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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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Those are actually great questions.
These are my answers and I definitely could be wrong.
Andreas Mertens wrote: Suppose you start rolling as described. If the first or second roll isn't a 20, do you finish through to the third roll?
1) There is no need to roll the die two more times because each die roll is independent of each other (mutually exlusive outcomes). However, you would simply increment your count of tries and failures that have occurred.
Andreas Mertens wrote: 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?
2)Effectively no. Because each 20-sided die roll outcome is mutually exclusive to all others whether you roll them at the same time or individually. However, maybe there are some physics involved that affects the way the dice bounce against each other?? However, that isn't really counted in probability math. Instead it is the pure math of just the theoretical values (no physics involved).
Mutually exclusive meaning that if you roll a 20-sided die twice the second roll is in no way affected by the first roll. A non-mutually exclusive event might be like selecting one of three doors for a prize. Once you select a door it is then removed from the choices. So the next choice is only 1 out of 2.
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Won't you need an initial estimate for Bayes, and then improve that estimate on each result?
You can already guess the 1:20 for the initial roll you don't need to do.. then look to the next roll to get two in a row, (19 chances of failure), then a final 50:50 hope for the last '20' with a 1:10 chance.
Needs a few more unknowns that need determining by measurement and Bayes update. Maybe the dice is weighted? If it was, how much would you pay per throw to give confidence of the weighting?
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For my brother the accountant this Christmas, I got a big bag of receipts. I told him it was OK if he didn't like them, I'd kept all the presents…
"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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Did those from sales slip into that bag? Sometimes your good intentions don't quite register 'till later on.
| Ravings en masse^ |
|---|
| "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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This isn't even really enough for a tip but if you ever can't implement a real hashtable for reasons there's a pretty decent fake you can do for strings:
Store sz strings prepended by a size_t length (which includes trailing zero char) contiguously [len][string][len][string][len][string], like that
when you go to check for a match, start at the beginning.
get the size_t len from the current position
see if it matches your passed in string's length (don't forget to account for the trailing zero). if it doesn't, increment the current table pointer by len and continue
if it does, then in a loop check each character to see if they match. once you find one that doesn't match, you can increment the current table pointer by the remaining characters (computed from len and your current position)
this way you early out on non matches and you've got a remedial hash on top of that (the count)
all of that and you don't require an actual hashtable, and it's not bad in terms of space/time from my tests (currently on JSON fields in a document)
Real programmers use butterflies
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Hmmm... on a big-endian system -- if you limit each string's length to ensure that the high byte of the next string's length is always zero, you can avoid adding "trailing nulls" -- just terminate the array with a zero. Profit.
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