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XBear

a quick primer on probability when rolling dice

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I'm sure several players already know this, but I seem to see a lot of people blaming luck for rolling four blanks or if their opponent rolled four natural hits, as if it's something that's supposed to be rare, so I thought this might be useful to those people who are not just complaining but actually want to understand a bit more about probabilities.

 

the basic thing every player know is averages: the chance of rolling a hit or crit on a red dice is 50%, or 0.5 in math notation, so if you roll 4 dice the average is 2. not that great, right? we all know sometimes we'll roll 4 and sometimes it might be zero, but how do you quantify that? with the binomial distribution, which tells you that, for 4 dice you'll get:

 

6% of the times: 4 hits/crits

25% of the times: 3

38% of the times: 2

25% of the times: 1

6% of the times: 0

 

ok so, 6% of the times isn't that much, or is it? well lets say you play 10 rounds and shoot for 10 rounds. now that's a 60% chance (6% times 10 rounds) you rolled 4 natural hits/crits in one round (or 4 natural blank/focus in one round). so it's not surprising it happened. it could be the first roll you made, or the last you made, it's still about a 60% chance you make that roll.

 

so what if I have a TL and focus? OK the binomial distribution tells you that:

 

78% of the times: 4 hits/crits

20% of the times: 3 

2% of the times: 2

0.1 % of the times: 1

tiny % of the times: 0

 

see, even assuming you shoot with a TL+focus each round for 10 rounds, the chance of rolling only 2 hits/crits in one of those 10 rounds (even after rerolling and spending a focus)  is about 20%. it's still a pretty significant chance! in fact, if you look at the number of rounds you play in a tournament, or across multiple tournaments, you'll see there's a pretty significant chance you'll roll zero or one hits even with a TL and focus.

 

 

as an example, in a tournament I played a few days ago, my opponent had a choice to move Vader in an aggressive position to focus fire on my Poe, and a more defensive position where he would receive decreased fire from my ships. he went for the risky choice, rolled below average, Poe survived, my ships rolled above average, he lost Vader and with it the game.

 

it's not that he necessarily made the wrong choice. if he had killed Poe it would have given him an advantage. but several times he complained about that roll. I got the impression he thought it was improbable. what I meant to show is that yes it was a deviation from average, but if you look at the binomial distribution and consider the number of rolls you're making, you can understand that those rolls do happen with pretty significant chances. it's not luck or lack thereof, but rather normal and expected probability distributions.

 

 

if you happen to like distributions, you could even take the 6% chance of rolling 4 natural hits and ask, how many times am I going to roll 4 hits with 4 dice in 10 rounds? well, you can make another binomial distribution, and see that you have about:

 

54% chance of rolling 4 natural hits zero times in 10 rounds,

34% chance to roll 4 natural hits 1 time in 10 rounds

10% 2 times

2% 3 times

 

so you see, rolling 4 natural hits several times in 10 rounds is not so unusual. and just as well, since the chance to get 4 natural hit is the same as the chance to get 4 natural non hits, the chance of rolling 4 natural zero hits several times in 10 rounds is pretty significant as well.

 

 

PS: I tried to keep the explanation simple and so the above is not intended as a rigorous statistical treatment, for example I have not discussed conditional probabilities (the probability for a roll to happen based on a previous roll).

Edited by XBear

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I am a professional poker player.  I see the same things in poker that I see in X-Wing.

 

Human beings see patterns where none truly exist, and rarely do they understand what "the long run" really means.  It just appears to them that they have taken a "horrible beat", when in reality, it is often a probable event that truly occurs.

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We all tend to focus a lot more on the poor dice rolls, when they occur, than on the positive ones. Like I said in another thread -- you know every story you've ever heard about the poor sap who just rolls terribly? For every one of those stories, there's a guy sitting across the table who just rolled really freakin' well. But how many of those stories do you hear?

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ok so, 6% of the times isn't that much, or is it? well lets say you play 10 rounds and shoot for 10 rounds. now that's a 60% chance (6% times 10 rounds) you rolled 4 natural hits/crits in one round (or 4 natural blank/focus in one round). so it's not surprising it happened. it could be the first roll you made, or the last you made, it's still about a 60% chance you make that roll.

This is not how statistics work. The dice have no memory of the previous roll. Each roll is a unique chance at 4 chits/crits. You really have to look at the chance of NOT getting 4 hits/crits over 10 rolls.

There is a 6.25% chance rolling 4 dice that all 4 will come up with hits/crits (50% chance on each individual die). There is a 93.75% chance they won't do that. There is a 52.45% chance you will not to roll at least one of those combos in a sting of ten rolls, or a 47.55% chance that at least one roll out of a group of ten rolls will be 4 hits/crits.

tl;dr probability and combinatorics are weird.

Edited by Hexis

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We all tend to focus a lot more on the poor dice rolls, when they occur, than on the positive ones. Like I said in another thread -- you know every story you've ever heard about the poor sap who just rolls terribly? For every one of those stories, there's a guy sitting across the table who just rolled really freakin' well. But how many of those stories do you hear?

while I generally agree with you, it's possible in theory sometimes there's something going on. say they left the dice in the sun and now they roll below average because they are defective

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ok so, 6% of the times isn't that much, or is it? well lets say you play 10 rounds and shoot for 10 rounds. now that's a 60% chance (6% times 10 rounds) you rolled 4 natural hits/crits in one round (or 4 natural blank/focus in one round). so it's not surprising it happened. it could be the first roll you made, or the last you made, it's still about a 60% chance you make that roll.

This is not how statistics work. The dice have no memory of the previous roll. Each roll is a unique chance at 4 chits/crits. You really have to look at the chance of NOT getting 4 hits/crits over 10 rolls.

There is a 6.25% chance rolling 4 dice that all 4 will come up with hits/crits (50% chance on each individual die). There is a 93.75% chance they won't do that. There is a 52.45% chance you will not fail to roll at least one of those combos in a sting of ten rolls, or a 47.55% chance that at least one roll out of a group of ten rolls will be 4 hits/crits.

tl;dr probability and combinatorics are weird.

 

 

Gah Hexis, you beat me to it.  Except you were nicer about it than I was going to be...  I was going to be obnoxious and say something like "a quick primer on how probabilities work" or something... 

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ok so, 6% of the times isn't that much, or is it? well lets say you play 10 rounds and shoot for 10 rounds. now that's a 60% chance (6% times 10 rounds) you rolled 4 natural hits/crits in one round (or 4 natural blank/focus in one round). so it's not surprising it happened. it could be the first roll you made, or the last you made, it's still about a 60% chance you make that roll.

This is not how statistics work. The dice have no memory of the previous roll. Each roll is a unique chance at 4 chits/crits. You really have to look at the chance of NOT getting 4 hits/crits over 10 rolls.

There is a 6.25% chance rolling 4 dice that all 4 will come up with hits/crits (50% chance on each individual die). There is a 93.75% chance they won't do that. There is a 52.45% chance you will not fail to roll at least one of those combos in a sting of ten rolls, or a 47.55% chance that at least one roll out of a group of ten rolls will be 4 hits/crits.

tl;dr probability and combinatorics are weird.

 

I think you miswrote something there, you seem to say there is a 52% chance to roll 4 hits at least once in 10 tries, and then you say there's a chance of 48% for the same thing?

 

if you look at the last binomial distribution you'll see that I give a 54% chance of rolling 4 hits zero times in 10 tries, and therefore 46% chance of rolling 4 hits at least once in 10 tries. if you average all the possible outcomes for 1 or more rolls of 4 hits in 10 tries you get 60%, which agrees with the simple average of multiplying 6% by 10, as it should.

 

while it's possible I made some mistakes and in that case I'm grateful for any corrections, you posting something merely stating I'm wrong without any real explanation it's just going to be confusing to readers who don't know much about probabilities distributions.

 

by the way, if you don't round the digits like I did, the binomial distribution more correctly says the chance of rolling 4 hits zero times in 10 rounds is 52.45% (not 54%) so you are saying the same thing I did (except you put an extra 'not' that made if confusing to read). our math agrees, and you had no reason to say my math was wrong.

Edited by XBear

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We all tend to focus a lot more on the poor dice rolls, when they occur, than on the positive ones. Like I said in another thread -- you know every story you've ever heard about the poor sap who just rolls terribly? For every one of those stories, there's a guy sitting across the table who just rolled really freakin' well. But how many of those stories do you hear?

It is human nature to take credit for the good and blame the bad on outside forces.

If I win, it is because I am an awesome player. My retelling of the game doesn't focus on the dice rolls but on my superior skills.

If I lose, it must be some other factor. My retelling of the game focus on the random factors rather than the better play of my opponent.

These alternate retellings aren't dishonest, but rather how we subconsciously interpret events.

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Another thing when people compare results is that they really forget about those times of overkill.  When looking at overkill the Green dice are probably the better example because while you have a standard range there are times when part of that range is no better than a lesser part.  If you roll 4 defense dice and have to cancel two "good" attack dice you only pay attention to three result: 0 success, 1 success, and 2+ successes although this last range also includes all the 3 and 4 good result times which really gives you a bias when you say you have having bad rolls.

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Never tell me the odds. Lol. I try to use the force on the dice. No luck. :(

XBear crew upgrade: say out loud the expected roll according to the binomial distribution. if you DON'T roll the expected result, add 1 hit or evade 

Edited by XBear

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Chance of 2 damage with autocorrector autoblaster: 100%

That's the only kind of probability worth calculating

I love how it works even when you get the crit that makes you roll 1 less die, because AC still lets you add 2 hits.

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ok so, 6% of the times isn't that much, or is it? well lets say you play 10 rounds and shoot for 10 rounds. now that's a 60% chance (6% times 10 rounds) you rolled 4 natural hits/crits in one round (or 4 natural blank/focus in one round). so it's not surprising it happened. it could be the first roll you made, or the last you made, it's still about a 60% chance you make that roll.

This is not how statistics work. The dice have no memory of the previous roll. Each roll is a unique chance at 4 chits/crits. You really have to look at the chance of NOT getting 4 hits/crits over 10 rolls.

There is a 6.25% chance rolling 4 dice that all 4 will come up with hits/crits (50% chance on each individual die). There is a 93.75% chance they won't do that. There is a 52.45% chance you will not fail to roll at least one of those combos in a sting of ten rolls, or a 47.55% chance that at least one roll out of a group of ten rolls will be 4 hits/crits.

tl;dr probability and combinatorics are weird.

I think you miswrote something there, you seem to say there is a 52% chance to roll 4 hits at least once in 10 tries, and then you say there's a chance of 48% for the same thing?

 

if you look at the last binomial distribution you'll see that I give a 54% chance of rolling 4 hits zero times in 10 tries, and therefore 46% chance of rolling 4 hits at least once in 10 tries. if you average all the possible outcomes for 1 or more rolls of 4 hits in 10 tries you get 60%, which agrees with the simple average of multiplying 6% by 10, as it should.

 

while it's possible I made some mistakes and in that case I'm grateful for any corrections, you posting something merely stating I'm wrong without any real explanation it's just going to be confusing to readers who don't know much about probabilities distributions.

 

by the way, if you don't round the digits like I did, the binomial distribution more correctly says the chance of rolling 4 hits zero times in 10 rounds is 52.45% (not 54%) so you are saying the same thing I did (except you put an extra 'not' that made if confusing to read). our math agrees, and you had no reason to say my math was wrong.

You said there's a 60% chance, or 6% times 10, chance of coming up with 4 successes in 10 rounds. That's demonstratably not true, and it's what Hexis was objecting to.

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ok so, 6% of the times isn't that much, or is it? well lets say you play 10 rounds and shoot for 10 rounds. now that's a 60% chance (6% times 10 rounds) you rolled 4 natural hits/crits in one round (or 4 natural blank/focus in one round). so it's not surprising it happened. it could be the first roll you made, or the last you made, it's still about a 60% chance you make that roll.

This is not how statistics work. The dice have no memory of the previous roll. Each roll is a unique chance at 4 chits/crits. You really have to look at the chance of NOT getting 4 hits/crits over 10 rolls.

There is a 6.25% chance rolling 4 dice that all 4 will come up with hits/crits (50% chance on each individual die). There is a 93.75% chance they won't do that. There is a 52.45% chance you will not fail to roll at least one of those combos in a sting of ten rolls, or a 47.55% chance that at least one roll out of a group of ten rolls will be 4 hits/crits.

tl;dr probability and combinatorics are weird.

I think you miswrote something there, you seem to say there is a 52% chance to roll 4 hits at least once in 10 tries, and then you say there's a chance of 48% for the same thing?

 

if you look at the last binomial distribution you'll see that I give a 54% chance of rolling 4 hits zero times in 10 tries, and therefore 46% chance of rolling 4 hits at least once in 10 tries. if you average all the possible outcomes for 1 or more rolls of 4 hits in 10 tries you get 60%, which agrees with the simple average of multiplying 6% by 10, as it should.

 

while it's possible I made some mistakes and in that case I'm grateful for any corrections, you posting something merely stating I'm wrong without any real explanation it's just going to be confusing to readers who don't know much about probabilities distributions.

 

by the way, if you don't round the digits like I did, the binomial distribution more correctly says the chance of rolling 4 hits zero times in 10 rounds is 52.45% (not 54%) so you are saying the same thing I did (except you put an extra 'not' that made if confusing to read). our math agrees, and you had no reason to say my math was wrong.

You said there's a 60% chance, or 6% times 10, chance of coming up with 4 successes in 10 rounds. That's demonstratably not true, and it's what Hexis was objecting to.

 

 

you and him didn't understand the math of my post. honestly, take more than 5 seconds to reply before saying I'm wrong, and if you think I'm wrong (hey, I could be wrong) then EXPLAIN it, not just merely state it. 

 

I took the time to read his post and see through his grammar mistake. if you take into account my rounding down, the last (more advanced) binomial distribution I show in my original post says 54% of zero (52.45% more accurately), 34% of 1, 10% of 2, 2% of 3. if you average the chances for 1, 2, and 3 successes of 4 hits in 10 rounds, you'll get an average of 60%. this means that, on average, you roll 4 hits 0.6 times (on average) in 10 rounds. this agrees with the basic calculation that if something has a chance of 6% and you have 10 tries, you get a 60% "chance of it happening", which is a simpler and less accurate way to say, I was giving the expected average.

 

basic probabilities have to agree with those found through binomial distributions, otherwise statistics makes no sense. finally, the bit of 6% times 10 was to make it simple to understand, as I said at the end of my original post, it was not meant as a rigorous treatment. the more rigorous treatment is the last binomial distribution which agrees with what he was saying (and agrees with the expected average given by the more basic calculation).

Edited by XBear

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you and him didn't understand the math of my post. honestly, take more than 5 seconds to reply before saying I'm wrong, and if you think I'm wrong (hey, I could be wrong) then EXPLAIN it, not just merely state it.

Here's what I'm objecting to:

ok so, 6% of the times isn't that much, or is it? well lets say you play 10 rounds and shoot for 10 rounds. now that's a 60% chance (6% times 10 rounds) you rolled 4 natural hits/crits in one round (or 4 natural blank/focus in one round). so it's not surprising it happened. it could be the first roll you made, or the last you made, it's still about a 60% chance you make that roll.

Then you did it again:

so what if I have a TL and focus? OK the binomial distribution tells you that:

 

78% of the times: 4 hits/crits

20% of the times: 3 

2% of the times: 2

0.1 % of the times: 1

tiny % of the times: 0

 

see, even assuming you shoot with a TL+focus each round for 10 rounds, the chance of rolling only 2 hits/crits in one of those 10 rounds (even after rerolling and spending a focus)  is about 20%. it's still a pretty significant chance!

You're using distributions, and it happens that the likelihood of seeing an event with a probability of 0.0215 at least once in 10 trials is 0.1953, or about 20%. But in the context of those two quotes, it looks like you're saying you can take an event with a probability p and multiply it by the number of trials n to get the probability of seeing that event at least once.

That's the problem. It's what Hexis and Khyros objected to, and it's what I objected to.

The problem is not that I don't understand your math. I've taught statistics and physics at the college level, I work with statistics and random variables on a daily basis at my job, and I'm completing a Ph.D. in social science with a strong emphasis on psychometrics, sampling, and modeling. The problem is that you got the quantities roughly right, but in a way that feeds into a strong and common misconception about how random variables work--and you did it under the headline "a quick primer on probability when rolling dice".

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you and him didn't understand the math of my post. honestly, take more than 5 seconds to reply before saying I'm wrong, and if you think I'm wrong (hey, I could be wrong) then EXPLAIN it, not just merely state it.

Here's what I'm objecting to:

ok so, 6% of the times isn't that much, or is it? well lets say you play 10 rounds and shoot for 10 rounds. now that's a 60% chance (6% times 10 rounds) you rolled 4 natural hits/crits in one round (or 4 natural blank/focus in one round). so it's not surprising it happened. it could be the first roll you made, or the last you made, it's still about a 60% chance you make that roll.

Then you did it again:

so what if I have a TL and focus? OK the binomial distribution tells you that:

 

78% of the times: 4 hits/crits

20% of the times: 3 

2% of the times: 2

0.1 % of the times: 1

tiny % of the times: 0

 

see, even assuming you shoot with a TL+focus each round for 10 rounds, the chance of rolling only 2 hits/crits in one of those 10 rounds (even after rerolling and spending a focus)  is about 20%. it's still a pretty significant chance!

You're using distributions, and it happens that the likelihood of seeing an event with a probability of 0.0215 at least once in 10 trials is 0.1953, or about 20%. But in the context of those two quotes, it looks like you're saying you can take an event with a probability p and multiply it by the number of trials n to get the probability of seeing that event at least once.

That's the problem. It's what Hexis and Khyros objected to, and it's what I objected to.

The problem is not that I don't understand your math. I've taught statistics and physics at the college level, I work with statistics and random variables on a daily basis at my job, and I'm completing a Ph.D. in social science with a strong emphasis on psychometrics, sampling, and modeling. The problem is that you got the quantities roughly right, but in a way that feeds into a strong and common misconception about how random variables work--and you did it under the headline "a quick primer on probability when rolling dice".

 

if you multiply p by n you get the expected average number of events. since I was not writing a textbook, to simplify things I: 1. called the expected average the chance of it happening (you added 'at least once', I didn't say that) 2. I wrote the correct method to calculate it as binomial distribution later in the post 3. I put a disclaimer the above treatment was not rigorous.

 

there are plenty of rigorous treatments anyone can google but obviously several players don't go there and read it, so I tried to simplify it. 

 

for example, the chance of rolling a hit for 1 die is 50%, if you roll 4 you get 200% "chance of it happening", which is not a rigorous term, and technically speaking, it's totally wrong, since no chance can be higher than 100%. but hey, 0.5 times 4 gives you the expected average number of successes, which is 2. and you can calculate the expected average of successes also through the binomial distribution for 4 tries with a 50% chance of success, by averaging the binomial chances for 1, 2, 3 and 4 successes, and it agrees with the simple math as it should. that's what I did, and I didn't get it "roughly right", I simply rounded to simple digits, so 6% instead of 6.25%, which for example leads to 54% in the binomial instead of 52.45%.

Edited by XBear

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I'd like to know how you came up with the numbers using both TL an focus. Remember us poor non-science people. The hardest math I had to do in the last 18 years was using Hardy Cross and the Hazen-Williams formula.

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you and him didn't understand the math of my post. honestly, take more than 5 seconds to reply before saying I'm wrong, and if you think I'm wrong (hey, I could be wrong) then EXPLAIN it, not just merely state it.

Here's what I'm objecting to:

ok so, 6% of the times isn't that much, or is it? well lets say you play 10 rounds and shoot for 10 rounds. now that's a 60% chance (6% times 10 rounds) you rolled 4 natural hits/crits in one round (or 4 natural blank/focus in one round). so it's not surprising it happened. it could be the first roll you made, or the last you made, it's still about a 60% chance you make that roll.

Then you did it again:

so what if I have a TL and focus? OK the binomial distribution tells you that:

 

78% of the times: 4 hits/crits

20% of the times: 3 

2% of the times: 2

0.1 % of the times: 1

tiny % of the times: 0

 

see, even assuming you shoot with a TL+focus each round for 10 rounds, the chance of rolling only 2 hits/crits in one of those 10 rounds (even after rerolling and spending a focus)  is about 20%. it's still a pretty significant chance!

You're using distributions, and it happens that the likelihood of seeing an event with a probability of 0.0215 at least once in 10 trials is 0.1953, or about 20%. But in the context of those two quotes, it looks like you're saying you can take an event with a probability p and multiply it by the number of trials n to get the probability of seeing that event at least once.

That's the problem. It's what Hexis and Khyros objected to, and it's what I objected to.

The problem is not that I don't understand your math. I've taught statistics and physics at the college level, I work with statistics and random variables on a daily basis at my job, and I'm completing a Ph.D. in social science with a strong emphasis on psychometrics, sampling, and modeling. The problem is that you got the quantities roughly right, but in a way that feeds into a strong and common misconception about how random variables work--and you did it under the headline "a quick primer on probability when rolling dice".

if you multiply p by n you get the expected average number of events. since I was not writing a textbook, to simplify things I: 1. called the expected average the chance of it happening (you added 'at least once', I didn't say that) 2. I wrote the correct method to calculate it as binomial distribution later in the post 3. I put a disclaimer the above treatment was not rigorous.

 

there are plenty of rigorous treatments anyone can google but obviously several players don't go there and read it, so I tried to simplify it.

The expected number of successes is NOT the same thing as the chance of seeing a success. If you're applying the not-rigorous method to a coin toss, do you have a 500% chance of seeing a heads result after 10 tosses?

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you and him didn't understand the math of my post. honestly, take more than 5 seconds to reply before saying I'm wrong, and if you think I'm wrong (hey, I could be wrong) then EXPLAIN it, not just merely state it.

Here's what I'm objecting to:

ok so, 6% of the times isn't that much, or is it? well lets say you play 10 rounds and shoot for 10 rounds. now that's a 60% chance (6% times 10 rounds) you rolled 4 natural hits/crits in one round (or 4 natural blank/focus in one round). so it's not surprising it happened. it could be the first roll you made, or the last you made, it's still about a 60% chance you make that roll.

Then you did it again:

so what if I have a TL and focus? OK the binomial distribution tells you that:

 

78% of the times: 4 hits/crits

20% of the times: 3 

2% of the times: 2

0.1 % of the times: 1

tiny % of the times: 0

 

see, even assuming you shoot with a TL+focus each round for 10 rounds, the chance of rolling only 2 hits/crits in one of those 10 rounds (even after rerolling and spending a focus)  is about 20%. it's still a pretty significant chance!

You're using distributions, and it happens that the likelihood of seeing an event with a probability of 0.0215 at least once in 10 trials is 0.1953, or about 20%. But in the context of those two quotes, it looks like you're saying you can take an event with a probability p and multiply it by the number of trials n to get the probability of seeing that event at least once.

That's the problem. It's what Hexis and Khyros objected to, and it's what I objected to.

The problem is not that I don't understand your math. I've taught statistics and physics at the college level, I work with statistics and random variables on a daily basis at my job, and I'm completing a Ph.D. in social science with a strong emphasis on psychometrics, sampling, and modeling. The problem is that you got the quantities roughly right, but in a way that feeds into a strong and common misconception about how random variables work--and you did it under the headline "a quick primer on probability when rolling dice".

if you multiply p by n you get the expected average number of events. since I was not writing a textbook, to simplify things I: 1. called the expected average the chance of it happening (you added 'at least once', I didn't say that) 2. I wrote the correct method to calculate it as binomial distribution later in the post 3. I put a disclaimer the above treatment was not rigorous.

 

there are plenty of rigorous treatments anyone can google but obviously several players don't go there and read it, so I tried to simplify it.

The expected number of successes is NOT the same thing as the chance of seeing a success. If you're applying the not-rigorous method to a coin toss, do you have a 500% chance of seeing a heads result after 10 tosses?

 

yes exactly. 500% chance, which I agree is technically wrong, but in colloquial terms it's something people say (sorry I edited my previous post to clarify that before I saw your reply). 

Edited by XBear

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I'd like to know how you came up with the numbers using both TL an focus. Remember us poor non-science people. The hardest math I had to do in the last 18 years was using Hardy Cross and the Hazen-Williams formula.

you use the binomial distribution with a probability of 0.5 for natural hits/crits, but with TL and focus you do the binom. distribution with a probability of 0.94.

 

why 0.94? because the chance of rolling a blank is 0.25, so the chance of rolling anything but a blank is 0.75.

 

so if you can reroll the die, you get a chance of either a success on the first try, 0.75, or a blank on the first try 0.25, times a success on the reroll, 0.75

 

so it's 0.75+0.25*0.75=0.94

 

you add chances that are x OR y and multiply chances that are x AND y

Edited by XBear

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I'd like to know how you came up with the numbers using both TL an focus. Remember us poor non-science people. The hardest math I had to do in the last 18 years was using Hardy Cross and the Hazen-Williams formula.

you use the binomial distribution with a probability of 0.5 for natural hits/crits, but with TL and focus you do the binom. distribution with a probability of 0.94.

 

why 0.94? because the chance of rolling a blank is 0.25, so the chance of rolling anything but a blank is 0.75.

 

so if you can reroll the die, you get a chance of either a success on the first try, 0.75, or a blank on the first try 0.25, times a success on the reroll, 0.75

 

so it's 0.75+0.25*0.75=0.94

 

you add chances that are x OR y and multiply chances that are x AND y

Ok, now I'm really lost. If the first two numbers in the equation represent the OR part and the third number represents the AND doesn't that come out to 0.75? (1*0.75) Please excuse this Math Challenged Fool.

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I used to know this guy that was a math professional, specifically Statistics and Probability.  One day I was telling him I was flying the family to Orlando to go to Disney World.  He freaked out.  He was telling me about the odds of there being a bomb on a plane and how dangerous it was and that he always would drive even if the trip was days instead of hours.

 

About a year later, I'm on a plane and here he comes walking down the aisle.  I said to him, "Hey what's up?  I though you said the odds of a bomb being on a plane were too high and you would never fly."

 

He said, " I know, I know, but then I recalculated the odds of there being TWO bombs on a plane and the number is infinitesimal."

 

I said, " What does that have to do with anything?"

 

He smiled and replied, "Well now I carry my own bomb."

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