# about probablity of a card game with replacement allowed

I am thinking of a card game using standard 52 card and randomly deal 5 cards as initial hand. It is easy to calculate the probability of each winning hand (straight, flush, full house, etc.) dealt to the player. But if we allow the player to replace any number of cards in the initial hand ONCE (i.e. the player could pick up N (1 or 2 or 3 or 4 or 5) cards in the initial hand to be replaced by N cards in the rest of the deck, than how can I estimate the probability of those winning hands? This problem is not straightforward because the player could choose any number of any cards in the hand to be replaced. I have been thinking on this for days but I can only think of I way to work out the problem but I am not sure if this works or not.

Consider the 5 cases in which the player would like to replace 1, 2, 3, 4 or all 5 cards. We analyze them one by one

1) Consider the player would like to replace only 1 card, so 1 more card will be drawn from the deck. It seems to be the same as we randomly draw 6 cards out the 52-card deck initially and consider how many winning hands could be obtained from that.

2) Similarly the 1), when the player would like to replace 2 cards, we randomly draw 7 cards out of the 52-card deck and count how many winning hands could be made from that.

3) 4) 5) Similarly to 1) but we need to draw 8, 9 and 10 cards out of the 52-card deck separately and count the number of winning hands out of those hands.

I use computer to simulate 1) to 5) and count how many winning hands possible, but here comes to another question. Taking straight as example, if I get K winning hands by the way mentioned above, to what I should compare to so to find the probability? Should I compare to $P_{52}^{5}$ (i.e. permutation of choosing 5 out of 52) or $52!$.

I can only think of the way to use computer program to count the winning hands but I would like to know any neat mathematical way to calculate the probability for this issue. Thanks.

• Replacing $1$ is not the same as winning hands from $6$, since the discarded card is unavailable. – André Nicolas Mar 26 '14 at 18:13
• thanks. Ok, that makes sense. But then I have no clue how to approach this problem. – user1285419 Mar 26 '14 at 18:24
• It is a very difficult problem, since strategy is involved. One can solve more restricted types of problems, but they are mostly too easy. For example, you have a couple of $8$'s and a couple of Kings, and a useless card. If you discard it, what is the probability of a full house? – André Nicolas Mar 26 '14 at 18:28
• Hi Andre, your comment lead to two questions. I wonder why strategy is involved in this game? Is it because the way we discard the cards will affect the result? What I am thinking is even though the way (number of cards and which cards) to discard the cards will change the result, but the replacing cards from the deck are random also, so even that shall we still count this as strategy game? – user1285419 Mar 26 '14 at 18:41
• So if this is stragey based, can I do the analysis in this way? I start from a random hand, try all possible actions (i.e. replace all possible number of cards) one at a time so I build up a tree of decision and I will count the path leading to all possible winning hands. Do you think this way works? If each initial hand give $N_i$ total paths in the game and there are total C(52, 5) initial hands (i.e. total C(52, 5) trees), and if finally I find that $M_i$ path leading to a straight, can I say that the probability of winning a straight is $P=\sum_i M_i / (\sum_i C(52,5)*N_i)$ – user1285419 Mar 26 '14 at 18:41

I would calculate this with the fundamental strategy of hand improvement: e.g.

calculate the probabilities for each of the standard 5 card poker hands, and then for each of these calculate the probability of improvement.

Hands with little or no probability for improvement, and so will not draw:
Straight Flush (includes royal)
Full House
Straight
Flush
4 Kind

You wanted the probability of the resulting hand, so you would ignore the facts that given a straight the player could possibly discard 1 or more to try for the straight flush... Or calculate it and see how crazy the player is being.

Hand with draw 1 for improvement

2 Pair: Basic Hand 13C2 * 4C2 * 44 / 52C5
Only 4 cards improve the hand out of the 43 cards in the deck so that the improvement probability is (4 / 48) multiplied times the basic probability and added to the final full house probability as what you are really saying is "What is the probability of the player final hand being a full house by drawing 0, 1, 2, 3, 4 OR 5 cards. The OR implies summation of the probabilities

4 Card contiguous Straight: 8 cards complete the straight, and 12 cards make 1 pair
4 card inside straight: 4 cards complete the straight, and 12 cards make 1 pair
4 Card flush: Only 9 cards complete the flush, and 12 cards make 1 pair.

wash, rinse, repeat down to draw 5.

Regarding the proposed simulation, you would not compare K to either 52P5 or 52!, you would calculate K / Runs e.g. you deal 1E6 hands, and it results in K Flushes, then it is K / 1E6 showing the occurrence frequency of Flushes. As the number of runs increases, this ratio tends towards the probability. Be aware however about the total number of runs required to simulate poker hands with a high degree of certainty is an extremely large number, starting around 1E9 or better. Remember a royal flush occurs approximately 1/6E5 hands.