Math Problem!

[Perplexed]

You could just buy 6 plants ...

What are the odds of getting berries from a group of six plants?

 

[trekrok]

What are the odds of getting berries from a group of six plants?

A little over 99%, right? That's the smart play.

 

[Perplexed]

A little over 99%, right? That's the smart play.

I figured out a formula for this particular situation. There are two possible outcomes - male and female - and N is the number of plants in the group. The total number of possible permutations then is 2^N; you then subtract the two extreme outcomes (all male or all female) from the total, and divide the remainder by the total, and multiply by 100 to get the probability:

[(Total-2)/Total] * 100

So in the case of six plants, the formula then becomes:

2^6 = 64

[(64-2)/64] * 100 = 96.875

Thus, the probability of getting berries from a group of six plants is close to 97%.

By the way - you will never achieve 100% probability of getting berries, no matter how big the group of plants becomes. You can add so many plants to the group that the odds of getting all males or all females becomes vanishingly small, but you’ll never get rid of that possible outcome

 

[HillsideDesolate]

Using chicken math I would say at least a dozen.

Wouldn't be the first time I've lost the straight run lottery.

 

[HillsideDesolate]

I’m just curious about what kind of berries you’re trying to produce ???

Obviously Sea Berry,

 

[turkeyrun]

Using chicken math I would say at least a dozen.

Wouldn't be the first time I've lost the straight run lottery.

Chicks are easily sexed, 100% can be achieved.

Mixed will always be mostly roosters.

With bushes and getting, 3 bushes has 8 possibilities, with 6 having opportunities for berries.6/8 = 75% POSSIBILITY of berries.

With chicks, assume 50 - 50 mix on 100 chicks. An order for 100 chicks, 25 straight run and 75 mix pulled. Results in 25 and 50 - 25 mix. So, 2 in 3 are rooster. Order comes in for 35 and 65, nets 50 - 15.
I bought 48 mixed, at TSC and got 4 hens, with 42 roosters. 2 didn't survive.
Luckily, I was wanting meat birds, not layers. IF you want hens, pay the extra for straight run, unless you can get a helpful employee that will sex them as you go.

Son ordered 100 fertilized eggs. Hatched 54 hens and 45 roosters.

Dingleberries is a whole nuther subject, I WILL NOT get into.

Time for coffee, my head is starting to hurt.

 

[Two Gun Warrior]

After some heated debate, my gf and I realized we were arriving at the same conclusion, only in radically different ways. I have to admit, I still don’t fully understand her thought process!

And yes, we figured out the 75% chance of getting berries out of a group of three plants. I used the same method as @Okie4570 with the eight permutations, she used a 25% method of some kind and a process of elimination which somehow seemed faster.

A wise man admits he does not understand a woman's thought process.

 

[Bixby Sooner53]

Too many are way overthinking this. All you need to do is ask the plants what they identify as and just buy two. We have all been taught that you are what you identify as. It has to be the same in the plant world too, right?

 

[Chuckie]

So many trying to make a mountain out of a molehill. The probability of getting berries is not that complex of a problem and is 50% . . . 'ya either will or 'ya won't - that's 50%

 

[Perplexed]

With bushes and getting, 6 bushes has 8 possibilities, with 6 having opportunities for berries.6/8 = 75% POSSIBILITY of berries.

How do you figure 8 possibilities with 6 plants? The permutations would start with:

MMMMMM
FFFFFF
MFFFFF
FMFFFF
FFMFFF
.
.
.
MMFFFF
FMMFFF

And so on. There should be 64 possible permutations (2^6). Three plants would result in a 75% probability of getting berries from the group, six plants would result in a 96.875% probability.