i need to work out some probability
dont want to just post the problem here, as i want to understand the 'how' behind it

would appreciate a pm or conversation by aim

should be reasonably straightforward



im looking for the probability of drawing the value of X from a standard pack of cards with the cards holding their respective values
the draw being done with 3 cards, but also want to understand how to change that number too

thanks

This post was edited by trollen on Jun 2 2015 12:23pm

Post your problem, you'll probably get an answer with a 'how' part.

In this problem can the card values add up to x as well or is it just x that will be drawn from 1 card?

Assume your deck of cards consist in 4 cards of value 1, 4 cards of value 2, ... 4 cards of value 10. That's an unusual 40 cards deck, but anyhow I don't know what value you'd give to Jacks, Queens or Kings.

Assume you pick up 3 cards from this deck, without replacing anyone of them before you pick the next (otherwise tell me).

Possible values for X, as the sum of the values of your 3 cards, are integers from 3 (picking 3 Aces) to 30 (picking 3 Tens).

Possible draws for each value of X :
3 = 1 + 1 + 1
4 = 2 + 1 + 1
5 = 3 + 1 + 1 = 2 + 2 + 1
6 = 4 + 1 + 1 = 3 + 2 + 1
7 = 5 + 1 + 1 = 4 + 2 + 1 = 3 + 3 + 1
8 = 6 + 1 + 1 = 5 + 2 + 1 = 4 + 3 + 1 = 4 + 2 + 2

and so on, until :
28 = 10 + 10 + 8 = 10 + 9 + 9
29 = 10 + 10 + 9
30 = 10 + 10 + 10

That's a lot of work to have all possible decompositions, but you'll find repeated patterns.

Now you want to know the probability of getting value X. You must work on each decomposition... but don't worry, results come quick at this point :

if 3 different values are involved in the decomposition (as in : 6 = 3 + 2 + 1) :
4 chances among 40 to pick up a Three at first,
4 chances among 39 to pick up a Two as second card,
4 chances among 38 to pick up an Ace as last card.
Multiply those chances, and multiply by 6, because the order doesn't count ( 3+2+1 = 3+1+2 = 2+3+1 = 2+1+3 = 1+3+2 = 1+2+3 ) :

A = (4/40) * (4/39) * (4/38) * 6 ~ 0.0064777

if only 2 different values are involved (as in : 4 = 2 + 1 + 1) :
4 chances among 40 to pick up a Two at first,
4 chances among 39 to pick up an Ace as second card,
3 chances among 38 to pick up a second Ace as last card.
Multiply by 3, because 2+1+1 = 1+2+1 = 1+1+2

B = (4/40) * (4/39) * (3/38) * 3 ~ 0.0024292

if only 1 value is use, as in : 3 = 1 + 1 + 1 :
4 chances among 40 to pick up an Ace at first,
3 chances among 39 to pick up a second Ace then,
2 chances among 38 to pick a third Ace last :

C = (4/40) * (3/39) * (2/38) ~ 0.0004049

Put that together to get the probability of getting a final value X.

For example, if X = 8 :
X=8 has 2 decompositions with 3 different values (8=5+2+1=4+3+1), and 2 decompositions with 2 different values (8=6+1+1=4+2+2) :

p(X=8) = 2*A + 2*B ~ 0.017814
p(X=8) ~ 1.78 %

Feel free to adapt with a different deck of cards, or if you pick a different number of cards from it, I hope you get the point.

In this context, what are the odds for the opponent and him ( The person who is hosting )

I did this manually by hand(could be slightly inaccurate), I have the probability of winning this game as 2433/5525 = .44036 or ~44% which gives an approximate 'house edge' of 12%.

Alright,thanks alot !