
Hopefully. It adds 9 pills to her morning pill loadout, and I've got to watch for side effects (depression / euphoria, Diabetes, Osteoporosis , ...) so I've ordered a blood glucose tester and we'll both stab ourselves in the name of science (so we have baseline to tell if she is increasing or decreasing given we both eat the same food at the same time, pretty much).
This thing just doesn't stop ...
"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 Tshirt
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OriginalGriff wrote: Hopefully.
OriginalGriff wrote: we'll both stab ourselves
That's clever.





Good luck and best wishes.
But I do have to ask  why all the precautions for someone who has already had it?
If the answer is you can get it again  then the question becomes  How will a vaccine work?  since it is based on the bodies reaction to curing itself from the disease in the first place?
( I will duck and cover now... )
If you can't laugh at yourself  ask me and I will do it for you.





Having had it should mean that you can't get it again  not that you can't come in contact with it and become infectious for some period (just like you have a infectiousbutnosymptoms period).
And this isn't something I want to start spreading to anyone, even antivaxxers don't deserve that ...
"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 Tshirt
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I disagree. Regardless, how many of them do you think will take the vaccine?
"They have a consciousness, they have a life, they have a soul! Damn you! Let the rabbits wear glasses! Save our brothers! Can I get an amen?"





That was my spine reaction as well. But you can't blame anyone for being cautious, and after thinking about it I came to the conclusion that I would probably do the same.
Reason being, while most people get long lasting immunity after an infection, this is not the case for everyone. So after having had it badly, would you really want to take a chance on that?
DRHuff wrote: If the answer is you can get it again  then the question becomes  How will a vaccine work?
Most probably better actually, that's why they give you two shots 21 days apart, to make sure your immune system remembers the virus.
The only reasonable safety is achieved when enough people are vaccinated and the virus goes endemic.
Wrong is evil and must be defeated.  Jeff Ello
Never stop dreaming  Freddie Kruger





wishing you and your wife the best, and to better health soon.





OriginalGriff wrote: It's not cancer, it's not a fungal infection
Well, there's a little bit of good news along with the bad. I hope your wife makes a full recovery from this.
Freedom is the freedom to say that two plus two make four. If that is granted, all else follows.
 6079 Smith W.





«One day it will have to be officially admitted that what we have christened reality is an even greater illusion than the world of dreams.» Salvador Dali





Bon rétablissement, Michelle ! Take care !





I'm reading the book, Bayesian Statistics the Fun Way, (No Starch Press)[^].
It's perfect for my level of mathiness. And it actually is a fun book so far. Good book for thinking different.
Anyways, at the end of chapter 3 there is a quiz with a question:
quiz What is the probability of rolling a 20 three times in a row on a 20sided die?
As you probably already know this is
P(X) and P(X) and P(X)  each die roll is a mutually exclusive events
1/20 * 1/20 * 1/20 = (.05 * .05 * .05) = 0.000125
Flip it upside down 1  0.000125 = 0.999875 (99.9875% chance you won't succeed).
Turn that into a fraction and you get: 1/8000
Here's the Q
Does that mean, on average you would have to try 3 rolls of a 20sided die 8,000 times (24,000 die rolls) before you would achieve this feat?
With odds like that, I can probably beat The House in Vegas.





raddevus wrote: What is the probability of rolling a 20 three times in a row on a 20sided 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 Tshirt
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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.





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.





Nope, even Google translate doesn't like it.
Quote: Our tip: If a 20piece 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 Tshirt
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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.





Not quite. Assuming that you take each triad of throws as a unit (i.e. you throw 3 20sides 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 P_{2} = (7999/8000)^{2}
...
The probability of NOT throwing a triple 20 in N throws P_{N} = (7999/8000)^{N}
We want to find N for which P_{N} == 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.





Thanks for taking time to comment...but I'm really mathslow.
I'm going to stare at this some more and see if my brain can absorb it further.





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.





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.





That depends, did you blow on the die before rolling?
Everyone knows that greatly boosts your chances of success





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.





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^ 

"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 





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!





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.



