How to resolve that easy game ?
Join Date: Sep 2011
Posts: 30
Hi,
I got a question that is not poker related, but for another card game (Hearthtones for the curious).
We got 30 cards in our deck, it's a 2 players game.
When we start first we draw 3 cards, when we play second we draw 4 cards (3+1).
It's not standard poker card like As, King etc. it's specific text cards and we have a lot of different cards in a pool for build our deck (It's a game where we build our 30 cards deck and then play with it).
I would like to know how many of the same card i should play for draw it in the first 3 cards or 4 cards.
We also get the choice, when we draw our 3 or 4 cards , to selection any number of these cards (if we don't like them)and put them back in the deck at random and re draw the same number of cards that we put back in the deck.
I would redraw all the cards in my hands who is not the one i want.
How would you calculate this?
Thanks for your insight
Join Date: May 2005
Posts: 4,240
The rows below show the probability of getting at least 1 of a certain card given the number in your deck. The first number is how many of the cards you have in your deck (1 to 5), the second (10% on the first line) is the probability of drawing at least one of them when you go first and draw three cards, the third (13% on the first line) is the probability when you draw 4.
1 10% 13%
2 19% 25%
3 28% 36%
4 36% 45%
5 43% 54%
With a redraw option, the probabilities increase to:
1 19% 25%
2 35% 44%
3 48% 59%
4 59% 70%
5 68% 79%
Join Date: Sep 2011
Posts: 30
Thanks a lot !
This is very helpfull for me.
Can you write down the equation ?
I would like to be able to do it by myself :
Join Date: May 2005
Posts: 4,240
If you have k copies of a card in a deck of N cards, and you draw d cards then:
There are C(N,d) possible hands.
There are C(N - k, d) possible hands that do not contain any of the k special cards.
Therefore, the chance is 1 - C(N - k,d)/C(N,d) that you get at least one of the special cards.
If you can discard and draw again it just squares the last term since to fail to get one of the cards you have to fail on the initial draw, then discard your entire hand and replace it in the deck, and fail again. So that's 1 - [C(N - k,d)/C(N,d)]^2.
C is the "choose" function, COMBIN in Excel. C(N,d) is N!/[d!(N-d)!].
! is the factorial function, N! = N*(N-1)*(N-2)*. . .*2*1.
Join Date: Sep 2011
Posts: 30
I got some difficulties to di it myself, and would be sure about my result.
Really thanks you Aaron to take your time for this, i really enjoyed it.
I would just need, theses columns to be extended to 80/90 %
example :
The rows below show the probability of getting at least 1 of a certain card given the number in your deck. The first number is how many of the cards you have in your deck (1 to 5), the second (10% on the first line) is the probability of drawing at least one of them when you go first and draw three cards, the third (13% on the first line) is the probability when you draw 4.
1 10% 13%
2 19% 25%
3 28% 36%
4 36% 45%
5 43% 54%
With a redraw option, the probabilities increase to:
1 19% 25%
2 35% 44%
3 48% 59%
4 59% 70%
5 68% 79%
I needed to see the result with more cards played for make it simple.
Thanks guys
Join Date: May 2005
Posts: 4,240
Okay, here is the first table. The first column is the number of cards out of 30, the second is the probability of drawing at least one in three cards, the last number is the probability of drawing at least one in four cards.
1 10% 13%
2 19% 25%
3 28% 36%
4 36% 45%
5 43% 54%
6 50% 61%
7 56% 68%
8 62% 73%
9 67% 78%
10 72% 82%
11 76% 86%
12 80% 89%
13 83% 91%
14 86% 93%
15 89% 95%
16 91% 96%
17 93% 97%
18 95% 98%
19 96% 99%
20 97% 99%
21 98% 100%
22 99% 100%
23 99% 100%
24 100% 100%
And with redraw.
1 19% 25%
2 35% 44%
3 48% 59%
4 59% 70%
5 68% 79%
6 75% 85%
7 81% 90%
8 86% 93%
9 89% 95%
10 92% 97%
11 94% 98%
12 96% 99%
13 97% 99%
14 98% 100%
15 99% 100%
16 99% 100%
17 100% 100%