Dice game problem within the game "Kingdom Come: Deliverance"
First of all, hello guys!
So I was recently playing Kingdom Come: Deliverance and there's a cool little dice game within the game.
Now I wanted to device a few strategies and calculate a few probabilities for that game but unfortunately after tinkering with the numbers over a night for a bit I came to a complete stop and can't advance any further.
The game itself works as follows for those of you who haven't played it:
Each player has 6 dice. On each roll they have to throw either a 1, a 5, or a triplet, quadruplet, quintuplet or 6x the same number to get another roll.
After each roll the player has to remove at least one dice if he gets a second roll. Let's say I roll 5 4 2 3 2 6 then I get a new roll since I got a 5, I have to take the 5 away and reroll with the remaining 5 dice. I would also get a reroll if I got 4 2 2 2 3 3 and I would remove the 3 dice I rolled a 2 with, giving me a reroll with the remaining two dice.
There are point values associated with all of that but that isn't really relevant for the problem that I encountered.
So first off I started to calculate what the probability of rolling either a 5 or a 1 is with 1 (33%), 2 (56%), 3 (70,38%), 4 (80,24%), 5 (86,83%) and 6 (91,22%) dice. So far so good (hopefully). I simply calculated the outcome of having a "no hit" on each of these dice numbers and multiplied the probabilities for it, meaning (2/3)^n with n being the number of dice. So with 6 dice we get a probability of 1 - (2/3)^6 for not getting a hit, aka ~91,22%.
Now what we have is the probability of rolling either a 1 or a 5 on our throw but here is where my trouble began. Since you also get a reroll if you get a triplet, quadruplet and so on, the probabilities of these cases have to also be added to the existing 91,22% probability to get a full accurate picture on the probability of getting a reroll.
And this is where my math kinda failed hard.
I tried simplifying the problem looking at 3 dice first and looking at their probability of rolling either 3 x 2, 3 x 3, 3 x 4 or 3 x 6.
With three dice I only have 4 possibilities of rolling a triplet of 2s, 3s, 4s and 6s:
222
333
444
666
That gave me 4/216 (the total number of possibilities for 3 dice to land, aka 6^3, which is roughly 1,85%. Or, simpler put, for the first throw you have a 2/3 chance of hitting a 2, 3, 4 or 6, for the two consecutive throws it's a 1/6 chance to hit the same number again. So 2/3 x (1/6)^2 = ~0,0185 which is the probability.
Now when I tried expanding this model to 4 dice I kinda got stuck.
I tried writing out all the different possibilities for triples, of which there are 4x4=16 variants:
X X X -
X X - X
X - X X
- X X X
The x represents a number 2, 3, 4 or 6. So for each line there's 4 combinations, equaling up to 16 possible solutions for the triplets.
However, now that there are 4 dice there are also 4 possibilities for quadruplets, which would be 4x2, 4x3, 4x4 or 4x6.
So there are 20 possibilities that we get a reroll via triples or quadruplets on 4 dice, right?
But if I try to calculate the number of possibilities for triplets using my math from earlier, I get to a sticking point.
So we could still use the formula 2/3 x (1/6)^2 in theory but we would have to expand it with a 4th dice, right? So the 4th dice cannot be the same as the first three and it cannot be 1 or 5 since we already calculated that. So if our triplet would be 3 2s we would have to get either a 3, a 4 or a 6, aka 3/6, aka 1/2.
So if I expand this formula then I get this:
2/3 x (1/6)^2 x 1/2= 0,00925...
And 0.00925 x 1296 (number of possible throws aka 6^4) = 12?
But that can't be the number of possible triplets for 4 dice as I've shown above?
So does anybody know how I could progress from that sticking point and maybe even move all the way up towards the probability of 6 dice giving you a reroll on your throw?
Where is my mistake in that calculation? Where am I taking a wrong turn? Is there a more simple way of calculating this, a more easy way that I can't see at the moment?
Cause I've been trying to solve this sticking point for a few hours now and I can't seem to progress any further.
If there are any questions regarding anything, feel free to ask me.
Cheers!
First of all, hello guys!
So I was recently playing Kingdom Come: Deliverance and there's a cool little dice game within the game.
Now I wanted to device a few strategies and calculate a few probabilities for that game but unfortunately after tinkering with the numbers over a night for a bit I came to a complete stop and can't advance any further.
The game itself works as follows for those of you who haven't played it:
Each player has 6 dice. On each roll they have to throw either a 1, a 5, or a triplet, quadruplet, quintuplet or 6x the same number to get another roll.
After each roll the player has to remove at least one dice if he gets a second roll. Let's say I roll 5 4 2 3 2 6 then I get a new roll since I got a 5, I have to take the 5 away and reroll with the remaining 5 dice. I would also get a reroll if I got 4 2 2 2 3 3 and I would remove the 3 dice I rolled a 2 with, giving me a reroll with the remaining two dice.
There are point values associated with all of that but that isn't really relevant for the problem that I encountered.
So first off I started to calculate what the probability of rolling either a 5 or a 1 is with 1 (33%), 2 (56%), 3 (70,38%), 4 (80,24%), 5 (86,83%) and 6 (91,22%) dice. So far so good (hopefully). I simply calculated the outcome of having a "no hit" on each of these dice numbers and multiplied the probabilities for it, meaning (2/3)^n with n being the number of dice. So with 6 dice we get a probability of 1 - (2/3)^6 for not getting a hit, aka ~91,22%.
Now what we have is the probability of rolling either a 1 or a 5 on our throw but here is where my trouble began. Since you also get a reroll if you get a triplet, quadruplet and so on, the probabilities of these cases have to also be added to the existing 91,22% probability to get a full accurate picture on the probability of getting a reroll.
And this is where my math kinda failed hard.
I tried simplifying the problem looking at 3 dice first and looking at their probability of rolling either 3 x 2, 3 x 3, 3 x 4 or 3 x 6.
With three dice I only have 4 possibilities of rolling a triplet of 2s, 3s, 4s and 6s:
222
333
444
666
That gave me 4/216 (the total number of possibilities for 3 dice to land, aka 6^3, which is roughly 1,85%. Or, simpler put, for the first throw you have a 2/3 chance of hitting a 2, 3, 4 or 6, for the two consecutive throws it's a 1/6 chance to hit the same number again. So 2/3 x (1/6)^2 = ~0,0185 which is the probability.
Now when I tried expanding this model to 4 dice I kinda got stuck.
I tried writing out all the different possibilities for triples, of which there are 4x4=16 variants:
X X X -
X X - X
X - X X
- X X X
The x represents a number 2, 3, 4 or 6. So for each line there's 4 combinations, equaling up to 16 possible solutions for the triplets.
However, now that there are 4 dice there are also 4 possibilities for quadruplets, which would be 4x2, 4x3, 4x4 or 4x6.
So there are 20 possibilities that we get a reroll via triples or quadruplets on 4 dice, right?
But if I try to calculate the number of possibilities for triplets using my math from earlier, I get to a sticking point.
So we could still use the formula 2/3 x (1/6)^2 in theory but we would have to expand it with a 4th dice, right? So the 4th dice cannot be the same as the first three and it cannot be 1 or 5 since we already calculated that. So if our triplet would be 3 2s we would have to get either a 3, a 4 or a 6, aka 3/6, aka 1/2.
So if I expand this formula then I get this:
2/3 x (1/6)^2 x 1/2= 0,00925...
And 0.00925 x 1296 (number of possible throws aka 6^4) = 12?
But that can't be the number of possible triplets for 4 dice as I've shown above?
So does anybody know how I could progress from that sticking point and maybe even move all the way up towards the probability of 6 dice giving you a reroll on your throw?
Where is my mistake in that calculation? Where am I taking a wrong turn? Is there a more simple way of calculating this, a more easy way that I can't see at the moment?
Cause I've been trying to solve this sticking point for a few hours now and I can't seem to progress any further.
If there are any questions regarding anything, feel free to ask me.
Cheers!
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