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raddevus wrote: What is the probability of rolling a 20 three times in a row on a 20-sided die?
Zero on a sensible die: the faces will be numbered 0 to 19 inclusive.
"I have no idea what I did, but I'm taking full credit for it." - ThisOldTony
"Common sense is so rare these days, it should be classified as a super power" - Random T-shirt
AntiTwitter: @DalekDave is now a follower!
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OriginalGriff wrote: Zero on a sensible die: the faces will be numbered 0 to 19 inclusive.
Your Devish is showing. Better hurry, cover it up. 
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Take this[^] and roll a save against reality. And no, it's not from the popular VB roleplaying game where normal users pretend to be smart code monkeys. 
I have lived with several Zen masters - all of them were cats.
His last invention was an evil Lasagna. It didn't kill anyone, and it actually tasted pretty good.
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Nope, even Google translate doesn't like it.
Quote: Our tip: If a 20-piece cube and a school cube are used together, the big 1 x 1 can be practiced excellently.
Are they saying that a cube can have 20 sides (in a normal 3D Space that's difficult to manage) but combining it with a school gives you a huge one sided die ... I've never noticed extra spatial dimensions in Germany, but it would explain a lot of things ... 
"I have no idea what I did, but I'm taking full credit for it." - ThisOldTony
"Common sense is so rare these days, it should be classified as a super power" - Random T-shirt
AntiTwitter: @DalekDave is now a follower!
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My dice are all blank - I remember where the dots are.
Freedom is the freedom to say that two plus two make four. If that is granted, all else follows.
-- 6079 Smith W.
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Not quite. Assuming that you take each triad of throws as a unit (i.e. you throw 3 20-sides dice at one time):
The probability of NOT throwing a triple 20 in ONE throw P = 7999/8000
The probability of NOT throwing a triple 20 in TWO throws P2 = (7999/8000)2
...
The probability of NOT throwing a triple 20 in N throws PN = (7999/8000)N
We want to find N for which PN == 0.5, or 0.5 = (7999/8000)N.
Taking the log of both sides, we have log(0.5) = N * log(7999/8000)
N = log(0.5) / log(7999/8000) = 5544.83
After 5545 rolls (16635 individual rolls), your chances are 50% of having thrown three 20s in a single roll.
Freedom is the freedom to say that two plus two make four. If that is granted, all else follows.
-- 6079 Smith W.
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Thanks for taking time to comment...but I'm really math-slow.
I'm going to stare at this some more and see if my brain can absorb it further. 
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Only because the probability says you should have done it in X times... it doesn't mean you will.  
M.D.V. 
If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about?
Help me to understand what I'm saying, and I'll explain it better to you
Rating helpful answers is nice, but saying thanks can be even nicer.
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Nelek wrote: Only because the probability says you should have done it in X times... it doesn't mean you will.
Right! That is an interesting part of this. It's funny because you could go on rolling dice for eternity but never have it happen.
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That depends, did you blow on the die before rolling?
Everyone knows that greatly boosts your chances of success
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Sander Rossel wrote: That depends, did you blow on the die before rolling?
Everyone knows that greatly boosts your chances of success
I knew I forgot some of my maths.
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In your latter scenario, you are excluding that if you try more than one than if your last throw (for example, of the first three rolls) is a 20 you are already beginning the next string.
I don't feel like the math, but there are 400 way you can have a 20 for your last roll, 20 of these are for 20 in the last two rolls (subtract 1 if you have already succeeded).
So, whilst it 1:8000 for the first three rolls you may succeed if your fourth roll is a twenty once in 400 times and and if the last two were twenty, then your 4th roll has a 1:20 chance of you winning. These added opportunities will happen roughly one throw in 20 if you chain.
Nothing in your description precludes appending the rolls - any 3 in a row fulfills its requirement.
| Ravings en masse^ |
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| "The difference between genius and stupidity is that genius has its limits." - Albert Einstein | | "If you are searching for perfection in others, then you seek disappointment. If you seek perfection in yourself, then you will find failure." - Balboos HaGadol Mar 2010 |
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I interpreted it in the same way as Daniel (above) and upvoted his answer.
The average number of rolls before any 3 in a row are 20 is tougher.
And that's the second whilst from you in a short while! You're hanging around too many Brits!
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Using a "moving window", you have the following cases:
(Last throw was NOT 20): occurs 19/20 times, chance of 1/8000 of throwing three 20s in the next three moves
(Last throw was a 20, next to last throw was something else): Occurs 19/400 times, chance of 1/400 of throwing two 20s in the next two moves
(last two throws were a 20, previous throw was something else): occurs 19/8000 times, chance of 1/20 of throwing a 20 in the next move
(last three throws were a 20): occurs 1/8000 times.
Total incidence: 19/20 + 19/400 + 19/8000 + 1/8000 = (7600 + 380 + 19 + 1) / 8000
Any of the 4 cases may occur, so P is additive.
P = 19/20 * 1/8000 + 19/400 * 1/400 + 19/8000 * 1/20 + 1/8000 * 1 = 19/160000 + 19/160000 + 19/160000 + 20/160000 = 77/160000 ~ 1/2000- (slightly under 1/2000).
The difference from the triples result is because each 20 has a chance to participate multiple times. In this case, N(50%) would be 1440.
Freedom is the freedom to say that two plus two make four. If that is granted, all else follows.
-- 6079 Smith W.
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W∴ Balboos, GHB wrote: Nothing in your description precludes appending the rolls - any 3 in a row fulfills its requirement.
That is a good point. Let's say there is a rule that says the game is a set of three rolls independent from all others. 
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I got into the special math class exactly over such debate (with normal dice). I stated that there is only two outcoms - you either throw a triple or not. So it is 50-50...
Three different teachers spen four years tontey and explain to me why I was wrong...
I was rigth...
"The only place where Success comes before Work is in the dictionary." Vidal Sassoon, 1928 - 2012
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Kornfeld Eliyahu Peter wrote: you either throw a triple or not. So it is 50-50
Yeah, that's true...if you only try once.  So make sure you only try this once and you greatly increase your odds.
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I rolled 3 double sixes in a row, sitting on the bar, with the 1-5 points open, at the time leading, in a Las Vegas Backgammon tournament final.
"What are the odds!!!" 
It was only in wine that he laid down no limit for himself, but he did not allow himself to be confused by it.
― Confucian Analects: Rules of Confucius about his food
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Gerry Schmitz wrote: What are the odds!!!"
Well, I could calculate that...
Also, are you serious? I think you are. I actually play backgammon so I can see the pain in what happened.
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Fun with statistics!*
If you are playing "Let's make a Deal" and you select 1 door from the choice of 3 doors. The host then reveals the crap prize behind one of the other doors and offers you a chance to switch.
Should you switch?
(* - there is only about an 1 in 2000 chance** that you will actually have fun with statistics!)
(** 87.4% of statistics are made up on the spot!)
If you can't laugh at yourself - ask me and I will do it for you.
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Absolutely, more information changes things. This is a classic Bayesian problem. Thanks for reminding me of it. Posterior probabilities. 
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Fantastic article. Thanks!
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Thanks a lot for your vote, and kind remarks.
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Just a thought...
Suppose you start rolling as described. If the first or second roll isn't a 20, do you finish through to the third roll?
What if instead of one 20-sided dice, you have 3 of these dice and roll all three at once? Would that change the odds?
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