This weeks puzzle

@blood-on-the-tracks said
Well, the ways of picking r things from N is given by N!/(N-r)!r!

Damn.
Right formula but on RHS reduced "N" to (N-1) thinking number of flavours was reduced by one too.

Nice little problem.
Well done BOT

@blood-on-the-tracks said
@venda
No problem. Have missed your weekly 'puzzle', though you did say the quality dwindled!

Yes .
They've mostly been next number in the sequece type things.
I look every week and occasionally there's an interesting one.

This weeks puzzle is an interesting one for a change!!
It concerns a lone Amoeba!!
On each hour there is an 8% probability it will die,with 56%probability it will split into two identical copies living independently and with 36% probability it will remain the same.
What are the chances that any resulting colony will eventually die out if left to it's own devices?

@venda said
This weeks puzzle is an interesting one for a change!!
It concerns a lone Amoeba!!
On each hour there is an 8% probability it will die,with 56%probability it will split into two identical copies living independently and with 36% probability it will remain the same.
What are the chances that any resulting colony will eventually die out if left to it's own devices?

Hmm. Probability is not my strong area.

There is an 8% chance I will figure this one out. 🙂

@venda said
This weeks puzzle is an interesting one for a change!!
It concerns a lone Amoeba!!
On each hour there is an 8% probability it will die,with 56%probability it will split into two identical copies living independently and with 36% probability it will remain the same.
What are the chances that any resulting colony will eventually die out if left to it's own devices?

I expect, that it won't die out statistically:

after one hour we have:

8 cases dead Amoeba

36% one Amoeba

56 % two Amoeba

After two hour we have:

8 + 3 11 cases of no Amoeba

13 (36% of 36) + 9 =22 cases of one amoeba (total amount of two Amobea case times 8% not taking into account that there is a non-Zero probablity for two Amoeba dying in the same time)

21 (36% of65) + 21 (56% of 36)= 42 cases of two Amoba

25 cases of three and more Amoeba.

Since only in cases of one Amoeba is a reasonable Chance of dying out and the other populations will grow I don't Thing that the case of Zero Amoeba will reach 50

@ponderable said
I expect, that it won't die out statistically:

after one hour we have:

8 cases dead Amoeba

36% one Amoeba

56 % two Amoeba

After two hour we have:

8 + 3 11 cases of no Amoeba

13 (36% of 36) + 9 =22 cases of one amoeba (total amount of two Amobea case times 8% not taking into account that there is a non-Zero probablity for two Amoeba dying in the sam ...[text shortened]... ing out and the other populations will grow I don't Thing that the case of Zero Amoeba will reach 50

The answer is quoted as a "probability" i.e 1:x
The explanation is quite involved.
I'll leave it a day or two and then copy the explanation from the paper

@venda said
The answer is quoted as a "probability" i.e 1:x
The explanation is quite involved.
I'll leave it a day or two and then copy the explanation from the paper

This is the given answer:-
1/7 Any new amoeba formed will have the same probability of it's colony dying out so if this probability is "p" then by considering the position at hour one,p=0.08 +0.36p +0.56p squared.From this equation p could be 1/7 or 1 but the answer 1 can be excluded as the expected number on amoebas increases each hour.