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Home » Forums » Off-Topic » General Discussion » The hoax that is the 1.41% house advantage on pass line bets
The hoax that is the 1.41% house advantage on pass line bets
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| March 9th, 2010 at 11:37:31 AM permalink | |
| RaleighCraps Member since: Feb 20, 2010 Threads: 29 Posts: 601 | Alan, For comparison sake, can you put up the distribution for a passline bet that has ~$164 at risk? Perhaps $30 pass line with $120 odds. If you had the same seed as you just used for tutti, that would be awesome. Always borrow money from a pessimist; They don't expect to get paid back !
Be yourself and speak your thoughts. Those who matter won't mind, and those that mind, won't matter! |
| March 9th, 2010 at 2:02:03 PM permalink | |
| tuttigym Member since: Feb 12, 2010 Threads: 5 Posts: 176 |
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| March 9th, 2010 at 2:14:54 PM permalink | |
| tuttigym Member since: Feb 12, 2010 Threads: 5 Posts: 176 |
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| March 9th, 2010 at 2:25:15 PM permalink | |
| SPOOS1 Member since: Mar 9, 2010 Threads: 0 Posts: 3 | tuttigym: The house edge is calculted on the result from all 36 combintions of the dice. Since the lowest common denomintor is 160 the formula is as follows 36x160= 5760 to figure the house edge from the pass line it is: 2 3 4 5 6 7 8 9 10 11 12 160 320 480 640 800 960 800 640 480 320 160 It is a perfect bell shaped curve. To plot the edge -160 -320 -160 -128 -72 +960 -72 -128 -160 +320 -160 = -80 -80/5760= -1.388% For the don't pass +160 +320 +160 +128 +72 -960 +72 +128 +160 -320 PUSH = -80 -80/5760 = -1.388% The house edge on all other bets is greater than this. |
| March 9th, 2010 at 2:45:52 PM permalink | |
| tuttigym Member since: Feb 12, 2010 Threads: 5 Posts: 176 |
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| March 9th, 2010 at 2:48:27 PM permalink | |
| goatcabin Member since: Feb 13, 2010 Threads: 4 Posts: 495 |
First, you say that the only two numbers relevant are the six and seven. That is correct. But the probability of winning or losing that bet is dependent on the probabilities of getting a six and seven RELATIVE TO EACH OTHER. So you cannot use 36 as the divisor. There are five ways to roll a six and six ways to roll a seven, so the divisor is their sum - 11. You don't care about any other rolls, right? So, p(6 before 7) is 5/11, p(7 before 6) is 6/11. I still don't see how you ever came up with .167, in any case. And, of course, the ways are not 6 and 7 at all. Your confusion is incredible!
The payoff does not affect the probability of winning a bet. Since you think the only relevant thing is the probability of winning a bet, I ask again: "Why do you bother to bet $12 on place 6/8, while you bet $10 on 5/9/4/10?"
Who would play a lesser FO bet on the 4/10? Not I, unless I am taking max odds at a 3,4,5X table. Do you understand why 3, 4, 5X odds are now more-or-less standard? It's because the payoff is always 6 units, i.e. it's easier on the dealers. Also, there is more variance on the outside numbers, due to the increased, 2:1 payoff, and the casinos do not like to give too much variance for free. I guess you simply are not intelligent enough to understand that the evaluation of a bet must include BOTH its probability of being won and the payoff relative to the amount bet. A coin-flip game: heads you pay me $5 tails I pay you $5 OR heads you pay me $6 tails I pay you $5 Are these equally fair? If not, why not? Cheers, Alan Shank |
| March 9th, 2010 at 2:54:48 PM permalink | ||||||||||||||||||||||||||||||||||||||||||||||||||||
| goatcabin Member since: Feb 13, 2010 Threads: 4 Posts: 495 |
These are not correct. I'm not sure how you came up with the values for each number, but: In the first place, you can express all the possible outcomes as integers with 1980 bets as below:
If you add them all up, you get -28 / 1980 = -.01414. For the don't pass, it's -27 / 1925 = -.01403 (some books consider the 55 ways for the 12 as "risked", and show -27 / 1980 = -.01364). Cheers, Alan Shank | |||||||||||||||||||||||||||||||||||||||||||||||||||
| March 9th, 2010 at 2:57:42 PM permalink | |
| tuttigym Member since: Feb 12, 2010 Threads: 5 Posts: 176 |
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| March 9th, 2010 at 4:52:26 PM permalink | |||||||||||||||||||||||||||||||||||||||
| goatcabin Member since: Feb 13, 2010 Threads: 4 Posts: 495 |
Keep in mind that the odds bet is only made 2/3 of the time, so in order to get an average bet of around $164, you have to go up substantially from $30 plus $120. I ended up trying $40 pass with 5X odds, which results in an average of over $173 for each passline bet. Of course, that is going to be a LOT more variance than tutti's game, where he stops betting after 3 rolls. The results indicate this clearly:
So, this has almost three times the volatility as tutti's $164 method. In my simulation of tutti's method, in about the save average time (# of rolls), about 225 bets were resolved, but those bets were for a maximum of $30 of some of them only $4. In this sim, an average of 93 bets were resolved per session, but they averaged $173. Note that the mean bet handle was almost 2 1/2 times higher with PL/FO at this level. OTOH, despite the huge bet handle, the mean loss was over four times higher with tutti's method, and the probability of a winning session was .403, compared to .477. That is how the relationship between expected loss and variance works: the lower the ratio of ev / SD, the higher the probability of breaking even or better. In tutti's case, 179 / 636 = .281, while 43.4 / 1718 = .025. Of course, the higher the variance, the higher the "penalty" for having bad luck. There is no "magic bullet". If you squeeze a balloon on one side, it puffs out on the other. Gambling is the same. Here's something else to consider. If you know the expected loss and SD for two strategies, you can figure out the degree of luck where one method comes out better than the other, IOW the "breakeven point". In this case, suppose both players have exactly average luck, i.e. each player experiences the expected value from a session (like the 244-251 that's been beaten to death here). tutti: -179 PL/FO: -43, so the advantage at average luck is $136 to PL/FO Now, the difference in the SDs is 1718 - 636 = 1082 This means that, for every one SD of "good luck", the PL/FO gets more ahead by $1082. If tutti has one SD of good luck, he comes out -179 + 636 = +$457. If PL/FO has one SD of good luck, which is equally likely, he comes out -43 + 1718 = +$1675. However, what if they have equal degrees of bad luck? If tutti has one SD of bad luck, he comes out -179 - 636 = -$815 If PL/FO has one SD of bad luck, which is equally likely, he comes out -43 - 1718 = -$1761. So at -1 SD, PL/FO is more badly off. Somewhere in between +1 SD and -1 SD, then, there must be a "crossover point" where the heavy variance in the PL/FO method makes it come out worse. We can find it by dividing the "initial advantage", i.e. the advantage of the PL/FO at average luck, by the difference in the standard deviations. So: 136 / 1082 = .1257 So, if they have average or better luck, the PL/FO comes out ahead, the more good luck the farther ahead (note the $7420 biggest win, compared to $1821 for tutti). However, if they both come out .1257 SD WORSE than average, then tutti is going to lose LESS, the worse the luck, the more less, so to speak. This is why the bust rate is much higher for the PL/FO. Let's check our figures: -179 - (.1257 * 636) = -259 for tutti -43 - (.1257 * 1718) = -259 for PL/FO Below that, the PL/FO at that level gets hammered worse and worse. The probability of coming out .13 SD OR MORE WORSE than the mean expectation is .448. So, the PL/FO has a .552 probability of doing better than the "tutti method", with a much, much higher "up side" but a much, much lower downside. Each individual's tolerance for volatility is different; if you can afford (and are willing) to lose $2500 in a quest for winning $1000 or more (PL/FO has a probability of .28 of doing that), then I am happy for you. >:-) For my taste, tutti's method has too much expected loss and too much variance, both, because of the relatively high HA and the amount of the bets. The key, in my view, is to understand what the session-outcome probabilities are, just like one (except tuttigym) understands the probabilities of winning the different bets and the probabilities of the dice rolls. Then you can come up with a strategy that gives you the best chance to achieve what you want to achieve, whether it's to minimize losses and stay at the table, to win a certain amount of money, or whatever. Cheers, Alan Shank | ||||||||||||||||||||||||||||||||||||||
| March 9th, 2010 at 8:12:30 PM permalink | |
| boymimbo Member since: Nov 12, 2009 Threads: 12 Posts: 2533 |
House advantage, with 30 ways to win, 6 ways to lose, as specified. Your expected loss, per roll: HA per roll =(1/36 * 30 - 35/36 + 2/36 * 15 - 34/36) * 2 + 25 * 3/36 * 9/5 * 2 + 25 * 4/36 * 7/5 * 2 + 30 * 5/36 * 7/6 * 2 - 160 * 1/6 = -$2.166667 PER ROLL, (HA per roll, 1.321%). = YOU LOSE. No betting system can beat the house. Why? Because the house always pays you less than the true odds on the dice. Here's a bet for you. I have ONE unbiased dice with six sides. You bet $6 and get to pick 5 numbers. If one of the numbers hit, I pay you $1. This is why craps works in the house advantage. Here's another bet for you. I have a roulette wheel with 37 numbers. You get to pick every number but zero and I'll pay you $1 each time a number hits. Of course if the single zero hits you lose everything. Is the hoax that the single zero roulette house advantage is 2.702703%? Should we open another thread? Craps is the same. Just because there are two dice doesn't mean there aren't 36 combinations. The house advantage is easily calculable. It doesn't require any sampling at all. Just because you don't get the expected result on 495 rolls doesn't mean it isn't so. That's what variance is all about. Edit: Tutti, a word of advice to help you out on your $160 across bet. Find a casino where you only buy the 4 and 10 and pay the commission only when you win. A $25 buy 4 or 10 will pay you $49 while a $25 place 4 or 10 will pay you $45. They are both the same bet. "HA =(1/36 * 30 - 35/36 + 2/36 * 15 - 34/36) * 2 + 25 * 3/36 * 49/25 * 2 + 25 * 4/36 * 7/5 * 2 + 30 * 5/36 * 7/6 * 2 - 160 * 1/6 = -$1.50 PER ROLL, (HA per roll, 0.915%). = YOU LOSE LESS. " More advice: Drop the $4 horn bet. HA = 25 * 4/36 * 7/5 * 2 + 30 * 5/36 * 7/6 * 2 - 160 * 1/6 = -$1.00 PER ROLL, (HA per roll, 0.625%). = YOU LOSE EVEN LESS. -----
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