# Statistically speaking, the probabilities of a d4 aren't affected by the shape, i.e. whether it's 4-sided, 8-sided, 12-sided etc.

This is basic statistic knowledge, but you can alternatively also verify it using sites such as anydice.com. See Sdjz's answer for details.

# In real-world terms, ignoring statistics, the shape of your "d4" won't affect the result either, assuming that you use e.g. a dice shaker. If you roll with your hands only, though, it is easier to cheat with a standard d4 as opposed to 8- or more-sided d4s.

This is basic logic - if you try to throw a die in a certain way so that it lands on a specific number, you will have an easier time the fewer times the dice rolls / the fewer sides it has. Therefore, it's easier to cheat with a 4-sided d4 as opposed to an 8- or 12-sided d4, and it would be easier to cheat with a 20-sided d20 than with an 80-sided one.

# Even if you don't intend to cheat, 8- or 12-sided d4s are probably a good choice.

Why, you ask? Because 4-sided d4s, especially metal ones, are notoriously murderous weapons of deadly vicious torturous vile evil death, and are probably better of being called carpet caltrops (credit goes to Korvin Starmast). These little f***ers are worse than ~~Legos~~ Lego bricks.

For illustration, I found the following image of a d4 on boardgamegeek.com:

I don't personally have d4s with more than 4 sides, since I have pretty cool metal d4s with rounded corners that are not as murderous. However, a few of my fellow players have 12-sided d4s, which can be found here.

Note: I am not in any way affiliated with the linked shop, nor do I mean to advertise it (in fact, their 12-sided d4 are about their only item I would buy). It's just that, in my experience, it's not easy to find 12-sided d4 at all, and this site has them, so I want to save OP (and others) the time in finding them.

## Best Answer

## Yes it does.

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

some1s.If I'm reading you right you're just interested in whether

any1s 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

ndice, we have the probability ofn"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

somedie/dice being a 1 is the complement ofall nd10notbeing 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}