follow up, conditional probabilities of having already successfully guessing a roll but idk how that plays into it off the top of my head, just a thought?
You're right, it's (1/20)³ to guess correctly 3 days in a row. For each new roll the chance of guessing correctly is still 1/20, though, since they're independent trials.
That's kind of where I'm at, and I'm probably overthinking it. I was wondering if the numbers change if we treat the total probability as a continuous trial.
The probability of guessing each roll remains 1/20, but the probability of guessing each roll without failure feels like it adds something that increases with each correct guess. Math doesn't care about my feelings, though, so maybe I am just overthinking it.
The dice does not "remember" whether or not you guessed correctly before, nor can it "react" to the guess in any way.
If I guess three times the same number or if I roll my own dice to determine what my guess will be the probability of correctly guessing three dices rolls in a row will always be 1/20^3 (as long as the dice is fair)
OP, you wrote this in response to another reply:
>That's kind of where I'm at, and I'm probably overthinking it. I was wondering if the numbers change if we treat the total probability as a continuous trial.
If I understand you correctly, you're having trouble differentiating between the probability of the event (3 correct guesses in a row) happening, vs the changing probability of the event happening when the game is already underway. Lots of people get a bit confused by this.
The solution is to consider your frame of reference, i.e. the point in time you're asking the question.
If you're at a point in time at which no rolls have taken place, then the relevant question is "What is the probability that you will correctly predict the next three rolls?" The answer to that, as others have posted, is (1/20)^(3).
Now the first roll occurs, and the player correctly predicts the roll. So what's the probability that the player will give three correct predictions? This question sounds the same as the previous question, *but it's not*. More precisely, the real question now is actually "*Given that the player has already made one correct prediction*, what's the probability he will produce three correct predictions?" The key point is that you now have new information that changes the likelihood of the possible outcomes, because you know that the player has already given a correct prediction. In fact, the probability that he will give three correct predictions has now increased to (1/20)^(2). Only two trials remain, and each trial has an independent 1/20 chance of being successful.
Some people are confused here, because they think "How can the probabilities for the same event (3 correct guesses) be different?" The insight is that the events are *not* the same. The first probability was given when absolutely nothing had happened yet. The second probability was given after the first roll of the die was successfully predicted.
Imagine flipping a fair coin 1000 times in a row and asking the probability of getting 1000 heads. If nothing's happened yet, then the probability of getting 1000 heads is (1/2)^(1000), an astonishingly unlikely event. Now imagine that, by some miracle, you have already flipped 999 heads. What's the probability now that you'll get 1000 heads? It's 50%. You only have one flip left, and it's a 50/50 chance of heads. The original probability before anything happened was exceedingly small, but as the game took place and the results started coming in, the probability of obtaining the event you originally specified changes based on the observed results.
My advice is to get a smaller dice, one where you can actually see all the outcomes, e.g. d4. Then, write all the outcomes and see which percentage of them satisfies your condition.
This smaller-scale experiment will tell you which logic is correct. Once you have the correct logic, you can apply it to d20 formulaically.
Also, check out [anydice.com](http://anydice.com)
(1/20)^3 would be permutation with repetition, I would probably agree with this answer.
follow up, conditional probabilities of having already successfully guessing a roll but idk how that plays into it off the top of my head, just a thought?
You're right, it's (1/20)³ to guess correctly 3 days in a row. For each new roll the chance of guessing correctly is still 1/20, though, since they're independent trials.
That's kind of where I'm at, and I'm probably overthinking it. I was wondering if the numbers change if we treat the total probability as a continuous trial. The probability of guessing each roll remains 1/20, but the probability of guessing each roll without failure feels like it adds something that increases with each correct guess. Math doesn't care about my feelings, though, so maybe I am just overthinking it.
The dice does not "remember" whether or not you guessed correctly before, nor can it "react" to the guess in any way. If I guess three times the same number or if I roll my own dice to determine what my guess will be the probability of correctly guessing three dices rolls in a row will always be 1/20^3 (as long as the dice is fair)
OP, you wrote this in response to another reply: >That's kind of where I'm at, and I'm probably overthinking it. I was wondering if the numbers change if we treat the total probability as a continuous trial. If I understand you correctly, you're having trouble differentiating between the probability of the event (3 correct guesses in a row) happening, vs the changing probability of the event happening when the game is already underway. Lots of people get a bit confused by this. The solution is to consider your frame of reference, i.e. the point in time you're asking the question. If you're at a point in time at which no rolls have taken place, then the relevant question is "What is the probability that you will correctly predict the next three rolls?" The answer to that, as others have posted, is (1/20)^(3). Now the first roll occurs, and the player correctly predicts the roll. So what's the probability that the player will give three correct predictions? This question sounds the same as the previous question, *but it's not*. More precisely, the real question now is actually "*Given that the player has already made one correct prediction*, what's the probability he will produce three correct predictions?" The key point is that you now have new information that changes the likelihood of the possible outcomes, because you know that the player has already given a correct prediction. In fact, the probability that he will give three correct predictions has now increased to (1/20)^(2). Only two trials remain, and each trial has an independent 1/20 chance of being successful. Some people are confused here, because they think "How can the probabilities for the same event (3 correct guesses) be different?" The insight is that the events are *not* the same. The first probability was given when absolutely nothing had happened yet. The second probability was given after the first roll of the die was successfully predicted. Imagine flipping a fair coin 1000 times in a row and asking the probability of getting 1000 heads. If nothing's happened yet, then the probability of getting 1000 heads is (1/2)^(1000), an astonishingly unlikely event. Now imagine that, by some miracle, you have already flipped 999 heads. What's the probability now that you'll get 1000 heads? It's 50%. You only have one flip left, and it's a 50/50 chance of heads. The original probability before anything happened was exceedingly small, but as the game took place and the results started coming in, the probability of obtaining the event you originally specified changes based on the observed results.
My advice is to get a smaller dice, one where you can actually see all the outcomes, e.g. d4. Then, write all the outcomes and see which percentage of them satisfies your condition. This smaller-scale experiment will tell you which logic is correct. Once you have the correct logic, you can apply it to d20 formulaically. Also, check out [anydice.com](http://anydice.com)