18 Dec 01 There are books all over the place talking about how to "beat the casino" at particular games like poker, blackjack, or craps. The most intelligent (and least suspicious) of these talk about the optimal strategy -- the strategy that will give you the best results in the long run. These books will often note that the best strategy will still make you =lose= money, but more slowly than you would if you hadn't used the best strategy. I visited a few gambler's sites recently, specifically to check out video poker, and for most games, with optimal strategy, they promised a return of 99+%. In one case, it was a return over 100%! I was suspicious of that, seeing as there's =scads= of ads and links to online casinos at that site. I will have to check my own calculations, but I won't do all of them here. So, how to check out what kind of money you expect to make, or lose, in the long run? One calculates the =expected value= of your profits/losses, which tells you what you expect to make/lose per game if you play it many, many times. So here's a payoff scheme for a particular video poker game: Royal Flush 250 Straight Flush 50 Four of a kind 25 Full House 9 Flush 6 Straight 4 Three of a Kind 3 Two Pairs 2 Pair of Jacks or better 1 And you pay 1 coin to play, so basically you're even if you get a pair of jacks, queens, kings, or aces. Now, we've calculated the probabilities for most of these occurances, except for Royal Flush (10, J, Q, K, A all in one suit -- only 4 ways to do this) (and this affects the counting for straight flush) and one pair, Jacks or better. So, there are 4 ways to make a Royal Flush, out of a total of 2598960 - probability: .00000154 The straight flush probability now changes - from 40 ways to 36 ways - probability: .0000139 And to get a pair of jacks or better, one must pick a face value from (J, Q, K, A), then pick two of the 4 cards with that face value, then pick three other face values, and 1 of the 4 from each of those. number of ways = 4_C_1*4_C_2*12_C_3*4*4*4 = 4*6*220*4*4*4 = 337920 probability: 0.130 So to figure out expected value, one takes a weighted average of the values - the weights being the probability with which one expects to get that value. hand payoff prob product Royal Flush 250 .00000154 .000385 Straight Flush 50 .0000139 .000695 Four of a kind 25 .00024 .006 Full House 9 .00144 .01296 Flush 6 .001965 .011790 Straight 4 .003925 .0157 Three of a Kind 3 .0211 .0633 Two Pairs 2 .0475 .0950 Pair of Jacks or better 1 .130 .130 total: .335830 Only 34% return! And the site claims 99.5%! Okay, calm down, we're not done yet. No one would play video poker with odds as bad as these -- any fool would notice how quickly their pile was going down and how rarely they won anything. One way to fix the above problem would be to give bigger payouts for the rarer hands (or even a bigger payout for the most likely hand in the list). When playing to a payout scheme like this, the probability of each outcome =and= the payouts must be taken into account. Suppose the payoffs were ten times what is given here. Surefire money-maker! Heck, even if only the payoff for royal and straight flushes were increased enough, you'd have a very positive value on your hands. Well, the casinos are not about to be playing losing games, they want to make money after all. So why not add a little choice to the mix? So video poker is almost always a 5-card draw poker game. In this game, you are dealt a hand, and you can turn in any number of cards to get different cards to replace, in order to try to better your hand. =However=, though certain strategies in a regular 5-card draw game against other people simply look for the strategy which will maximize the probability of getting a better hand, in video poker you have to maximize your expected payout. This is an optimization problem, but since it's discrete, it's a relatively easy optimization problem. For any 5-card hand, you have 2^5 = 32 possible strategies (for each card in your hand, you can either keep it, or discard it). So, given a particular hand, there's only a maximum of 32 possibilities to try out. Let's do that for a particular hand: Jack of hearts Queen of hearts Queen of spades King of diamonds Ace of hearts Now, we are sitting here with a pair of Queens, sure to get us our 1 coin back if we don't do anything. However, it might be to our interest to discard some of those cards to see if we get a better hand. For example, if we discard the J, K, and A, we could possibly get one or two more queens. We could also could get one queen, and an extra pair for full house. However, we've almost got a straight. Perhaps we should discard one queen and try for the straight. As well, we've got three hearts - perhaps we should throw out the spade and the diamond and go for the hearts-flush. You may think you know the proper strategy in 5-card draw, and perhaps it extends to video poker, but optimal strategy in a payoff game like this is very dependent on payout schemes. I have no desire of calculating the expected payout on all 32 possible discards, because some of them are obviously not going to achieve a maximum expected value (like discarding all 5 cards, or just discarding the queens). Let's try a few possibilities: Keep the queens, discard the 3 others: You're taking 3 cards out of the remaining 52 - 5 = 47 cards, so there's 47_C_3 = 16215 possible outcomes. -can get four-of-a-kind by getting the other two queens, and 1 other card from the remaining 45 -- 45 ways - prob = .002775 -can get full house by getting one other queen, and an extra pair, or by getting three-of-a-kind of some other value. This is a little tricky, because the particular Jack, King, and Ace aren't available. So let's break it up: Q + non-J/K/A pair = 2_C_1*9_C_1*4_C_2 = 2*9*6 = 108 Q + J/K/A pair = 2_C_1*3_C_1*3_C_2 = 2*3*3 = 18 non-J/K/A triple = 9_C_1*4_C_3 = 9*4 = 36 J/K/A triple = 3_C_1*3_C_3 = 3 so a grand total of 165 ways -- prob = .0102 -can get a three-of-a-kind by getting another queen, and two cards that aren't a pair. Easily enough, rather than go through the whole J/K/A razzmatazz, let's just note that it's the number of ways of getting a queen, two other card, minus the number of ways of getting a full house: 2_C_1 * 45_C_2 - 165 = 2*990 - 165 = 1815 ways - prob = .112 -can get two pair by getting another pair of cards, and one extra card: non-J/K/A pair: 9_C_1*4_C_2*41_C_1 = 9*6*41 = 2214 J/K/A pair: 3_C_1*3_C_2*42_C_1 = 3*3*42 = 378 for a total of 2592 ways, prob = 0.160 -then you can get nothing extra - just have the one pair of queens you started with: # ways = 16215 - 2592 - 1815 - 165 - 45 = 11598 prob = .715 so our expected value is: 4-of-a-kind 25 .00278 .0695 full house 9 .0102 .0918 3-of-a-kind 3 .112 .336 2 pair 2 .160 .32 1 pair 1 .715 .715 total: 1.532 Yeah! Much improved over our original 1, and note we are guaranteed to get at least 1 in return. Let's try a different strategy now. Suppose I try to go for the royal flush: I'll throw out the queen of spades and king of diamonds, seeing what I can get. Again, there are 47 cards left to pick from, and we're getting 2. So that's 47_C_2 = 1081 possible outcomes. - can get a royal flush by getting the 10 of hearts and the King of Hearts -- there's only one way to do this. prob = .000925 - can get a flush by getting two hearts, but not the 10 and king: 10_C_2 - 1 = 45 - 1 =44 ways. prob = .0416 - can get a straight by getting a 10 and a king, but not both the 10 and king of hearts: 3_C_2*3_C_2 - 1 = 3*3 - 1 = 8. prob = .0074 - can get a three-of-a-kind by getting a pair of aces, jacks, or queens: 3_C_2 + 3_C_2 + 2_C_2 = 3 + 3 + 1 = 7. prob = .0065 - can get two pair, by getting two of one ace, one jack, or one queen: 3_C_1*3_C_1 + 3_C_1*2_C_1 + 3_C_1*2_C_1 = 3*3 + 3*2 + 3*2 = 21 ways. prob = .0194 - can get one high pair, by getting: pair of kings or one ace, jack or queen and something else: 3_C_2 + 8_C_1*39_C_1 = 3 + 8*39 = 315 ways. prob = .291 - can get absolutely nothing: 1081 - 315 - 21 - 7 - 8 -44 - 1 = 685 ways. prob =.634 let's do our chart: payout prob product royal flush 250 .000925 .23125 flush 6 .0416 .2496 straight 4 .0074 .0296 three-of-a-kind 3 .0065 .0195 two pair 2 .0194 .0388 one pair 1 .291 .291 nothing 0 .634 .000 total: .85975 Obviously, this strategy is less than optimal - it doesn't even achieve the payoff of 1 you're sure to get if you just held on to all your cards. However, if the payoff scale were different - say a royal flush were worth 500 and a flush were worth 18, then the total would be 1.59 -- higher than our previous optimal strategy. So, keep this in mind should you ever play video poker - optimal strategy is contingent on the payouts provided. As long as you're not actually tampering with machines, most people will leave you alone as you play video poker, but sitting down and calculating the optimal strategy for each of the possible 2.6 million original hands can get to be a bit too slow as one sits there. Luckily, there are standard payoff schedules and one can go to gambler's sites to take a look at them and sometimes even optimal strategy. As well, one need not look at =every= individual possible hand, as they naturally group themselves into such things as "4 card flush" and the like. In any case, you could see how easy it would be to program a computer to determine the optimal strategy for any payoff scheme; 2.6 million isn't really that much for a montecarlo program. If you wonder where expected value might come in handy in other places other than gambling, consider that ever-legal and ever-distinguished form of gambling: investing in stocks and bonds. Investment corporations have complicated models meant to calculate the probabilities of various outcomes so that they can make optimal choices which will provide optimal expected payoff. Of course, the market isn't exactly controlled by pure probability. But that's beyond my realm of expertise.

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