# Probability Combinations and Conditions

I have very limited math knowledge and might make little sense but here we go:

If I wanted to substitute rolling a 4 sided die with "flipping" coins and adding a modifier, would it be unfair?

Ex. Take 3 coins and shake them in your hands like dice and lay them down. If there are no heads the result is a 1. If 3 are heads, the result is a 4. (X = N + 1)

I initially thought this was balanced since the combinations of coins would only give 4 unique combinations (HHH, HHT, HTT, TTT). While I understand that coin flipping is associated with standard odds, I thought that by adding that specific rolling setup and condition would make it fair, because it isn't "flipping" 3 coins in a row (more like a singular rolling) and the coins don't need to be distinguishable.

I have gotten feedback that it would still be more likely to get 2 and 3. I then tried to compare it to drawing from a bag of 6 stones with black and white stones. Draw three of them and the number of black stones (heads) decides the result. My logic was that similar to 3 coins with 2 outcomes, drawing 3 stones of 2 colors are comparable.

Are the two methods equivalent? Is there a flaw in the logic that specifically if the combinations cannot be variations, that the first method is similar in probability to the second? I don't know enough to know what I'm missing.

Edit- What if instead of "flipping" them all at once, they were one at a time, thus shrinking the pool of results with each "flip." If the first was a heads, the pool then loses TTT, and the probability is calculated with one less option. Thank you for your time, I know this one's a weird one. Any suggestions on how you would do it differently would also be appreciated.

Edit- I looked into it a little but might be misunderstanding. Is the bag version a Hypergeometric Distribution calculation while the coin is not (because the coins' positions are not predetermined)?

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• You can do the simulation, just look at pairs. Toss a fair coin twice. Now each (ordered) pair of outcomes is equally probable. So, let $HH=1$, $HT=2$, $TH=3$, $TT=4$. For example.
– lulu
Jun 23 at 11:17
• Note: it only comes out this neatly because $4=2^2$ is a power of $2$. If you wanted to simulate a $6$ sided die, then you could take $6$ of the $8$ possible ordered triples of three tosses, but that would mean "ignoring" two of the $8$ possible outcomes. Of course, in those $2$ cases you could just retoss, but this process could (in theory) take a long time to yield a result.
– lulu
Jun 23 at 11:19

Flipping a coin has two outcomes and is represented by the generating function $$1+x$$.
Throwing a four-sided dice has four outcomes, and is represented by the generating function $$1+x+x^2+x^3$$.
If you flip a coin $$3$$ times, the generating function becomes $$(1+x)^3=1+3x+3x^2+x^3$$, which has four outcomes, but they are not distributed in the same way as with the dice, and we say that they have different distributions.