# [RPG] Does the chance of rolling a 1 in a pool of d10s increase as the number of dice increases

dicestatistics

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.

## 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:

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}