D-Day Dice Strategy Guide | D-Day Dice: Free Trial Version

Michael Yang

(CapNClassic)

United States
Fresno
California

I just started reading your strategy guide, and noticed an error. (great game by the way)
In the starting example about which dice to hold, you note that you could hold a single soldier Red, and single soldier White die, and then reroll the two BLUE dice giving you a 1/3 chance to get a RWB bonus on your 2nd roll, and a 2/3 chance to get a RWB bonus by your 3rd roll (final tally).
This is not the right way to calculate odds. The proper answer to this question is an 11/36 of a chance on your 2nd roll, and a 671/1296 to complete a Single Soldier RWB bonus by your final tally.

The way you calculate the odds is to figure out all possible outcomes. In your example of rolling the two BLUE dice (ignoring the other non single soilder Red and White dice, since they don't factor into this question), there are 36 possible outcomes. I have a small table below showing both die pair results:
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

What you will notice is that you only have 11/36 resulting pairs that give you at least one Blue Single Soldier (instead of 12/36, or 1/3 as you said in the guide). This is because the 3rd row where the first die is 3 overlaps with the 3rd column where the second die is 3. This (3,3) result can only be counted once as a successful RWB bonus, even though either die could be used to complete the bonus.

To figure out the odds of having rolled a RWB single soldier bonus by your final tally (3rd throw, ignoring Corporal/Noncom specialists), is slightly harder to figure out. It is probably easier to think about it in this respect, the odds of getting the RWB bonus is 1 minus the odds of not getting it after your 3rd throw. (all possible outcomes should equal 1 or 100%. There are only two possible outcomes, you either get the RWB bonus, or you don't, therefore the combined probabilty of both outcomes should equal 1).
The odds of you not getting the RWB bonus is the odds of not getting it after your 2nd roll (we already know this. It is 1 - (11/36), or (25/36)), multiplied by the odds of not getting on your 3rd roll (this is also 25/36, since you are still rerolling the two BLUE dice.). This gives you a final answer for getting a RWB by your final tally of 1 - (25/36) * (25/36), which is approximately 52%, not 66.67% as your guide claims.

Edited Tue Nov 15, 2011 5:38 am
Posted Tue Nov 15, 2011 5:31 am
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Emmanuel Aquin

(Amiral)

Canada
Mont-Saint-Hilaire
Quebec

And D-Day Dice, too!

I'm an "A"vatar.

Thanks for the math lesson, Michael (and for taking the time to write this). This proves without a doubt that I'm a an artist, not a number cruncher...

E

Posted Tue Nov 15, 2011 5:39 am
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Michael Yang

(CapNClassic)

United States
Fresno
California

You are welcome. I only did a poor mans job of explaining the odds (now looking over my answer, I notice I forgot to explain why you multiply the odds of the 2nd and 3rd rolls together). Even as a non mathematician (this isn't to say that I am, I only know a little about probability because I have been enjoying Delve, Barbarian Prince, and other dice games recently), you seem to have done pretty good job balancing the difficulty on the free maps. I look forward to the official release version.

I just noticed your second error (I wonder if I should just compile all these errors into a single post, or just PM you the results so that you can fix the math in a later release of the strategy guide).

Quote:

Why? Simple: since you have 1 in 6 chances to obtain a “2 Soldiers”
result on every other die, your chances of obtaining this result in red and white are doubled (1 in 3), since you roll 2 dice for each color. This means you have 2 in 3 chances of obtaining at least one other “2 Soldiers” result on your second roll (which ain’t bad!), and another 1 in 3 chances of obtaining your third result in the right color on the third roll (plus a bonus 1 in 6 chance if you have a Noncom Specialist or a Backpack).

The odds of getting at least 1 "2 Soldiers" results on your second roll isn't calculated by adding the probabilities of the WHITE and RED dice together. There are two general rules that are good to know about statistics. Stolen from here.
http://www.stat.berkeley.edu/~bradluen/stat2/lecture16.pdf

The Addition Rule: If two (or more) events are mutually exclusive, we can find the probability that one or the other of them happens by adding the probabilities of each of them.
The Multiplication Rule: to find the probabilities of successive events, multiply conditional probabilities given the previous events.
Mutually Exclusive: Two events are mutually exclusive if they can't both happen.
You cannot add the probability of the WHITE and RED dice together to get the probability of rolling at least 1 "2 Soldier", because the two results are not mutually exclusive (i.e. it is possible to get a "2 Soldier" result on a WHITE and RED die in your 2nd roll.)

Another way to think about getting "at least 1" 2 Soldier result, is that this is the opposite of getting no "2 Soldier" results. The odds of not getting a "2 Soldier" are 5/6 for each die.
The odds of getting no "2 Soldier" results rolling 4 dice is (5/6)*(5/6)*(5/6)*(5/6).
So the odds of getting at least 1 "2 Soldier" result is 1 - (odds of getting none), or 1 - (625/1296).

If you want to get really elaborate, you can figure out the odds of getting each of the possible outcomes. You should be able to add all these outcomes together (because each outcome is mutually exclusive) and get 1296/1296.
The odds of getting exactly 0 "2 Soldier" results = (5/6)*(5/6)*(5/6)*(5/6) = 625/1296
You can think of each of these independent die rolls as successive events, with each event resulting in anything other than a "2 Soldier" result. This means we should multiply the individual probabilities together to get our result.

The odds of getting exactly 4 "2 Soldier" results = (1/6)*(1/6)*(1/6)*(1/6) = 1/1296
This can also be thought of successive die rolls (since each die is independent), with every result being "2 Soldier".

The odds of getting exactly 3 "2 Soldier" result = (5+5+5+5)/1296
333(1,2,4,5,6) or (i.e. the 1st, 2nd, 3rd die are 3 and the 4th is not)
33(1,2,4,5,6)3 or (i.e. the 1st, 2nd, 4th die are 3 and the 3rd is not)
3(1,2,4,5,6)33 or ...
(1,2,4,5,6)333 ...
We can add these probabilities together, because they are mutually exclusive (if three of the dice are 3, their is only one die left that has to be anything other than a 3. This is helpful, because you can avoid double counting a result). There are 5 possible outcomes for each of the four situations where you have 3 threes.

The odds of getting exactly 1 "2 Soldier" results = (125+125+125+125)/1296 = 500/1296
3(1,2,4,5,6)(1,2,4,5,6)(1,2,4,5,6) (i.e. the 1st die is 3, and the rest are not)
(1,2,4,5,6)3(1,2,4,5,6)(1,2,4,5,6) (i.e. the 2nd die is 3, and the rest are not)
(1,2,4,5,6)(1,2,4,5,6)3(1,2,4,5,6) ...
(1,2,4,5,6)(1,2,4,5,6)(1,2,4,5,6)3 ...
We can add these probabilities together, because they are mutually exclusive. This is very similar to the exactly 3 "2 Soldier" calculation, because you don't have any overlap to worry about. In this situation, it is as if you are moving the "2 Soldier"/"3" die to each of the possible positions (instead of the anything but a 3 die). This gives you many more total outcomes, but the calculation is very similar. You have four situations where the anything but "2 Soldier" die results can each be five different results. (5/6)*(5/6)*(5/6) outcomes on each of the non-"3" dice.

The odds of getting exactly 2 "2 Soldier" results = 150/1296
I put this last, because you have to think of the possible permutations that allow for two "3" results. The answer you get here, should add up with the others to get 1296 (so we should get 150). Two of the dice are 3, which allows the other two dice to be any other number but "3" (25 possibilities ((1,(1,2,4,5,6)),(2,(1,2,4,5,6)), ...). All that you need to do is make sure that correctly identify each of the permutations where two of the dice are 3. This is not the best way of figuring out these probabilities, but it is sometimes what I do as a layman. The proper way i believe looks at binomial probabilities.
33(5/6)(5/6) - 25
3(5/6)3(5/6) - 25
3(5/6)(5/6)3 - 25
(5/6)33(5/6) - 25
(5/6)3(5/6)3 - 25
(5/6)(5/6)33 - 25

As for the second part,

Quote:

and another 1 in 3 chances of obtaining your third result in the right color on the third roll (plus a bonus 1 in 6 chance if you have a Noncom Specialist or a Backpack).

The best way to answer this question is to use a probability tree. What we want to know, is what is the probability that on the third roll we will have obtained the RWB bonus, and what is the probability that on our 4th single die roll we will obtain the RWB bonus.
The first part of the tree contains the dice we are going to keep. A BLUE "2 Soldier" (B2), and BLUE "Courage" die (BC). The next part of the tree contains possible outcomes for our 2nd roll (W#/R# are any other result besides W2/R2 respectively). We only care about "2 Soldier" results. There are 4 outcomes, you roll a "2 Soldier" on both White and Red ((11/36)*(11/36)=121/1296), you get no "2 Soldier" results ((25/36)*(25/36)=625/1296), or you get either the White or Red two soldier (since we know that getting the White or Red "2 Soldier" result is just as likely, we can just divide the remaining possible outcomes by two and assign half to White and the other half to Red (1 -(121+625)/1296=550 or 275W and 275R). This part of the tree tells us our odds about how likely we have already obtained our RWB bonus, and how close we are to obtaining it during our third roll.

(B2,BC)
/
(W2,W#,R2,R#),(W2,W#,R#,R#),(W#,W#,R2,R#),(W#,W#,R#,R#)
121/1296, 275/1296, 275/1296, 625/1296
/ , / , / , /
(RWB Success),(Snake Eyes-Loss),(Snake Eyes-RWB Success),
(120/1296),1/1296*(1/36),(1/1296)*(35/36),
The 3rd roll is a little more complicated, because we forgot about skulls during our 2nd roll. Imagine if our roll was (W2,WSkull,R2,RSkull). Normally this would fit in our first branch of the tree (121/1296) where we have already completed our RWB bonus and could stop rolling, but since we have 2 Skulls that cancel two dice we have to reroll the 2 Skulls during the 3rd roll and not roll snake eyes again (1/36). The second branch of the tree has similar problems. If the other white die was a Skull (W2,WSkull,R#,R#), you would have to reroll it or increase the risk canceling another Red die result.
(keep W2,W1 = 9/36 = 25% chance of getting RWB bonus. (1,3) and (3,1) are no longer a success because you now have two skulls that need to cancel a die.

RED
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

(keep W2, reroll W1 = (Losses + Successes) = (9+55)/216 = ~29.6%
(W1,(!R1, R2)), (1/6)*( 9/36)= Win (White Skull cancels other Red roll)
(W1, (R1, R2)), (1/6)*( 2/36)= Loss (White Skull & Red Skull)
(W1, (R#, R#)), (1/6)*(25/36)= Loss (Didn't roll a "2 Soldier" on Red)
(!W1,(R2, R#)), (5/6)*(11/36)= Win (Rolled a Red "2 Soldier" and no White Skull)
(!W1,(R#, R#)), (5/6)*(25/36)= Loss (Didn't roll a "2 Soldier" on Red)

So, as you can see the odds are quite complicated. The only way to properly figure out the odds of completing a RWB bonus at the end of your final tally is to figure out the odds at each branch of the tree. I have already partially completed the 2nd branches of the tree, and the start of the 3rd branches (I separated the one chance of the 121 where you roll two Skulls and need to reroll, from the 120 where you have already succeeded.). I will leave the rest of the calculations for the 3rd level of the tree (and the separation of the 2nd level one White "2 Soldier" result with/without a White Skull) for another time.

Edited Tue Nov 15, 2011 4:09 pm
Posted Tue Nov 15, 2011 5:51 am
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