For the game mastermind, I need to work out the probabilities of certain combinations of a randomly generated secret code.
Using 5 colours of pegs and 4 peg-holes, the maximum number of combinations that there can be are 625.
What is the probability that of these 625 the secret code contains:
1. 4 duplicate colours
2. 3 duplicate colours, 1 unique
3. 2 duplicates, 2 uniques
4. 2 duplicate colours, 2 other duplicates
5. 4 uniques
Gratitude will be enormous.
Many thanks
>>8693817
this is trivial counting problam, you can probably robotically apply formulas you have in your book
im a biology person and can solve it, you should probably end it
>>8693836
i cba to remember if i need to use binomial distribution to solve this or not
can u pls explain
>>8693840
just count it nigga
1) pick any one color, theres 5, next one must be the same, next one must be the same, next one muxt be he same, thus 5/625 are 4 duplicates
2) first one can be anything, second one can be anything, third one must be same as firs tor second, last one must be same as third, (5*5*2*1)/625
and so on
>>8694077
This doesn't take the other colors into account. The problem asks for 4 duplicate colors, e.g. blue-blue-blue-blue, but that isn't the only possible duplicate color.
You'll need binomial distribution for this.
>>8693817
>1
5 colours, 1 four-combo for each. Order doesn't affect them.
Answer = 5*1 = 5
>2
5 colours, 4 three-combos (in different places remember), then 4 colours could be the one left.
Answer = 5*4*4
I remember I figured something else similar for non-ordered combinations (e.g. dice throws), but I never bothered figuring out ordered.
Non-ordered was
>No. possible combinations = (no. possible 'states')^(no. possible 'digits')
With two D6s, that means (6 sides)^(2 dice) = 36 combos,
three D6 = 216 combos, etc.
There's somewhere to start off, I can't be asked to factor in the ordered part. Have fun
>>8693817
I'm a real mathematician so I don't use numbers other than 0 and 1. I think these kind of elementary children problems are so boring. You already got your answer! You said there's about 1000 combinations, so answer to every option you have is "<=1000". Solved.
No go solve some real problems.
At a glance seems to be:
5, 80, 720, 60, 120
tfw played this game yesterday with my 7 yo niece
>>8696319
I have the game, but no one ever played it with me.
Always excuses about
>I dont want to think when playing a game
>its too difficult
>Im too dumb to play this with you anon
Same with cluedo, i loved the game but got the same responses as above when recommending it, then in the end everyone just watches TV, I decide fuck it and go play PC.
>>8693836
What this guy said
In this problem, the probability for an event is :
(number of combinations that fulfill the event) / (total number of combinations)
Just count