Holy hell, Dice Poker

[NovaPrime1985]

Is it just me or is this s*** heavily geared to screw you over majority of the time, most of these opponents have some of the hottest hands at dice that ive ever seen, they should take up professional gambling by the numbers they throw! -______-

[Ron AF Greve]

First off 3 * 1/6 = 3/6 = 0.5. However, somewhat counterintuitive, you cannot just add them together. Say that you throw with 7 dice you would get 7/6 which is a probability greater than 1.

The correct formula is P = P(A) + P(B) - P( A and B ). So say that you want to calculate the chance that you throw 6 with at least one of two dice it wouldn't simpy be 1/6 + 1/6 but 1/6 +1/6 - 1/36

The reason is simple to see by just enumerating:
FIrst dice 6 combinations:
6 1
6 2
6 3
6 4
6 5
6 6

Second dice 6
1 6
2 6
3 6
4 6
5 6
6 6

i.e. chance of 12 /36 which would be 1/3 BUT it is not correct because we counted 6 6 twice:

So it would become 6/36 + 6/36 - 1/ 36= 11/36 with more than two dice you would subtract all combinations so it is easier to calculate the chance you won't throw 6 and then subtract that from 1 which becomes 1 - 5/6 * 5/6 = 11/36

[RobotParty]

No, it's 3/18(you multipy both the top and bottom in fractions). 3/6 is the chance for ONE dice. You're counting 3 dice, so the total chance is 18 because there are 18 possibilities total. 3/6 woud be if you were going for odds or evens on a dice, or 50-50 chance. On two dice, it would be 6/12 chance to get odds / evens. Also, you count each number as individual. 3 ones on 3 dice is still an individual chance, so it's still 3 / 18 to get 3 1's, a 1/6 chance on each dice.

I 'unno, I'm no expert. I'm going to go and look this up.

[Ron AF Greve]

No you do NOT multiply both numerator and denominator (3/18 equals 1/6) . 3 * 1/6 = 3/6

Consider pieces of a pie. Say you have cut it in 6 i.e. each part is 1/6 now you take 3 parts you have half the pie right :-)

Edit: Just verify it on your calculator if you don't believe me.

[RobotParty]

Right, so this is ripped from http://gwydir.demon.co.uk/jo/probability/calcdice.htm#more under "more than two dice".

"The easiest is the probability of a number of dice being a particular number. For n dice, this is 1/(6n)[n = exponent]. For example, the probability for throwing 5 dice and getting them all sixes is 1/(65) = 1/(6x6x6x6x6) = 1/7776 = 0.000128 or 7775 to 1."

So, for the probability of 3 dice being the same particular number, it's 1/(6x6x6) = 1/216 and in this case, it's if you roll a pair on your first roll, which is 1/(6n2) = 1/36 - a pretty good chance for 2 dice and getting a pair. [Actually, this one is wrong in the context of the game, but I did 2 dice to give an idea of it, I can't find an easy way of finding out "at least 2 of x number on x dice rolls" although the pair still applies to the chance of getting the same number from 5 dice, it's still very low]

Let's say you rolling on the first dice roll, then it's 1/(6n5) = 1/7776 which... Is a very low chance.

That's simply the chance of getting the same number, since that's what you want to go for with a 5-of-a-kind or a 3-of-a-kind, in the case of you getting a pair straight off.

Man, this just brings me back to data management. I don't want to remember that stuff. Too much work. Way too much work.

EDIT: Also, yeah, my bad. I was confused. 1/6 is the same as 3/18. but isn't 1/6 multiplied by 3. 1/18 is 1/6 x 3 I think.

[Ron AF Greve]

What I calculate was one or more of the same number so you already had 2 then the chance you throw one or more of the ones equal to the first two.

Additionally you could throw three of a kind so for instance 6'es with the first two and three times 5 with the remaining together totalling 0.44.

What you do in your latest calculation is the chance you throw 5 of a kind. For anything in between you cannot just count 1/(6^n) because you also need to multiply that with all the combinations you can make, say you try to throw 2 6's then you can make the following combinations ( 3! /2!) = 3
x 6 6
6 x 6
6 6 x
My calculation is a lot simpler by just inverting and then subtracting it from 1

(Still thinking of creating a table that would tell me exactly what to do in every situation :-) ).

[Ron AF Greve]

No 1/6 * 3 is really 3/6 = 1/2 (try it on your calculator and you will see).

[RobotParty]

I give up. I never retain most of my math. I'll stick to steam-rolling the basics.

[Ron AF Greve]

Originally posted by Robot Death Party:

I give up. I never retain most of my math. I'll stick to steam-rolling the basics.

:-) You might also want do a little refresh course like for instance Khan academy https://www.khanacademy.org/math/pre-algebra/fractions-pre-alg/understanding-fractions-pre-alg/v/numerator-and-denominator-of-a-fraction

[RobotParty]

Originally posted by Ron AF Greve:

Originally posted by Robot Death Party:

I give up. I never retain most of my math. I'll stick to steam-rolling the basics.

:-) You might also want do a little refresh course like for instance Khan academy https://www.khanacademy.org/math/pre-algebra/fractions-pre-alg/understanding-fractions-pre-alg/v/numerator-and-denominator-of-a-fraction

I was too lazy for that when it was actually useful!

[Yes, I have a Mic.]

Dice poker is pure luck. Seriously don't get why they didn't just have real poker.

[Ron AF Greve]

Pure luck for the dice itself maybe but you still need to make a decission which dice to reroll for instance rolling 6 different numbers one of them being 6 do I reroll all dice or all but the 6?

[Yes, I have a Mic.]

Originally posted by Ron AF Greve:

Pure luck for the dice itself maybe but you still need to make a decission which dice to reroll for instance rolling 6 different numbers one of them being 6 do I reroll all dice or all but the 6?

Yeah. Deep strategy involved right there. -.-

[Lobstersaurus]

Try changing ur gambling tactics if you are having trouble. Example One:

If I roll a two pair with 2's and 4's and my opponent gets a pair of 3's. Even though my hand is better, i'll reroll everything but my 4's to try and get 3 of a kind.

The reasoning for that is because even if I don't get three of a kind, my pair of 4's will still beat his pair of 3's assuming he doesn't get three of a kind. If I get three of a kind and he gets three of a kind my hand will still beat his.

...............................................................................................................

Another example would be if I roll two pair with 2's and 4's and my opponent gets a pair of 6's. Instead of going for three of a kind like last time, i'll keep my 2 pair and roll my single die for a full house instead.

The reasoning for that is because if I don't get my three pair it'll lose to the opponent. If I get three pair and he gets three pair, i'll still lose. By keeping my hand, the opponent will lose if he doesn't get three pair. Also i'll have the chance of beating his three pair with a potential full house.
................................................................................................................

The same exact hand for both examples, but different tactics were used depending on my opponents cards. While a lot of it is luck, there are still tactics you can use to better your odds. Good luck...

[Alpha-methyl]

Originally posted by Dog:

Nah... it's pure luck.

I disagree, I am very sure it's heavily scripted and the opponents hand is usually going to be better, even if you have a flush or a straight or whatever, they get a higher one almost every time, especially if you bet the higher amounts. I have, literally, won 1 game out of 50 in The Witcher: Enchanced Edition betting mid / max - a real RNG is never so one sided.

If it were luck and probability, I wouldn't lose with my super good hands that should give the opponent a VERY miniscule chance to win. That's definitely not the case in this game. The NPC logic is far luckier than the player character and if you score a good hand and bet max, the opponent will beat you with the one higher hand. It's happened numerous times to me now.

Of course, I could just have the worst luck out of anyone ever, but considering this doesn't happen in other games... well...

Originally posted by JamT:

I'm with everyone in this so I think it's pure dumb luck. At first I hated dice poker because I always lose, but over time I started to like it.

Ahh, so you're an M?

[Dr.Fumetastic]

Originally posted by Alpha-methyl:

they get a higher one almost every time, especially if you bet the higher amounts.

That is actually how it works. If you can't beat an NPC just lower the initial bet to the lowest and you'll find yourself way more likely to win.

If it were luck and probability, I wouldn't lose with my super good hands that should give the opponent a VERY miniscule chance to win.

A miniscule chance is still a chance. You see this kind of salt everywhere on the forums if RNG is involved and the player cannot or will not be bothered to eliminate the opposing chance.

Of course, I could just have the worst luck out of anyone ever, but considering this doesn't happen in other games... well...

The reason why this might seem so is that the Dice AI indeed does have significant "luck" or "rolls" bonuses based on how "strong" they are. That bonus also increases with the amount used during the initial bet but does not during the other bets.