Probability of Yahtzee

I've recently read The Art of Game Design: A Book of Lens and highly recommend it. It covers how to calculate those odds, but I'll try my best to help.

Let's look at the odds of rolling a '1' on three dice
( 1 means the specific die was rolled, 0 means it was missed)
1 1 1 (three '1's)
0 1 1 (two '1's and 2-6 on the last die)
1 0 1
1 1 0
0 0 1
0 1 0
1 0 0
0 0 0

You calculate the odds for each of the results above by multiplying the odds for each die.
1 1 1 is three '1's. Rolling a '1' is 1/6. 1/6 x 1/6 x 1/6 = 1/216 (or 1/6 ^ 3). You add each other result to this.
0 1 1 is two '1's and one 2-6. Rolling 2-6 is 5/6 odds. The result would be 5/6 x 1/6 x 1/6 = 5/216
Interesting note here: 0 1 1, 1 0 1, and 1 1 0 all have the same odds. I this case, you just add the results (5/216) together. 5/216 + 5/216 + 5/216 = 15/216.

Ignore the rest of the rolls since they do not have two or more '1's

Add the first result (for the three '1's) together with the other rolls (15/216) for the final tally. 1/216 + 15/216 = 16/216= 2/27 = 7.04%

Let's do it again for four dice, looking for two-of-a-kind.
(I'm leaving out any results that do not matter)
1 1 1 1 = 1/6 x 1/6 x 1/6 x 1/6 = 1/1296
0 1 1 1 = 5/6 x 1/6 x 1/6 x 1/6 = 5/1296 x 4 = 20/1296
1 0 1 1
1 1 0 1
1 1 1 0
1 1 0 0 = 1/6 x 1/6 x 5/6 x 5/6 = 25/1296 x 4 = 100/1296
0 1 1 0
0 0 1 1
1 0 0 1

(4 1's) + (3 1') + (2 1's) = Total Ones
1/1296 + 20/1296 + 100/1296 = 121/1296 = 9.33%

Now to include rerolling? I'm going to have to guess on that one.

Odds of rolling '1' on a d6, with three rerolls: Count each roll as three dice, but only counting if the results come up with one '1' since you'd stop rolling at that point.

1 0 0
0 1 0
0 0 1 = 1/6 x 5/6 x 5/6 = 25/216 x 3 = 75/216 = 34.7%

Each die now has a 75/216 chance to hit a 1 instead of 1/6. Each miss now counts as 141/216 instead of 5/6.

Example: two hits on three dice with three rerolls.
1 1 1 = 75/216 x 75/216 x 75/216 = 421,875/10,077,696 = 4.19%
0 1 1
1 0 1
1 1 0 = 141/216 x 75/216 x 75/216 = 793,125/10,077,696 = 7.87%

If my math is right, that should provide you all the instructions to figure it out. It will take a bit of work though.

For these, the probability can get fairly complicated. You really want to calculate the chance of the event in question NOT happening, and then subtract that from one.

You're asking for a significant quantity of work for someone to give you the answers, even for people good at math, but I'll try and explain how to do the first option: Any 2 of a kind.

First off, you will need to understand how many possible results there are. Because each die is an independent event, and has a potential 6 results, to determine the number of possible results you take the number of outcomes of a single element (6 - one for each side on the die), and raise it to the power equal to the concurrent events (5 - one for each die). This would be different if, for example, the die did not have all the same faces or if subsequent dice had different numbers of sides.

So, we can determine that there are 7776 potential outcomes for rolling 5d6. (that's 5, 6 sided die). So how many of them will NOT result in 2 of a kind. Well abstractly, we can see that there are only (6) discrete outcomes that will not end in this result: [1,2,3,4,5]; [1,2,3,4,6]; [1,2,3,5,6]; [1,2,4,5,6]; [1,3,4,5,6]; [2,3,4,5,6]. However, it doesn't matter which of the 5 dice have which result within a particular set's results. If we consider each item in the set to have occurred on its die respectively, then we can determine the potential die sets that could occur by evaluating 5!. "!" is the mathematical operation that is equivalent to X*(X-1)*(X-2)... until the term is 1. 5! would therefore be 120. That means that each set that doesn't result in a pair (for example [1,2,3,4,5]) can occur in 120 different ways. This can be a slightly difficult concept, but you want to think of each die as a discrete value. So in the [1,2,3,4,5] example, the 1 can occur on die A,B,C,D, or E, and each of the subsequent results would obviously need to happen on the other dice.

So we have 6 discrete outcomes that don't result in two of a kind. And each of those results can occur 120 ways. This means there are 720 possible ways out of the total possible 7776 outcomes to NOT get a two of a kind. This gives us a 7056/7776 chance on the FIRST ROLL of getting a two of a kind (90.7%).

Now we get into the other rolls. Because you can choose to re-roll 0-5 dice (6 possible options) we have to now calculate each of the potential results. If you choose to re-roll 0, we know that's a failure (since the first result was a failure).

If you choose to re-roll 1, then you know you have 4 opposite values, so there's only 2 potential values that would give you a failure (because rolling any of the dice you've already rolled would obviously give you a success), so that's 2 outcomes out of 6 that result in failure or 1/3 chance of failure. [6 total potential outcomes]

If you choose to re-roll 2 dice, the first die rolled would have a 1/2 chance of failure (not rolling a double) and the second would have the same (1/3) chance as just rolling 1 die. - when you are looking at cumulative chance within the same event, you multiple the results. So when re-rolling 2 die, you have a 1/6 cumulative chance of failure (for the 2nd roll event). [36 total potential outcomes]

If you choose to re-roll 3 dice, you have a 2/3 chance for the first, 1/2 chance for the second, and a 1/3 chance for the third for a total of 1/9 chance of failure. [216 potential outcomes]

If you choose to re-roll 4 dice, you have a 5/6 chance for the first, 2/3 chance for the second, 1/2 chance for the third, and a 1/3 chance for the fourth. This results in a 10/108 chance of failure. [1296 potential outcomes]

If you re-roll all 5 dice, you have the same result as the original roll, or 720/7776.

So the chance for each result is distinct for each determinant. Because there are 6 potential options (rolling 0-5 dice), we need to look at the total failures out of the complete set of potential outcomes. When rolling 1-5 dice, there are a cumulative 9330 potential new outcomes. This means we have a cumulative quantity of 72550080 outcomes. Let's see the breakdown of the failures:

When re-rolling 0 dice - we know there are only 720 potential failure results.
When re-rolling 1 die - we know there are 720+ an additional 2 failure results. (722)
When re-rolling 2 dice - we know there are 720+ an additional 6 failure results. (726)
When re-rolling 3 dice - we know there are 720+ an additional 24 failure results. (744)
When re-rolling 4 dice - we know there are 720+ an additional 10 failure results. (730)
When re-rolling 5 dice - we know there are 720+ an additional 720 failure results. (1440)

5082 failures in 72550080 outcomes, which means we should see 72544998 successes in 72550080 outcomes, or 99.993% cumulative chance of success. However, there is another potential roll.

Luckily things don't change much with the second roll, the number of results are the same, and the potential failures are the same. But the number of potential results is vastly different. This is because there are 6 options for EACH of the previous options. I'm going to code them A-B-C-D-E-F below, so that BD would be re-rolling 1 die the first time, and 3 dice the second time. This means there are a staggering 47855048216844521080163520000000 potential results...

AA - 720 failures +0 failures (720)
AB - 720 failures +2 failures (722)
AC - 720 failures +6 failures (726)
AD - 720 failures +25 failures (744)
AE - 720 failures +10 failures (730)
AF - 720 failures +720 failures (1440)
BA - 722 failures +0 failures (722)
BB - 722 failures +2 failures (724)
BC - 722 failures +6 failures (728)
BD - 722 failures +24 failures (746)
BE - 722 failures +10 failures (732)
BF - 722 failures +720 failures (1442)
CA - 726 failures +0 failures (726)
CB - 726 failures +2 failures (728)
CC - 726 failures +6 failures (732)
CD - 726 failures +24 failures (750)
CE - 726 failures +10 failures (736)
CF - 726 failures +720 failures (1446)
DA - 744 failures +0 failures (744)
DB - 744 failures +2 failures (746)
DC - 744 failures +6 failures (750)
DD - 744 failures +24 failures (768)
DE - 744 failures +10 failures (754)
DF - 744 failures +720 failures (1464)
EA - 730 failures +0 failures (730)
EB - 730 failures +2 failures (732)
EC - 730 failures +6 failures (736)
ED - 730 failures +24 failures (754)
EE - 730 failures +10 failures (740)
EF - 730 failures +720 failures (1450)
FA - 1440 failures +0 failures (1440)
FB - 1440 failures +2 failures (1442)
FC - 1440 failures +6 failures (1446)
FD - 1440 failures +24 failures (1464)
FE - 1440 failures +10 failures (1450)
FF - 1440 failures +720 failures (2160)

For a total of 35064 failures out of a total of 47855048216844521080163520000000 potential outcomes. So that means we are looking at 47855048216844521080163519964936 success or 99.99999999999999999999999993% chance of success.

Now, when you're playing the game, your individual chances will be different because you would only apply the chance for the particular number of dice you are choosing to reroll. This is the chance of success if all choices are completely random and you stop on success.

That's just ONE of the problems. When you are looking for things that can have multiple values on a single die, it starts getting significantly more complex.

Does that help at all?

On a side note, I didn't double check your perelcentages. I have no clue where you got 0.08% on one roll for yatzee. Any single roll of 5 dice looking for 5 exact numbers is 1/7776 which is .012%. Additional rolls adjust that as above, but the base is always the same.

Am I missing something there?

Edit:
I'm an idiot. I forgot there are 6 cases of yatzee in one roll and 6 times .012 is .072 or about .08

This actually skews my above example percent as I forgot to take it into account, but only by like .06% each so not enough to really change the point.

Oh yea! You're right there. I was not factoring in getting ONLY the wanted amount. That does screw with the math. Thank you. Thought I was missing something.

Maybe I am missing something here, but...

Are you talking about rolling 3 dice in one go?
And you want to throw at least one 1? If so, then the "at least a result of [1]" calculation is incorrect.
Imagine throwing 6 dice. You think the chance of having at least a result of [1] is 100% then? I disagree on that.

At least a result of [1], so the three dice produce one, two or three [1]:
First dice has 1/6 or 36/216
Second dice has 5 * 1/36 = 5/36 or 30/216
Third dice has 25* 1/216 = 25/216
Adding up: 91/216 (42,1%, not 50%)

The other one isn't complete either. A result of one [1] only, so the three dice produce [1][not 1][not 1] and this [1] can be switch with a [not 1], so times 3, would be:
(1/6)*(5/6)*(5/6) + (5/6)*(1/6)*(5/6) + (5/6)*(5/6)*(1/6) = 75/216 (34,7%, not 11,6%)

But then again.
My apologies when I am missing the point.