I'm a maths dunce so need some help with dice and percentages.
Say I have a game where I roll a D6. On a roll of 6 I win. On a roll of 1 I lose. On any other number I reroll.
I know the percentage of me winning on the first roll is 16.67% But what are the cumulative percentages of me winning on subsequent rolls if a 1 or 6 isn't rolled?
1/6
You either win or lose or reroll until you either win or lose? Why not just flip a coin instead?
>>46518064
Because that's a different game.
>>46518079
It's got the same probability.
Since you always reroll on 2-5, those numbers are basically eliminated.
Reiteration is discounted when calculating probabilities, so you you have even odds of rolling a 1 or a 6. It's effectively a coin toss.
>>46518116
But it's not immediately to coin toss right?
First roll: 16.67%
Second roll: ?%
>>>
% roll: 50%
>>46518140
Since there is no limit to how many times you can reroll, the only two possible end results are win or loss and they have an equal chance of happening. It's a coin toss with an added level of redundant complexity.
This is a really dumb thing to argue about btw.
>>46518179
I couldn't give a shit about the end result. I'm only interested in the cumulative percentages of winning.
>>46517999
Are we assuming numbers are eliminated after they are rolled? Say, you roll a 3, then the next roll can only get 1, 2, 4, 5, or 6?
>>46518197
For example.
First roll: 16.67%
Second roll: ?%
Third roll: ?%
etc
>>46518200
No. If you roll a 2-5 then it's a straight reroll of the dice so you can roll anything.
>>46518201
You have the same odds of 1 or 6 each time no matter how many times you reroll. To think your odds go up at all is the gambler's fallacy. Look it up. Odds are odds, they don't change because you roll a lot. Like >>46518179 said.
>>46518217
that's not what the gambler's fallacy is anon
Look up Bernoulli's probability formula, OP.
Anyway, there is no such thing as "cumulative probability" when you are dealing with independent events such as dice rolls.
Dice rolls are independent events, so on every roll you have equal propability of 1/6 of wining and 1/6 of loosing.
Propability of reroll is 4/6, so almost surely
(whitch means with propability 1) this game will end on some roll.
So no mater how much you will be rolling, you still have 1/6 chance of winning
>>46518236
>>46518244
If I roll 25 dice the percentage chance of me rolling a 1 on any of those is clearly going to be much greater than if I was only rolling 1 dice. How is what OP is asking any different?
>>46518229
https://en.wikipedia.org/wiki/Gambler%27s_fallacy
Yes, it is. He thinks that the odds of him rolling a 1 or 6 go up merely because he's getting 2, 3, 4, and 5. Those may as well be non-factors entirely since the only results that matter are 1 and 6. His odds don't go up or down at all, and his faulty logic for assuming so is the gambler's fallacy because he thinks "oh, I've rerolled a lot, my odds of getting 1 or 6 must be way higher now!" and they're not. And Unless he's talking about an infinite time period, in which case it blows his math to pieces as all odds then fly off into infinity.
>>46518257
Because he is adamantly interested in his chance of winning when it's irrelevant as the chance of losing grows in proportion. The number literally does not matter if you got infinite rerolls and the odds of a loss or a win are always the same.
But he doesn't roll 25 dice. He is rolling 1 dice until he rolls 1 or 6
>>46518257
He's rolling one dice where the odds are 1/6. You'd be rolling 25, that's more die to factor in than one.
>>46518257
That's what Bernoulli's formula is for. It calculates the probability of rolling a 1 on any of those dice, for example. Every roll is still independent, and the odds of rolling a number on any single die are still the same.
>>46518274
OP here.
Fine. Then what is the percentage chance of me rolling a 1 one dice, then a 1 from rolling 2 dice, etc.
1 dice: 16.67%
2: ?%
>>46518268
He can't roll infininite number of rolls, the game will end at some point.
P(rolling infinite times) = lim (4/6)^n = 0
>>46518284
>1 from rolling 2 dice
The chances of rolling 1 on 2d6 are absolutely zero.
>>46518284
Odds for one die: 1/6
2 dice: 1/6 for each dice. You just double it, but the ratio does not change. You're clearly not understanding this.
>>46518284
https://en.wikipedia.org/wiki/Bernoulli_distribution
>>46518303
Clearly the overall percentage must change.
>>46518309
Is there an online calculator or something for this?
>>46518309
I mean biniomal distribution
https://en.wikipedia.org/wiki/Binomial_distribution
If you want the change that at least one 1 will be rolled at some point, what you want is the inverse of the change that no 1 ever is ever rolled. For n rolls, that probability is (5/6) to the nth power. Subtract that from 1 and you have the probability you want.
>>46518310
It does not. Because each die is individual and each dice has the same 1/6 chance.
You roll 1 and you get 1/6.
Roll 2 and you get 1/6 for each, because the ratio is 2/12. Simplified that is still 1/6 because it's split between two die. This is why you're a math dunce. You're being told the answer and still don't accept it.
You're wrong. Read the thread next time faggot.
>>46518197
You're asking, for instance, "What are the chances that by the second roll I've won? By the third?" and so forth?
16.667% chance to succeed on 1st roll
11.111% chance to succeed on 2nd roll
7.407% chance to succeed on 3rd roll
4.938% chance to succeed on 4th roll
3.292% chance to succeed on 5th roll
2.195% chance to succeed on 6th roll
1.463% chance to succeed on 7th roll
0.976% chance to succeed on 8th roll
0.650% chance to succeed on 9th roll
0.434% chance to succeed on 10th roll
Your chance to succeed on each given step decreases, but you have less chance of actually making it there (because there's only a 1/3 chance with every roll that you get either a 1 or 6, ending the process). Your cumulative chances of success or below:
16.667% chance you succeed by the 1st roll
27.778% chance you succeed by the 2nd roll
35.185% chance you succeed by the 3rd roll
40.124% chance you succeed by the 4th roll
43.416% chance you succeed by the 5th roll
45.610% chance you succeed by the 6th roll
47.074% chance you succeed by the 7th roll
48.049% chance you succeed by the 8th roll
48.699% chance you succeed by the 9th roll
49.133% chance you succeed by the 10th roll
50.000% chance you succeed given infinite rolls
>>46518344
Learn how to quote who you're talking to, newfag. Don't get mad because people aren't giving you the answer you want for dice rolls.
>>46518334
How would you calculate the odds of winning this game on turn n?
>>46517999
If you allow unlimited rerolls it is statistically impossible to fail a roll. Here's the formula fora limited number of rolls though:
100 - ( ( ( (die size - target number) / die size )^number of rolls) * 100 )
>>46518364
Read the fucking thread moron. Plenty of people are giving the correct answer but you're too much of a raging faggot to realise you're wrong. Now fuck off out of this thread, you're not helping.
>>46518376
And you are by bitching and expecting people to do your homework for you?
>>46518310
So you want to know the chances for getting a roll of 1 on one of two dice?
There are 36 different combinations
11 of them have at least 1, 1.
So 11/36 is around 30% The one factor that you didn't specify was if a roll of 6 overrides a roll of 1 in this situation, if it does then there are 10 possible losing combinations making it a 28% chance to lose roughly, in that scenario you would have a 30% chance to win.
If a one overrides a six those would be reversed, if a 1 and a 6 cause a reroll it would become.
9/36 to win or lose making the chance 25% for either.
>>46518361
THANK YOU
jesus fucking christ
>>46518376
>>46518364
Do you raging homosexuals even realize that math isn't the problem? OP just doesn't know how to properly phrase a question.
>>46518201
I think I get what you're asking now. You multiply the odds of getting a 1 by the odds of getting a 2-5 for each iteration. For example:
First roll: 1/6 (0.166666...)
Second roll: 1/6 x 2/3 (0.111111...)
Third roll: 1/6 x 2/3 x 2/3 (0.074...)
And so on. These numbers will eventually add up to 0.5.
>>46518390
OP was clearly asking for percentages, not singular fractional dice rolls. It's not rocket science.
>>46518386
Please don't go into round 2 thinking you have a 27,8% chance of winning. That's not what was calculated.
Your chances of winning every separate round are exactly one in six.
>>46518404
Yes I know. I'm making rules for an RPG and am currently trying to balance different abilities. I didn't realise this place would sperg out at the mention of winning or losing.
>>46518401
That's not the point. OP actually wanted to know about his chances of winning this game BY turn n. He asked about his chances of winning this game ON turn n.
There is a difference.
>>46518418
OP was clear, you're just a faggot.
>>46518423
He obviously wasn't, otherwise this thread would have been over by the second post.
>>46518429
It's just overrun by retards, that's all.
>>46518361
Basically, you have a 1/6 (16.667%) chance to succeed on any given roll. You also have a 4/6 chance (66.667%) to not get a result and have to roll again.
So, you have a straight 16.667% chance to succeed no the 1st (because you're always going to be making that roll).
You're only going to be making the 2nd roll 4/6 of the time (66.667%), so your chance to succeed on that one is your chance of getting there (66.667%) multiplied by your percentage chance of winning on that individual roll (16.667%) = 11.111%.
Now, your chance of making it to the 3rd roll is 66.667% (your chance of making it to the 2nd roll) times 66.667% (your chance of making it to the 3rd roll if you make it to the 2nd one) = 44.444%. So you multiply 44.444% by your chance to win on any individual roll (16.667%) and get a 7.407% chance you succeed on your 3rd roll.
And for the 4th, it's 66.667% x 66.667% x 66.667% (or 66.667%^3, if you prefer that notation) = 29.630% chance that you get there. Multiply that by 16.667% to come up with a 4.938% chance to succeed on your 4th roll.
Then it's 66.667%^4 x 16.667% for the 5th roll.
66.667%^5 x 16.667% for the 6th roll.
And so on.
To get your cumulative chance of success, just add the numbers together. So your chance to succeed by the 3rd roll = your chance of succeeding on the 1st roll + your chance of succeeding on the 2nd roll + your chance of succeeding on the 3rd roll.
>>46518361
>Your chance to succeed on each given step decreases, but you have less chance of actually making it there
Sorry, that should say: "Your chance to succeed on each given step decreases BECAUSE you have less chance of actually making it there."
>>46518418
OP was clearly asking about the cumulative effect of rerolls, stop being a douche.
If you roll 1 d6 with a target number of 2 and allow 5 rerolls there is roughly an 80% chance at least one die will fall under 1 or 2. There is also roughly an 80% chance at least one die will fall under 5 or 6 as well.
>>46518495
5 rolls or 5 RE-rolls? Because that makes a difference. Also, I think you're wrong either way.
33.333% chance things are resolved by the 1st roll.
55.556% chance things are resolved by the 2nd roll.
70.370% chance things are resolved by the 3rd roll.
80.247% chance things are resolved by the 4th roll.
86.831% chance things are resolved by the 5th roll.
91.221% chance things are resolved by the 6th roll
94.148% chance things are resolved by the 7th roll.
96.098% chance things are resolved by the 8th roll.
97.399% chance things are resolved by the 9th roll.
98.266% chance things are resolved by the 10th roll.
And by "98.266% chance things are resolved by the 10th roll" I mean that by the time you've rolled the dice 10 times, there's a a 98.266% chance you've gotten either a win or a loss by then and don't need to make an 11th roll.
>>46518600
No, my math is right. There is roughly an 80% chance of getting at least one success on the 5th reroll.
>>46518741
No. There's a roughly 20% chance you'll get *to* the 5th roll, meaning there's a roughly 80% chance you'll have resolved things by the time you've rolled 4 times. Look at my numbers in >>46518600 and tell me where I'm wrong.
>>46518415
>OP clearly can't do maths
>I'm making rules for an RPG
>>46519092
Well, there's Nobilis, Amber Diceless and Golden Sky Stories.
And no matter how good you are at maths, that's no guarantee for something even just resembling a good game, see Strike!
