For those who might be curious and want to see the math inherent in the game. This "table" shows the calculations involved setting the house advantage, on the pass line (or single come) bet. Code: Number Rolled Win probability with Probability number Product number rolled vs. 7-out is rolled (column 2 x 3) 2 0 1/36 0 3 0 2/36 0 4 3/9 3/36 1/36 5 4/10 4/36 2/45 6 5/11 5/36 25/396 7 1 6/36 1/6 8 5/11 5/36 25/396 9 4/10 4/36 2/45 10 3/9 3/36 1/36 11 1 2/36 2/36 12 0 1/36 0 As we know, we lose on the Pass or Come when the come out is 2, 3 or 12, and we win with a 7 or 11. The established point numbers win or lose, depending on whether they are repeated before the seven shows. Sometimes they win, sometimes they lose. When you multiply column 2 x column 3 for each of the numbers, the product gives you the contributing "win factor" for each number. Notice that the 3 craps numbers all contribute nothing --- they LOSE on the come out. Column 2 shows the 7 and 11 are guaranteed winners on the come out, and their probabilities determine the overall effect on the win when you multiply. When you add together all the numbers in the last column, you have the "math" for the game, a 49.292929........... 49.3% chance to win at the game playing one Do bet. Obviously nothing new here, just thought maybe a newbie or two might be curious.
The format of the table and its titles did not take properly into the html format on this page, even though when I try to edit, it appears exactly as I entered it. Sorry about the confusion - what shows above isn't very helpful as it shows here.
TY Jacob, but if I only knew what the heck you are talking about. As you can see, I'm a dinosaur with the computer stuff.
You are confusing the actual odds of making the point with the vig. They are 2 different things. They are calculated differently. 777
TDV, The calculated 0.4929292 or 49.3% chance of winning equates to a 50.7% chance of losing. The 1.4% comes from the total difference between winning and losing 50.7 - 49.3 = 1.4 If the damned table showed as planned, next would be the calculation of variance, and its square root, the standard deviation. Playing just single Do bets, without odds, the player can be within 3 standard deviations of the norm up to about 100,000 or so bets. Past this point, things calculate as essentially impossible to win. For example, at 1 million such bets, variance shows one would have to be 11 or so standard deviations better than the mean result in order to be ahead. How likely is this? From there, it's on to a table for betting the Don't, to come up with a slightly better 1.36% odds against, then the factoring in of weighted odds.
We have all had winning sessions at the craps table. Unfortunately, we also have lost (at least I think so). Variance is the square of the difference of a result from the mean/average result. At craps, this is the amount eventually won or lost with each bet placed. Assuming $1 bet (not a good assumption I know, but it simplifies the math), the result will be either +$1 or -$1. The variance of a single bet is either 1 squared or -1 squared. The expected value calculated in the table is -1.4% or -0.014 cents on the dollar that is bet. For multiple and similar bets (one pass bet only), the variance of the sums equals the sum of the variances. After 100 bets, -0.014 x 100 = -$1.40, the expected loss. Since variance of a single bet is 1, variance of 100 single bets is 100. Square root of variance , standard deviation is 10. After 100 $1 bets, the expected "result" is that the player will be out 100 x $-0.014 = -$1.40. There is a 67% chance he will be within $10 of this result, i.e., between -$11.40 and + $8.60. There is a 95% chance he will be within 2 standard deviations (10 x 2 =20) of the expected result (-$21.40 and +$18.60) and a 99% chance he will be within 3 standard deviations. At 10,000 bets placed, numbers are 10,000 x -0.014 = -$140. Variance is 10,000 and SD is 100. Here the likelihood of being within 3 SD's of the mean, results will be between -$440 and +$160, with 99% certainty. When you get to one million bets, you have -$14,000. SD is 1000. At this point, expectations are that you will be between -$16,000 and -$13,000, almost guaranteed - not a good place to be. As a mathematician would say, if you play the game long enough, you are assured to lose. Keep in mind that the numbers above apply to a SINGLE pass or come bet. Playing anything else, except ODDS on the bet, greatly decreases your chances. It is a mathematical marvel that anyone can win at this game over the long haul, but yet we all continue to play, and often enjoy doing so. Wish I was in the casino business.
Ever heard of the "Perfect 1980"? It takes 1980 decisions to reflect the probabilities of a pass line bet in integers, from which you can calculate most everything. Here goes: comeout win 440 comeout loss 220 point 6 or 8 550 (win 250, lose 300) point 5 or 9 440 (win 176, lose 264) point 4 or 10 330 (win 110, lose 220) wins: 440 + 250 + 176 + 110 = 976 losses: 220 + 300 + 264 + 220 = 1004 976 / 1980 = .492929 1004 / 1980 = .50707 1004 - 976 = 28; 28 / 1980 = .01414 660 / 1980 = .33333 = probability of decision on comeout roll 1320 / 1980 = .66667 = probability of a point being established For the Don'ts, it's a mirror image, except that comeout win is 165, push is 55. Cheers, Alan Shank
To calculate the average number of rolls per hand (i.e. until shooter sevens out and passes the dice): From the Perfect 1980: 300 losses on 6 or 8 264 losses on 5 or 9 220 losses on 4 or 10 ----- 784 seven-outs out of 1980 1980 / 784 = 2.5255 Cheers, Alan Shank
The odds of making the established PL point of 6 or 8 or 5 or 9 or 4 or 10 is the same as the PL vig of .014.......is that what you are saying ? 777
Whoops! I mis-read what you were saying. Of course, the win probability for the pass bet overall is a weighted average of the probabilities of winning on the comeout, on a 6/8 point, on a 5/9 point and a 4/10 point. The HA is the difference between the overall probability of winning and losing, i.e. .493 - .507 = -.014. Cheers, Alan Shank
These figures, while correct, don't seem to me to be very relevant to individual players; how many have made 10,000 bets in their lives? I think that the ev/SD is a good measure of how likely a player is to be ahead after x number of bets. When the SD gets to be the same as the ev, then the player has about a .16 probability of being even or better, i.e. the probability of being one standard deviation better than the mean. For a straight pass line bet, this point is reached after 5000 bets. When you get to 10,000 bets, the probability of being even or better goes down to about .08. Still roughly one in eight players would still be expected to "have their heads above water". When you start taking odds on your pass bets, the ev remains the same, but the SD goes up markedly, meaning that you don't have to be as lucky to be even or ahead. For 5000 $5 pass bets with 3, 4, 5X odds, the ev/SD is -0.2034, so just two tenths of a standard deviation of positive variance is required to be even or ahead, the probability of this being almost .42. Your +/- 1 SD figures are +$1384 and -$2091, so you pay dearly for negative variance. If you have 10X odds available, the SD per bet is $54, so the ev/SD at 5000 bets is -.0925, breakeven probability .4631, +/- 1 SD numbers ($5 line bet) +$3468 and -$4175. These numbers assume constant bet amounts. When you start getting into Tom's "tough craps", the variability gets huge. Cheers, Alan Shank
