# Personal Blogs

## Floral probabilities

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I’ve just bought 12 baby polyanthus plants via Amazon. When they flower they will be an assortment of 6 colours and I’m assuming there will be 2 plants of each colour. But at the moment they all look pretty much the same.

I’m planning to give 4 plants to a friend. Given we will be picking the 4 at random…

What is the probability that my friend will receive 4 plants all different colours?

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## Comments

### New comment

Hi Richard,

After the first colour has been picked the next colour has a 10 to I (on?) chance in favour of being different. The third colour has an 8 to 2 (on?) chance of being different and the fourth colour has a 6 to 3 (on?) chance of being different. All these figures are predicated on each  succeeding colour following the first choice being by turn different to the preceding choice. I use the word ‘on’ to say that it is much more likely that a colour will be different to the previous one.

I do not know what the overall probability is of four different colours being selected but I’d be very interested to know if my assessment of the individual odds for each choice is correct and to what degree, if any, this is helpful in assessing the overall probability of four plants of differing colours being chosen.

It is a very stimulating question and my reply is simply my attempt to try to clarify to what degree I can understand the odds for each choice.

I’ll be very interested to see the answer to the original question.

Joseph.

### New comment

I worked it out like this

We can choose 4 colours from the 6 in 6 choose 4 ways, which is (6x5x4x3) / (4x3x2x1) = 15 ways.

But there are 2 of each colour, so the number of selections that consist of 4 distinct colours is 15x2x2x2x2 = 240.

The total number of ways to choose 4 plants, of whatever colour combination, from the 12 is 12 choose 4, which is (12x11x10x9) / (4x3x2x1) = 495 ways.

So prob(4 distinct colours) = 240/495 = 16/33 ~ 48.5 %

I wrote a Python program to simulate the problem and after 1 million trials I got 48.54 % so this seem to confirm my working out.

### New comment

Very interesting solution and methodology. Would never have had the know-how to solve it but very intriguing nevertheless.