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Subject: A simple question of probability rss

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Tobias
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I have a normal D6. The die is not rigged, fixed or anything - just a standard D6 out of any boardgame. I am rolling it, trying to get a "4". The likeliness is exactly 1/6, so 16.666...%. That is what I know. Now two people begin a conversation next to me:

Person A: The probability for a "4" is always 16.67%, stop trying so hard.
Person B: No, if you roll more often, it will get more likely that you will get a "4".
Person A: That is not true. The die remains the same, so for each unique value you want to get, your probability will always be 16.67%. There are as many sixes on that die as fours - you will have to use a smaller die to increase the likeliness, e.g. a D4, which raises the 16.67% up to 25%.
Person B: Again, not entirely correct. You can roll the D6 a hundred times and most probably will get a "4" at some point eventually. The probabilities of each single roll add up in some way, there also is a formula for it. Sad one I forgot it right now.

So the question to clear this matter once and for all is:
is Person B correct and if yes, what is the formula? And if so, how often do I have to roll to get a "4" at, like say, "99%"?
Both standpoints seem kind of fair to me, each in their own way. But I am no mathematician and have never been good at math, so I am no help to the discussion here anyway.
Thanks for help and the most detailed but at the same time plain explanation you can deliver. Really, thanks millions in advance ♥
Also, have a great new years eve of course! meeple
Bacon
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Don Weed
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The difference is what you are looking at for a sample size. If you are rolling an unbiased die then, yes, the probability will be 16.667 all the time. If you look at more than one consecutive roll (i.e. the chances of getting a 4 in three rolls then the chance is cumulative (16.77 + 16.77 +16.77 or approximately 1/2, etc.). If you are looking to get at least one 4 in X number of rolls the equations get more complex as the probabilities will multiply (you may get 1, 2, 3 or 4 fours in four rolls). The argument seems to be one of independent events vs. sets of numbers.

Long ago Pirate baseball announcer Bob Prince called player B's argument the 'hidden vigorish'. The theory is that the probability of something that may happen (but hasn't) increases the longer that it hasn't happened. If this were true the Cubs would have won the World Series decades ago.
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Brent Gerig
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LookAtTheBacon wrote:
I have a normal D6. The die is not rigged, fixed or anything - just a standard D6 out of any boardgame. I am rolling it, trying to get a "4". The likeliness is exactly 1/6, so 16.666...%. That is what I know. Now two people begin a conversation next to me:

Person A: The probability for a "4" is always 16.67%, stop trying so hard.
Person B: No, if you roll more often, it will get more likely that you will get a "4".
Person A: That is not true. The die remains the same, so for each unique value you want to get, your probability will always be 16.67%. There are as many sixes on that die as fours - you will have to use a smaller die to increase the likeliness, e.g. a D4, which raises the 16.67% up to 25%.
Person B: Again, not entirely correct. You can roll the D6 a hundred times and most probably will get a "4" at some point eventually. The probabilities of each single roll add up in some way, there also is a formula for it. Sad one I forgot it right now.

So the question to clear this matter once and for all is:
is Person B correct and if yes, what is the formula? And if so, how often do I have to roll to get a "4" at, like say, "99%"?
Both standpoints seem kind of fair to me, each in their own way. But I am no mathematician and have never been good at math, so I am no help to the discussion here anyway.
Thanks for help and the most detailed but at the same time plain explanation you can deliver. Really, thanks millions in advance ♥
Also, have a great new years eve of course! meeple
Bacon


Perfect example of the Gambler's Fallacy, which is the incorrect belief that previous outcomes affect the probability of future outcomes. Every single die roll, taken individually, has a 16.67% chance of rolling a 4.

Now, that isn't to say that you can't also do probability on groups of die rolls. So yes, you can calculate how many times (on average) you'll have to roll in order for the probably of rolling at least one 4 to be 99%. But that doesn't change the fact that if you look at any single die roll, the probably of a 4 will still be 16.67%.

(Caveat: it's been a long time since I've actually taken a statistics course, so I may have said something wrong.)
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Brent Gerig
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To go into a bit more (hopefully correct) detail:

You have a 5/6 chance on a single die roll of not rolling a 4 (or any other given number). If you roll two dice (the same as rolling one die twice), then you have a 5/6 * 5/6 chance of not rolling a 4. As you add dice/rolls, you multiply by 5/6 each time. For a >99% chance of rolling what you want, you need to get to a <1% chance of not rolling what you want. So, put that together and you get 1 - (5/6)^n > .99, where n is the number of rolls. Some trial and error gets 26 for n. So, you want a 99% chance of rolling at least one 4? Roll 26 dice.
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Pete
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They're both right. A is right in the real world and B is right in fantasyland.

Pete (sums up)
 
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Brent Gerig
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And again, probabilities and statistics just talk about likelihoods, not certainties. So if you roll a die 25 times, and still haven't rolled a 4, then sorry, the chance of a 4 on the next die roll is still 1 in 6. Not 99%.
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Brent Gerig
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plezercruz wrote:
They're both right. A is right in the real world and B is right in fantasyland.

Pete (sums up)


You tell B he's right if you're the one betting against him.
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Alison Mandible
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Your person A and person B are not in disagreement. On any one roll, the chance of getting a 4 is 1/6, as person A said. But if you roll several times, the chance of getting at least one 4 is higher.

Look at it this way. Suppose you have three friends (X, Y and Z) coming to visit you at noon. Each one might be running late-- unlike with dice, you DON'T know the odds. But you still know for sure: the chance that *anyone* will be late is higher than the chance that *a given person* will be late.

After all, suppose you have this noon meeting every day for a year. List all the times Xavier (person X) was late. Now list all the times anybody, including Xavier, was late. Every thing on the first list will be on the second list; every time Xavier is late, that also goes into the category "someone, it doesn't matter who, is late". But the second list, the "anybody is late" list, also includes other listings-- times that person Y was late but not person X, and times that both Y and Z were late, etc. So, no matter what the odds for each person are (and, unlike die rolls, the odds don't have to be the same for each person), we see that *at least one person being late* happens more often than *one person we pick ahead of time being late*.

It's the same with dice. If I roll a die 10 times, the chance of *at least one die roll being a 4* is higher than *one specific die roll I pick ahead of time being a 4*.

Okay, so what's the formula? Well, let's do it the easy way, and find the percentage chance of NONE of the rolls being a 4. Obviously, if you add

chance of 0 rolls being a 4
plus
chance of 1+ rolls being a 4

together, you get 100%.

If you roll a die once, the chance of failure (not getting a 4) is 5/6, right?

What if you roll twice? Well, person A was right; the chance during any specific single roll is always 1/6 of getting a 4, and 5/6 of not getting a 4. So the chance of failing twice is 5/6 * 5/6 = 25/36.

Rolling the die and failing all three times? 5/6 * 5/6 * 5/6 = 125/216 (a little more than 50%).

So how many times do we have to roll for our chance of failing all the time to be about 1%, so that our chance of succeeding-- of rolling a 4 at least once is about 99%?

There are ways to find this more directly. But I'm lazy, so I just typed things into Google until I found the right answer.

I typed
(5/6) ** 3
into Google and got 0.578. That's the decimal form of 125/216, more or less. To try multiplying 5/6 by itself eleven times instead of three, I typed
(5/6) ** 11
and got 0.134-- about 13%. Still not low enough!

By just trying each possible number, I found that (5/6) ** 25 is almost exactly 0.01. That means, the chance of getting an unbroken streak of 25 failures-- 25 rolls of not-4 in a row-- is about 1%. So the chance of rolling a die 25 times and getting 4 at least once is about 99%.
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Ryan
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What has already been said is correct.

The probably of a single unbiased d6 throw will always be 1/6 for any outcome in the sample set (1 through 6 as integers). The probability of rolling at least one 4 during n unbiased d6 throws = 1-(5/6)^n.

The probability of rolling a 4 on a single unbiased d6 throw never changes, but the cumulative probability of rolling at least one 4 does increase as n throws gets larger.

Edit - btw, this is only true when the outcomes are independent, which they are with an unbiased d6.
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Lluluien
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Anyone looking for clarification on this should look up the central limit theorem and independence (probability theory) on Wikipedia. Person A is right that rolls are independent. Person B is right that rolling a number at least once happens more often the more you roll. However, Person B can't "cheat" by saying we're "partway" through a series of rolls, so the next roll is more likely to be a four. That's implying dependence where there isn't any. Person B can only say "well, we couldn't get a 4 on one roll, so we're going to roll ten times now instead, and were more likely to get a 4 that way."
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Thom0909
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plezercruz wrote:
They're both right. A is right in the real world and B is right in fantasyland.

Pete (sums up)


Or A is talking about individual rolls and B cumulative.

You roll six times, and there's a 99%+ chance you'll get a 4 in there somewhere (someone check my math).

Edit: Alison's post above seems to cover this better.
 
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Brent Gerig
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Prop Joe wrote:

You roll six times, and there's a 99%+ chance you'll get a 4 in there somewhere (someone check my math).


6 times gives you a probability of about 66.5% of rolling a 4.
 
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Thom0909
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flaquito wrote:
Prop Joe wrote:

You roll six times, and there's a 99%+ chance you'll get a 4 in there somewhere (someone check my math).


6 times gives you a probability of about 66.5% of rolling a 4.


Yeah, that's what I get for trying to do it in my head. Intuitively, I should have known six wasn't nearly enough.
 
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Brent Gerig
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Prop Joe wrote:
Yeah, that's what I get for trying to do it in my head. Intuitively, I should have known six wasn't nearly enough.


That's why I let Google do my math for me.
 
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Vanilla Ice
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Trial and error is for suckers (I kid)...

1-(5/6)^n=.99
(5/6)^n=.01
n=log(.01)/log(5/6)
n=25.3 (round up to 26)

Can generalize and find number of rolls (n) to have a probability (p) of getting at least one 4:

n=log(1-p)/log(5/6)
 
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Tobias
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Okay, so if I read all of you right, then both Person's opinions are correct in their own way, yes?

Or one could say that Person A did not know about the fact that there is a formula for "at least one X in Y number of rolls" or did not wanted to think that way (maybe to harden their own standpoint), correct?

In any event, thank you for all the quick and informative posts. So glad to be a member! meeple Happy new years eve!
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Ryan
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They are both correct in their own ways, but they're actually talking about two different probabilities.
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Brent Gerig
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mhuggins123 wrote:

n=25.3 (round up to 26)


Don't tell me what to do. I'll take my .3 dice roll, thank you very much.
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Peter Strait
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Ebyl wrote:
They are both correct in their own ways, but they're actually talking about two different probabilities.


Not exactly, but it comes down to semantics. Person B is incorrect in saying "it will get more likely", as the probability of a 4 on any specific dice doesn't change, but what Person B likely means is correct. Additionally, there's another aspect that acts to prove A's point (not that I want to over complicate things).

Say these two persons decided to buy a tub of a thousand d6s and dump them out (i.e., roll them all once) to settle their differences. The chance of none of the dice showing a 4 is indeed extremely, extremely low, so the odds of not getting at least one 4 are indeed smaller than if they only dumped a bin of 100 of them, or 10, or just 1.

However, the odds of any particular die landing on 4 are unchanged. In fact, and relatedly, it's extremely likely that 16.67% of the dice in the pile are showing a 4, and extremely unlikely that, say, 12% or 20% of the dice landed on 4. The more dice you roll, the less likely you are to see either more or fewer 4s than you would expect (this is called the law of large numbers - if I were better at foruming I would link to the Wiki article, as it's really the one you should read). So, Person B likely means something correct (the odds of seeing a specified result at least once goes up as you increase the number of tests / pool of samples), but has phrased it in a way that's incorrect (getting a 4 doesn't become "more likely", it's a direct consequence of the static likelihood of getting that result).

(Edited to fix a few subtle errors of phrasing of my own. )
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Ryan
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Yes, it is semantics. That is where most people stumble with statistics, and I didn't think it was necessary to go into a lesson in full statistical vocabulary. The meaning of both is clear enough to see they are both correct in what they are thinking, even if the wording is not technically correct - which is my point.
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Kelly Bass
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Tangent Empirical Story:
I was playing a game where I needed to get 9 hits on an enemy character. Each 6 sided die that had anything except a 1 would be a hit. So how many dice to roll to make fairly sure I got 9 hits?
Well, I ended up rolling 29 dice...

I rolled 21 ones! ...and only got 8 hits, failed to slay the creature. Lost the battle & the game. *sigh* shake
 
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Brent Gerig
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chockle wrote:
Tangent Empirical Story:
I was playing a game where I needed to get 9 hits on an enemy character. Each 6 sided die that had anything except a 1 would be a hit. So how many dice to roll to make fairly sure I got 9 hits?
Well, I ended up rolling 29 dice...

I rolled 21 ones! ...and only got 8 hits, failed to slay the creature. Lost the battle & the game. *sigh* shake


Wow! Well, if it makes you feel any better, I think the chances of that happening were (1/6)^21 * (5/6)^8 = 1.06*10^-17 = .000000000000000106. That's significantly worse than your chance of being struck by lightning. Twice. In fact, it's pretty close to your chances of being struck by lightning 4 times in your life. Moral: You should probably avoid storms. And play the lottery or something.

(Edit: based on a probability of 1:10000 chance of getting struck by lightning over 80 years, from a wikipedia article about a poor guy who got struck 7 times in his life.)
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Ryan
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That's not quite correct. To get exactly 8 out of 29 successes, where P(Success) = 5/6, the probability is 4.55E-11. It comes out to be {29 choose 8)*(5/6)^8*(1/6)^21.

If you want the probability of getting 8 or fewer successes (hits) out of 29, you sum the formula from 0 to 8. You can also approximate it with the Central Limit Theorem.

Needless to say, as you pointed out, it's an extremely unlikely set of rolls.
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Brent Gerig
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Ebyl wrote:
That's not quite correct. To get exactly 8 out of 29 successes, where P(Success) = 5/6, the probability is 4.55E-11. It comes out to be {29 choose 8)*(5/6)^8*(1/6)^21.

If you want the probability of getting 8 or fewer successes (hits) out of 29, you sum the formula from 0 to 8. You can also approximate it with the Central Limit Theorem.

Needless to say, as you pointed out, it's an extremely unlikely set of rolls.


Ah, thanks for the correction. Once I get beyond really basic statistics into permutations and combinations, things start to go bad.
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Ryan
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You had it correct for one possible combination of 8 successes out of 29 tries. All the 29 choose 8 does is include all the possible ways to get 8 successes in 29 tries. thumbsup
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