Great play through. I saw a preview of this from boardgamegeek during gen con. Looked fun. Nice to get a closer look. Looks like a good mix of luck/strategy and I really like the art/character designs.
"I don't know what my statistical odds are." You raaaang? Seriously though, these are kind of crazy to compute, so I'll just start five of a kind to illustrate how far off the deep end this is. The trouble is re-rolling. You have a 6/6^6 = 1/6^4 chance of rolling five of a kind on the first roll (five dice to roll, but you have 6 numbers that you could get five of a kind of). There is a 30/6^5 = 5/6^4 chance of hitting four of a kind of the first roll, but then you have a 25/36 chance of *not* filling it in on two rerolls, which leaves 11/36 probability that it does fill in, so rolling four of a kind then filling in has a 5/6^4 * 11/6^2 = 55/6^6 chance of working. There is a 150/6^5 = 25/6^4 chance of starting with three of a kind (six numbers to have "of a kind", then 25 ways to roll the remaining two dice to *not* have the tripled number), the chance of getting filling it in on the second roll are 1/36, getting nothing useful 25/36, so by elimination there is a 10/36 chance of getting one die of the two on the next roll. If you have four of a kind going into the last roll, there is definitely a 1/6 chance of filling, and if you still only have three of a kind you have a 1/36 chance to luck into a five of a kind. Thus starting with a 3 of a kind and completing has probability 25/6^4(1/36 + (10/36 * 1/6) + (25/36 * 1/36)) = 3025/6^8. The odds of starting with doubles get complicated, 6 numbers to have doubled, and 125 ways to roll the other three dice so they don't have that number on them. But of those 125 options, 5 have triples of numbers that aren't the doubled number, and we already dealt with those, so really we have 6*120/6^5 = 20/6^3 odds of starting with nothing better than a pair (although we may have two pairs, but we'd just choose one of them to try for in that case). There is a 1/216 chance of filling in five of a kind on the second roll, 15/216 of getting up to four of a kind, and 75/216 of picking up the third of a kind. Furthermore, there is 5/216 chance that you will get thee of a kind of another number, leading you to switch goals for the last roll, leaving a 120/216 chance that the second roll will not help in any way (surprisingly large to me!). If you start the third roll from two of a kind your odds of completing are 1/6^3, with three of a kind they are 1/6^2, and with four of a kind they are 1/6 of course, thus the odds of starting with at most two of any number and going on to get five of a kind are: 20/6^3(1/216 + (15/216 * 1/6) + (75/216 * 1/36) + (5/216 * 1/36) + (120/216 * 1/216)) = 4520/6^8 Last but not least, on five dice, there are only 6 ways to roll no doubles (if each die shows a different value, one of the six values must be missing, knowing which value is missing determines what the other values must be). Thus there is a 1/6^4 chance of not getting even a double. Now one would have to calculate if it is better to keep one of the dice arbitrarily, or to reroll all five. This boils down to calculating if it is easier to roll four of a specific number in two rolls, or five of an arbitrary number in two rolls (my guess is that it doesn't matter, so I'll proceed from there). The calculations for getting five of an arbitrary number in two rolls have already been completed, we just have to sum any of the previous cases that only used two rolls, remembering to include the unpleasant chance that we could again roll no doubles, but still hit the 1/6^4 chance to instantly finish on the last roll. This is as follows: 1/6^4(1/6^4 + (5/6^4 * 1/6) + (25/6^4 * 1/36) + (20/6^3 * 1/216) +(1/6^4 * 1/6^4)) = 3997/6^12 Summing these probabilities we have 1/6^5 + 55/6^6 + 3025/6^8 +4520/6^8 + 3997/6^12 = 12,628,333/6^12 or approximately .0058. Thus, starting with five dice you have about a 0.58% chance to hit five of a kind in Yahtzee. Remember, I thought this would be the easy calculation AND I didn't factor in any powers. Sometimes the most important reason to understand the math is to know when it makes sense to calculate something, and when you should definitely just go with your gut ;-)
![Pointers in C / C++ [Full Course]](http://i.ytimg.com/vi/zuegQmMdy8M/mqdefault.jpg)
Great play through. I saw a preview of this from boardgamegeek during gen con. Looked fun. Nice to get a closer look. Looks like a good mix of luck/strategy and I really like the art/character designs.
Looks like a Great Solo Game.
This is definitely a fun game, it's a little swingy, but its one of my favorite dice chuckers.
I have this game and I recommend it. It gets ⭐️⭐️⭐️⭐️⭐️
i really enjoy this type of dice game. I wanna try it out!
YAHTZEE! hey, this looks pretty fun, nice gameplay
Did you ever get a chance to play the expansion?
sadly no, but i'd like to!
@@rahdo Too bad, looks like a great game!
I want a Steamasaur.
The game looks neat, but the price seems a bit high. I'll have to debate getting it.
"I don't know what my statistical odds are." You raaaang?
Seriously though, these are kind of crazy to compute, so I'll just start five of a kind to illustrate how far off the deep end this is. The trouble is re-rolling. You have a 6/6^6 = 1/6^4 chance of rolling five of a kind on the first roll (five dice to roll, but you have 6 numbers that you could get five of a kind of). There is a 30/6^5 = 5/6^4 chance of hitting four of a kind of the first roll, but then you have a 25/36 chance of *not* filling it in on two rerolls, which leaves 11/36 probability that it does fill in, so rolling four of a kind then filling in has a 5/6^4 * 11/6^2 = 55/6^6 chance of working. There is a 150/6^5 = 25/6^4 chance of starting with three of a kind (six numbers to have "of a kind", then 25 ways to roll the remaining two dice to *not* have the tripled number), the chance of getting filling it in on the second roll are 1/36, getting nothing useful 25/36, so by elimination there is a 10/36 chance of getting one die of the two on the next roll. If you have four of a kind going into the last roll, there is definitely a 1/6 chance of filling, and if you still only have three of a kind you have a 1/36 chance to luck into a five of a kind. Thus starting with a 3 of a kind and completing has probability 25/6^4(1/36 + (10/36 * 1/6) + (25/36 * 1/36)) = 3025/6^8.
The odds of starting with doubles get complicated, 6 numbers to have doubled, and 125 ways to roll the other three dice so they don't have that number on them. But of those 125 options, 5 have triples of numbers that aren't the doubled number, and we already dealt with those, so really we have 6*120/6^5 = 20/6^3 odds of starting with nothing better than a pair (although we may have two pairs, but we'd just choose one of them to try for in that case). There is a 1/216 chance of filling in five of a kind on the second roll, 15/216 of getting up to four of a kind, and 75/216 of picking up the third of a kind. Furthermore, there is 5/216 chance that you will get thee of a kind of another number, leading you to switch goals for the last roll, leaving a 120/216 chance that the second roll will not help in any way (surprisingly large to me!). If you start the third roll from two of a kind your odds of completing are 1/6^3, with three of a kind they are 1/6^2, and with four of a kind they are 1/6 of course, thus the odds of starting with at most two of any number and going on to get five of a kind are:
20/6^3(1/216 + (15/216 * 1/6) + (75/216 * 1/36) + (5/216 * 1/36) + (120/216 * 1/216)) = 4520/6^8
Last but not least, on five dice, there are only 6 ways to roll no doubles (if each die shows a different value, one of the six values must be missing, knowing which value is missing determines what the other values must be). Thus there is a 1/6^4 chance of not getting even a double. Now one would have to calculate if it is better to keep one of the dice arbitrarily, or to reroll all five. This boils down to calculating if it is easier to roll four of a specific number in two rolls, or five of an arbitrary number in two rolls (my guess is that it doesn't matter, so I'll proceed from there). The calculations for getting five of an arbitrary number in two rolls have already been completed, we just have to sum any of the previous cases that only used two rolls, remembering to include the unpleasant chance that we could again roll no doubles, but still hit the 1/6^4 chance to instantly finish on the last roll. This is as follows:
1/6^4(1/6^4 + (5/6^4 * 1/6) + (25/6^4 * 1/36) + (20/6^3 * 1/216) +(1/6^4 * 1/6^4)) = 3997/6^12
Summing these probabilities we have 1/6^5 + 55/6^6 + 3025/6^8 +4520/6^8 + 3997/6^12 = 12,628,333/6^12 or approximately .0058. Thus, starting with five dice you have about a 0.58% chance to hit five of a kind in Yahtzee. Remember, I thought this would be the easy calculation AND I didn't factor in any powers. Sometimes the most important reason to understand the math is to know when it makes sense to calculate something, and when you should definitely just go with your gut ;-)
/boggle! :)