|
|
回复 3# 问渠
的确算错了，实际庄家优势接近2%。
下面是6副牌的算法，写得十分详细。1副牌的算法是一样的。
出和的概率是 = C(13,1)C(4, 2) / C(52,2) =13 X 4 X 3 /(52 X 51)=0.058823529
那么出庄/闲的概率是0.470588235
出和输一半，如果一直下龙/虎，因此庄家优势是0.058823529/2=0.029411764
Casino War
No, it's not a fight between Steve Wynn and Donald Trump. It's a casino game that was developed a few years ago by Dave Sugar, then of the Hacienda, and is marketed through Bet Technology of Carson City. It is based on the children's card game of War that many of us did indeed play as children. A child's card game in a casino? Well, lately I see Yahtzee, Monopoly, Battleship, Pin Ball, Three Stooges, Betty Boop and other ties to things of our youth being used as a basis for casino games. Why? I think it is because folks are more readily willing to try out a casino game that is based on something with which they are familiar. Anyway, there are a few tables of War around and people are playing it. When I attended the 11th Annual Conference on Gambling and Risk Taking at the MGM hotel in June, I noticed that the MGM casino had several War tables and they were occupied by players.
So what is involved in playing this simple game? Well, let me first describe the children's version to you. That is played with an ordinary 52-card deck and there are two players. The cards are shuffled and divided in two, each player receiving 26 cards. The cards are ranked as Ace low and King high; suits don't matter. Each player turns over a card from his/her pile. The higher card wins; both cards are placed face down on the bottom of the winner's deck. If the cards are of the same rank, a "war" is declared. This consists of each placing one card face down on his/her upturned card and another face up. The last up cards are compared and the winner takes all six cards. If another tie occurs, the process of face down, face up is repeated until the war is settled. This seemingly interminable game finally ends when one player gets all 52 cards in his/her pile.
The casino version of War is similar, except that (not surprisingly) money rather than cards are exchanged. The casino version is played on a standard Blackjack table, uses 6 decks, and is dealt from a shoe. The game begins with the player placing a one-unit bet. This done, dealer and player each receive one card. If the player's card is higher than the dealer's card, the player wins even money; if the dealer's card is higher than the player's card, the player loses the wager. In case of a tie, the player can either surrender for half the wager or can declare "war". The latter consists of the player placing an additional unit wager. Then the dealer burns three cards, deals another face-up card to the player and one face-up card to himself. If the player's card is of higher or equal rank to the dealer's card, the player wins one unit on the war bet and pushes the original bet, that is, the player wins one unit. If the dealer's card is ranked higher than the player's card, the player loses both the original and the war bet. There is also a side bet on ties, but it has such a huge house edge that I am going to omit any discussion of it and simply advise you to never take it.
War is a simple game and it is simple to analyze. Clearly there is no house advantage until a tie occurs; the player and dealer are each as likely to win the bet. How often does a tie occur? Recall from my August 1999 article Oh, New York, Bring Back Those Big Dippers that the number of ways of selecting a set k objects from n objects is C(n, k) where
C(n, k) = n!/[k!(n - k)!] (1)
and
n! = n(n - 1)(n - 2)(n - 3) . 1; 0! = 1 (2)
Since there are 13 packets of 24 identically ranked cards, the number of ways of selecting two identically ranked cards is C(13, 1), the number of ranked packets, times C(24,2), the number of ways of picking 2 of the 24. If we divide this number by C(312, 2), the number of ways of picking any two out of the 312 cards, we have the probability of a tie:
prob tie = C(13,1)C(24, 2) / C(312,2) (3)
or
prob tie = (13 x 24 x 23)/(312 x 311) =23/311 (4)
If one surrenders for one half a unit then the expected return on the surrender bet is
exp surr. = - 1/2 x 23/311 = - 0.0369775 (5)
so the house edge on the surrender bet is 3.69775%. On the other hand, to calculate the expected return for a war we have to calculate the probability of a second tie. There are now 310 cards in the deck, of which 12 packets of identically ranked cards each have 24 cards and one packet of identically ranked cards has 22 cards. (Note that 12 x 24 + 22 = 310.) The probability of obtaining the second tie is, therefore,
prob second tie = [C(12, 1) x C(24, 2) + C(22, 2)] / C(310, 2) (6)
which is
prob second tie = [12 x 24 x 23 + 22 x 21] / 310 x 309 (7)
This number is
prob second tie = 0.07397432 (8)
Subtracting the number in (8) from 1 we get 0.92602568, which is the probability of not tying. Since in no-tie situations the player will win half the time and the dealer half the time, the probability of a dealer win is just half this number or 0.46301284. The probability of a player win is, therefore, 0.46301284 + 0.07397432 or 0.53698716. We can now calculate the overall expected return for the game of War if we always go to war on ties:
exp = 23/311[1 x 0.53698716 - 2 x 0.46301284] (9)
or
exp = -0.02877 (10)
Some refer to the number 2.877% obtained from (10) as the house edge. Others divide this number by the expected bet in the game, which is 1 x 288/311 + 2 x 23/311 or 1.073995. The result is 2.67888%. However you figure it, it is better to go to war than surrender.
A couple of additional points. You might think that the above calculation and its results only apply to an "off the top of the shoe" game, that is, a full six decks. Let me convince you, in two ways, that this is not the case. First, suppose you feel that after 100 cards have been dealt that the situation is different than it was off the top. Very well, take a six-deck shuffled shoe and remove the first 100 cards. Now place them at the back of the shoe. We still have a 312-card shoe, except that we have changed our starting position. But, you say, what if we kept track of the 100 cards? Aha, that is a different matter, but let me put that off for the moment. What is the other way that I mentioned above? The answer is that I wrote a short program that plays War according to the above rules and keeps track of the player's stake and the amount wagered. I arranged it so that there was a cut card at 234 (last quarter of shoe cut off). I played 600,000,000 games. The amount lost was 17,259,828 units and the amount bet was 644,370,179. This works out to a house take per game of 2.8766% and a house edge per unit wagered of 2.67856%. Nufsed!
What about keeping track of cards played? Well, deck composition does matter. For example, if all but one packet of equally ranked cards were dealt out (you should live so long), you would have a lock on the next hand, since two matches in a row would be certain. The problem is that one would have to keep exact figures for each of the 13 ranks and look for a situation wherein the remaining deck consisted of a few sets of a large number of equally ranked cards. Not worth the effort, in my opinion. Keeping track of ties is useless. Suppose that from the start of the shoe, 100 ties in a row occurred. It is easy to see that with a few permutations of the player's cards one can produce a scenario with no ties and exactly the same remaining shoe composition.
I have heard some talk that slug cutters might be able to get an edge in this game by remembering a sequence of winning cards and cutting it to the top of the shoe. Maybe, but I would duck on the first bet just in case the slug is one card off. Anyway, I haven't heard of anyone doing this.
So, if you want to spend a few minutes playing a mindless casino game you don't have to sit in front of a slot machine punching buttons. You can play a live game with a reasonable house advantage, better than most slot machines. Just go to War. I'll have to admit, however, a few minutes of War is about all I can take. See you next month. |
|
发表于 2011-5-2 15:40
发表于 2011-5-2 22:00
楼主