.76\%\\ 3&27.10\%&13&74.58\%\\ 4&34.39\%&14&77.12\%\\ 5&40.95\%&15&79.41\%\\ 6&46.86\%&16&81.47\%\\ 7&52.17\%&17&83.32\%\\ 8&56.95\%&18&84.99\%\\ 9&61.26\%&19&86.49\%\\ &&20&87.84\%\\ \end{array}

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    \$\begingroup\$ Also note, as a good rule of thumb for making quick estimates, if the probability p of the event if reasonably small (like 10% here, or 5% with a d20) and the number of repetitions n is reasonably small, the probability of it happening in n tries is close to n×p. (See 19% vs 20% or 27.1% vs 30%.) It only deviates significantly as n gets bigger. (This is pretty much just a consequence of the binomial expansion for (1-p)^n.) \$\endgroup\$ – R.. Jun 25 '16 at 22:34
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Think of that way: if you roll a die, there's a 10% chance of it being a 1. If you roll two dice, it's the same as if you first rolled one of them (since both rolls are independent of each other), then the other. Say you make the experiment 100 times (and equate occurrences with percents).

With a single die, 10 times you lose, 90 times you win.

With two dice, 10 times you lose with the first die, 90 times you win with it. Now, we don't care about the second roll of the 10 losing times, you are losing regardless. Of the 90 times you won the first, you still have to roll a second die, 10% of which (9) will be ones, and 90% (81) non-ones. Total of 19 times you lose, 81 you win (vs 10/90 for 1 die).

Rolling more dice will always mean more chances to lose, because all of them except one might have been non-ones (so you would have won with one less die), and still the last one come up 1.

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    \$\begingroup\$ I like this answer, because it puts into words why the probability is the way it is. \$\endgroup\$ – Sumurai8 Jun 26 '16 at 12:58
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Yes, the odds of rolling at least one of anything increase as the number of dice goes up.

I don't know the actual function to determine how those odds increase, but as an example, here's an Anydice graph showing a 65.13% chance of getting at least one 1 on 10d10: http://anydice.com/program/8b09 and only 27.10% chance of getting at least one 1 on 3d10: http://anydice.com/program/8b0a

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  • \$\begingroup\$ Thanks Polisurgist. It certainly proves that more dice means more chances to hit a 1. \$\endgroup\$ – IronRadiant Jun 23 '16 at 23:06
  • \$\begingroup\$ In case you're familiar with the original World of Darkness rules, this is the reason that when a roll had a difficulty of 10, characters were equally likely to succeed and to botch (have a catastrophic failure). Only a 10 would count as a success, while a 1 would cancel a success out, and you were equally likely to roll 10s as 1s. For a deep dive into those stats: theonyxpath.com/dice-pools-difficulty-numbers-and-botching \$\endgroup\$ – Polisurgist Jun 23 '16 at 23:07
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Putting together a layman's answer/example.

To further simplify, I will use the example of a coin flip to emphasize the results.

Given a single coin, you have a 50% change of flipping a "heads".

Given a second coin (or flipping the same coin a second time), you now have a separate and independent 50% chance for each coin flip.

With two coin flips you now have four possible results, as shown in this graphic:2 coin flips

Each result has a 25% change of occurring, but 3 of the results include at least a single "heads" result. To get the likelihood of at least 1 "heads" you simply add together the likelihood of each result that includes a "heads". In this case, your chance of rolling at least 1 "heads" is 75%.

If you add a third coin (or flip the same coin a third time), you now have eight possible results, as shown on this graphic:3 coin flips

Each individual results has only a 12.5% chance of occurring, but 7 of the 8 results include at least a single "heads" result. Adding these together you get an 87.5% chance of getting a single heads.

Each additional coin-flip results in a greater and greater chance of getting at least a single heads.

Returning to using a 10-sided die, the results follow the same trend, though with a 10% chance of any single roll giving your wanted result, the increase is much less pronounced, though still significant.

A single die gives you your 10% chance.

Two dice (or a single die rolled twice) gives you 100 possibilities, of which 19 have at least a single 1 (19% chance).

Three dice (or 3 rolls) gives you 1000 possibilities, of which 271 have at least a single 1 (27.1% chance).

Source of images is math-prof.com

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There's another way of looking at this, which works particularly well with a ten-sided die.

Think of the single-digit numbers from 0-9. How many of them have a 1? One, out of ten, or 10%.

Now, out of all of the combinations of two digits, from 00-99, how many have any 1s? Well, there's 01, plus all ten of the numbers from 10-19, and then 21, 31, 41, 51, 61, 71, 81, and 91. That's 19, out of a hundred, or 19%.

Do this again for 3-digit numbers. You'll have your 19 from before, plus 100 through 199 (100 numbers), plus another 19 from each of the other eight sets of one hundred. That's 100+(19x9), or 100+171, so 271 out of 1000, or 27.1%.

Next is (271x9)+1000, and so on; the pattern is pretty clear.

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  • \$\begingroup\$ This appears to be a haphazard explanation of the concept reflected by combinations and permutations. \$\endgroup\$ – Drunk Cynic Jun 27 '16 at 20:20
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    \$\begingroup\$ @DrunkCynic Do you have any suggestions for improvement? \$\endgroup\$ – Dan Henderson Jun 27 '16 at 20:36
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    \$\begingroup\$ @DrunkCynic what you describe as a "haphazard explanation" might be just what someone who isn't conversant with algebra needs to "click." "How many of the numbers from 1-1000 have at least one 1" is an excellent discovery-based way to start to grasp the underlying mathematical concepts. (In my experience as a math educator.) \$\endgroup\$ – nitsua60 Jul 18 '16 at 0:11
  • \$\begingroup\$ @nitsua60 Mathematics BS, with courses in teaching secondary education. The answer presents the approach at the base level, fulfilling the adage about understand a thing enough to explain it simply. What is missing is the transition from this basic approach to permutations, explaining that from the set of all combinations of four elements that can number zero to nine, we want to know the proportion that contain a one. \$\endgroup\$ – Drunk Cynic Jul 18 '16 at 3:50
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Whenever you have an probability experiment where there are exactly two outcomes and you can assign a probably to the one of the outcomes, it's called a binomial experiment: True, false, yes, no, black, white, etc. In this case, you get a one or you don't.

The number of dice would constitute the number of trials, since it wouldn't matter if you rolled a single die three times or threw all three die at once.

In this case, the probability of success is .1. The number of dice you use would be number of trials. Success would be where at least one die came up with a one, or P(x >= 1).

The math for this can be daunting, so take the easy road. http://stattrek.com/online-calculator/binomial.aspx

If you have 9 dice, that would be nine trials. The probability for success of a single trial is .1, and you are looking to get at least one success.

You will find the P is just a little over 61%

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  • \$\begingroup\$ You may want to clarify that the "success" here, probabilistically, is equivalent to an automatic failure, in the game mechanics. \$\endgroup\$ – user17995 Jun 25 '16 at 23:11
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Here is a good way to intuitively understand your odds at this...

With one d10 there is a 90% chance of failure, that is, NOT rolling a 1.

With two dice there is a 90% chance of 90% chance of failure. That is .9 x .9 = .81 = 81% you will not roll a 1.

With three dice there is a 90% chance of 90% chance of 90% chance of failure. That is .9 x .9 x .9 = .729 = 72.9% you will not roll a 1.

Etc...

So as the pool of dice goes up the chances of failure goes down but you will never really go to zero percent. No matter how many dice are rolled, there is always some chance of not rolling a 1.

For example with 20 d10s, there is a .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 x .9 = 0.121577

Which is still a 12.1577% chance of not rolling a one with 20 dice!

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  • \$\begingroup\$ "Chance of X chance of failure" is mathematically incorrect. There is a 90% chance that each individual die will fail to produce a 1; for considering the chance a 1 won't appear among multiple dice, you multiply the failure probability for all die, as you've done. \$\endgroup\$ – Drunk Cynic Jun 27 '16 at 20:10

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