Probability maths question

[BlueBloodedBlue]

I'm always amazed how MNetters can answer these, so hoping so,done can help,

What is the probability that 3 randomly chosen letters of the alphabet will be in alphabetical order? (not necessarily consecutive, but alphabetical)

My attempt was:
Total possible combinations 26x26x26 = 17,576

Then alphabetical order letters will be 1x25x24 plus 1x24x23 all the way down to 1x2x1 which adds up to 5,187

Therefore the probability is 29.51%

Does this make sense or have I gone totally wrong?

Thinking this through as I write.
If you pick 3 letters:
If they are all different
There is 3x2x1=6 different orders those letters can come out, and only 1 will have them in alphabetical order, so 1 in 6.

But what is the chance they are all different? It is (1 x 25/26 x 24/26)

So chance you pick all different AND come out in alphabetic order is

1/6 x (25/26 x 24/26) = 25/169 = 14.79%

Will be higher if you allow doubles of the same letter.

The chance of all 3 being the same is
1x1/26x1/26 and they have a probability of 1 of coming out in alphabetical order (ie certain).

The chance of exactly 2 being the same is
1x1/26x25/26
Out of the 6 ways they can be drawn 2 will be correct.

So to my original 25/169 you have to add 1/676 and 25/676
= 63/338 = 18.63%

I'm not great on probability. Someone like @DadDadDad may like to correct me.

No letters can be the same (as then they wouldn't be in alphabetical order).

Your explanation sounds better than mine - just re reading to try and get my head around it.

Thanks for replying.

Ah mistake in second bit, I forgot the 2/6(=1/3) bit)
You have to add 1/676 and 1/3x25/676
So 82/507= 16.17%

Then alphabetical order letters will be 1x25x24 plus 1x24x23 all the way down to 1x2x1 which adds up to 5,187

This bit looks dodgy.
If you are counting you have to count
ABC,ABD,ABE,... Which would be 24
Then ACD,ACE,... which would be 23
To AYZ which would be 1
n/2x(n+1) = 156.
Then you have to count BCD, etc and onwards
So you are adding n/2x(n+1) for values of n from 24 to 2? and I have no idea off the top of my head how to do that in a formula.

0.5 x ( 24x25+23x24........1x2)
So it looks to me like maybe you didn't halve your answer to allow for your 2 letters coming out in either order?

(ignore the n/2x(n+1) = 156. line, I wandered off)

If I knew what I was doing I'd have spotted the missing 0.5 without having to work it through very slowly myself

0.0016% recurring (I think).

Assuming that it doesn't matter which letter you start with, and that YZA and ZAB count.

In which case it doesn't matter which letter is drawn first, but there's only 1 letter out of 25 that counts (so 1/25) then 1 out of 24. Multiply the two together and you get 1/600 or 0.00166666666.....7.

Time
I don't think that can be correct. Are you insisting they are consecutive?

Also I really don't think you can claim YZA are in alphabetical order!

It has to be nearish to 1/6 = 16.66% because any 3 tiles can come out in 6 different ways, only 1 of which will be alphabetical. Then (according to OP's update) you have to throw out doubles and triples which is how my first answer got down to 14.79%.

Skim-reading thread, I think @TeenMinusTests' method makes sense, but I'll work it out separately so we can see if agree:

26^3 ways of randomly drawing three letters with replacement.

26 of those will be three-letter repetition (AAA, BBB, ...)

26 x 25 of those will be two-letter repetition followed by a different letter (AAB, AAC, ... BBA, ...)

26 x 25 of those will be one letter followed by a two-letter repetition (ABB, ACC, ... BAA, ...)

26 x 25 of those will be the first and third letter being the same and the second letter different (ABA, ACA, ...)

That leaves 15600 draws where the letter will be all different. As TeenMinusTests explained, 1/6 of those will be in alphabetical order (three letters can be arranged in 3 x 2 x 1 ways, one of which will be alphabetical).

So 2600 will be alphabetical different letters.

So probability is 2600 / 26^3 = 14.8%

Where's all this smoke coming from?

Looks like 14.8% has some legs here...

Don't you just love how we get there using different methods?
Any method that gets the right answer is fine.
and stuff the NC which insists for certain methods for KS2 SATs

Actually, I don't know why I've made it so complicated. To draw three distinct letters there are 26 x 25 x 24 = 15600, so I can get to that point in my working straightaway without considering all the repetition options.

In general, for an alphabet of n letters with m drawn at random, the probability of them being m distinct letters in alphabetical order will be:

[n! / (n-m)! m!] / n^m

(so expression inside [ ] is just nCm).

You've lost me now. nCm?

The question works equally well with numbers ...

en.wikipedia.org/wiki/Binomial_coefficient#History_and_notation

Okayyyyyy - just picking my brain off the wall as my head has just exploded

I can follow the explanation at 16:24 but then you lose me

Thank you all, particularly @DadDadDad.