Covid-20 math puzzle
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5 members have voted
Once a microbe enters your lungs, what is the expected number of microbes will you have after seven days?
Quote: WizardDoes anyone not understand what's being asked?
The only thing I don't understand is, what is meant by "an average rate of one per day." Average in terms of what?
Even if you say, "If there are an infinite number of them created simultaneously, the mean time before they reproduce is 24 hours," what is the curve of time to probability of reproducing (or, alternatively, not reproducing) by that time?
Phrases that cause some ambiguity for me: "average rate", "same rate", "memory-less property"
I should have taken more mathematics classes in college.
Yes!Quote: WizardDoes anyone not understand what's being asked?
If a microbe always creates another one 24 hours after it was born, then clearly the number of microbes just multiplies by 2 every day. After 7 days there are 2^7=128 (assuming we're talking Monday 0h01 to Sunday 25h59).
However suppose a microbe either creates another one 12 hours or 36 hours, then the ones born earlier have a chance of multiplying earlier, which has a much bigger effect on the growth rate than the ones born later (average about 1154). As a silly idea suppose a microbe can either replicate after 1 minute (P=1/9) or about 8 days (Pr=8/9), then the average is 1 day but on average a new one will be born every 9 minutes.
btw I don't know how to work it out (I'm too lazy to do integrals etc.) if, as I suspect, it's an exponential distribution.
Me.Quote: WizardDoes anyone not understand what's being asked?
I’m sure the answer isn’t 2^6=64.....
I just changed the wording as follows:
Quote:A microbe, let’s call it Covid-20 can spawn a new microbe at any time. The probability of a particular microbe spawning at any given time is always the same, regardless of the time since the last spawning. The average time between spawnings from the same microbe is one day.
Once a microbe enters your lungs, what is the expected number of microbes will you have after seven days?
I'm trying to ask this without using mathematical jargon.
The time between spawnings is similar to the time between royal flushes in video poker if you averaged getting one per day. On any given day you might get zero or you could get 10, or anything. The difference is that with royal flushes there are only so many opportunities to get them in a day, about 40,000, but with the microbe there is no limit and they can happen anytime, even a nano-second after the last spawn.
dN/dt = + N/tm
where tm is the mean time to double/spawn
Integrating between any arbitrary zero time, No and a later time t gives us:
N = No eλt
where λ = 1/tm
So, starting with 1 microbe, after 7 days there should be e7 = 1096.6 microbes approximately.
Quote: gordonm888If N is the number of microbes at any time t, the instantaneous growth rate is given by:
dN/dt = + N/tm
where tm is the mean time to double/spawn
Integrating between any arbitrary zero time, No and a later time t gives us:
N = No eλt
where λ = 1/tm
So, starting with 1 microbe, after 7 days there should be e7 = 1096.6 microbes approximately.
I agree!
Here is my solution (PDF)
Gordon, I would be interested in your thoughts on how to word the problem in plain simple English. I am having a hard time getting across the point how the microbes grow at an exponential rate.
Quote: WizardQuote: gordonm888If N is the number of microbes at any time t, the instantaneous growth rate is given by:
dN/dt = + N/tm
where tm is the mean time to double/spawn
Integrating between any arbitrary zero time, No and a later time t gives us:
N = No eλt
where λ = 1/tm
So, starting with 1 microbe, after 7 days there should be e7 = 1096.6 microbes approximately.
I agree!
Here is my solution (PDF)
Gordon, I would be interested in your thoughts on how to word the problem in plain simple English. I am having a hard time getting across the point how the microbes grow at an exponential rate.
I thought you worded it almost perfectly. I immediately recognized from your write-up that this problem was identical to radioactive decay of an unstable atom, except that it involves growth -hence exponential growth.
Quote: gordonm888I thought you worded it almost perfectly. I immediately recognized from your write-up that this problem was identical to radioactive decay of an unstable atom, except that it involves growth -hence exponential growth.
Thank you! At least one person understood what was being asked. I have changed the wording several times so am not sure which version you approve of. Here is how I have it now:
Quote:A microbe, let’s call it Covid-20 can spawn a new microbe at any time. The probability of a particular microbe spawning at any given time is always the same, regardless of the time since the last spawning. The average time between spawnings from the same microbe is one day.
Once a microbe enters your lungs, what is the expected number of microbes will you have after seven days?
Quote: WizardQuote: gordonm888If N is the number of microbes at any time t, the instantaneous growth rate is given by:
dN/dt = + N/tm
where tm is the mean time to double/spawn
Integrating between any arbitrary zero time, No and a later time t gives us:
N = No eλt
where λ = 1/tm
So, starting with 1 microbe, after 7 days there should be e7 = 1096.6 microbes approximately.
I agree!
Here is my solution (PDF)
Gordon, I would be interested in your thoughts on how to word the problem in plain simple English. I am having a hard time getting across the point how the microbes grow at an exponential rate.
I agree with your solution.
The question clearly stated that a microbe spawns at rate rate of ONE per day, meaning it will be DOUBLE IN A DAY, and it should be 1, 2, 4, 8....... The question now is how to explain it to general people(in layman's term) that it should be 1, 2.718, 2.718^2, 2.718^3....and NOT 1, 2, 4, 8....... I find it is very HARD to explain it.
General people will argue that DOUBLE means 2 and NOT 2.718 ! LOL
Many people would understand in terms of compound interest. One dollar that earns 100% over 10 years without compounding becomes $2.00. Compounded once per year it becomes $1.00 * (1 + 1/10)^10 = $2.59. Compounded continuously it becomes $1.00 * (1 + 1/inf)^inf = e = $2.72Quote: ssho88The question clearly stated that a microbe spawns at rate rate of ONE per day, meaning it will be DOUBLE IN A DAY, and it should be 1, 2, 4, 8....... The question now is how to explain it to general people(in layman's term) that it should be 1, 2.718, 2.718^2, 2.718^3....and NOT 1, 2, 4, 8....... I find it is very HARD to explain it.
General people will argue that DOUBLE means 2 and NOT 2.718 ! LOL
Quote: Ace2Many people would understand in terms of compound interest. One dollar that earns 100% over 10 years without compounding becomes $2.00. Compounded once per year it becomes $1.00 * (1 + 1/10)^10 = $2.59. Compounded continuously it becomes $1.00 * (1 + 1/inf)^inf = e = $2.72
Not agree. 1, 2, 4, 8 . . .and 1, 2.718, 2.718^2, 2.718^3......BOTH are compound...
Quote: ssho88General people will argue that DOUBLE means 2 and NOT 2.718 ! LOL
"General people"?
I understand compound interest. I don't understand it in the context of the original question.
And, yes, I should have taken more mathematics in college.
Yes, both are compound. 1, 2, 4, 8 is 100% growth compounded ONCE per period. 1, e, e^2, e^3 is 100% growth compounded INFINITELY per period...call it a trillion times per period if that’s easier to grasp.Quote: ssho88Not agree. 1, 2, 4, 8 . . .and 1, 2.718, 2.718^2, 2.718^3......BOTH are compound...
Continuous growth is the same as continuously compounded interest
Quote: Ace2Yes, both are compound. 1, 2, 4, 8 is 100% growth per period compounded ONCE per period. 1, e, e^2, e^3 is 100% growth per period compounded INFINITELY per period...call it a trillion times per period if that’s easier to grasp.
Continuous growth is the same as continuously compounded interest
Agree
I find it interesting that the most important constant in math was discovered quite recently in human history.
Quote: ssho88The question clearly stated that a microbe spawns at rate rate of ONE per day, meaning it will be DOUBLE IN A DAY,
It said the average time to double was a day.
Quote: WizardIt said the average time to double was a day.
Meaning it can be doubled in nano sec, in 3 seconds, in 1.2 days......I am not good at English, I may not fully understand your question .
Quote: ssho88Meaning it can be doubled in nano sec, in 3 seconds, in 1.2 days......I am not good at English, I may not fully understand your question .
Yep.
Would it be fair to say that failing to factor in the "average per day" is what tripped up this general person? Using e in an expected-value problem allows one to factor in the possibility that sometimes multiple will be spawned in a day and sometimes zero will be spawned in a day; am I close?
Quote: IndyJeffreyWould it be fair to say that failing to factor in the "average per day" is what tripped up this general person?
That seems to have been the stumbling block, but I don't see why so many people interpreted an average of one per day to be exactly one per day.
Quote:Using e in an expected-value problem allows one to factor in the possibility that sometimes multiple will be spawned in a day and sometimes zero will be spawned in a day; am I close?
I'm not sure I see your point. e pops up because to get the answer you have to integrate 1/x.
