The Five Guys II
In order to illustrate and isolate the effects of different strategies on variance and skew, I wrote a program to simulate the play of five different players, all experiencing the same series of outcomes, for 40,000 sessions of 60 passline bets each (or until some other stop condition took place). I ran three sets of these, the first with just passline bets, the second taking double odds and third taking 3, 4, 5X odds. I used the same RNG seed for each set of sessions.
The five players had these characteristics:
1. constant $5 bet amount, no stop conditions other than 60 bets
2. constant $5 bet amount, but with a stop condition of a loss amount
3. constant $5 bet amount, but with a stop condition of a win amount
4. 50% progression after every win, no stop condition
5. 50% progression after every loss, no stop condition
I didn't combine any of the characteristics, in order to isolate their effects. Later, I will simulate some combinations.
First, for the passline without odds, for which the theoretical figures are:
60 bets: ev -$4.24, SD $38.73
player Guy 1 Guy 2 Guy 3 Guy 4 Guy 5
mean net : -$4.36 -$3.91 -$4.16 -$7.69 -$7.92
std. dev.: $38.91 $36.61 $37.94 $114.49 $136.59
+sessions: 16162 16058 16428 13135 23598
0sessions: 4111 3981 3964 241 225
-sessions: 19727 19961 19608 26624 16177
max win : $160 $160 $50 $3660 $185
max loss : $160 $50 $160 $188 $4615
busts : N/A 9248 N/A N/A N/A
win goals: N/A N/A 6890 N/A N/A
The graph for Guy #1 is the familiar bell-shaped curve. With a stop-loss of $50, you cut off the left tail and build a spike at -$50, with the opposite effect for a win-goal stop condition, except the spike is smaller. It's easier to lose $50 than win it. The progressions really kick in the "skew", as progressing on a win results in lots more losing sessions but large wins when the dice cooperate, a "right skew". Progressing on a loss creates more-or-less of a mirror image of progressing on a win, a "left skew". Naturally, since these "guys" were betting more, their mean loss is larger. When you're not taking odds, to get variance you have to progress your bets.
Now, for $5 passline (base bet), taking double odds, stop-loss and win goal each $100
60 bets: ev -$4.24, SD $110.67
player Guy 1 Guy 2 Guy 3 Guy 4 Guy 5
mean net : -$3.38 -$3.03 -$3.06 -$9.19 -$4.54
std. dev.: $110.48 $100.10 $102.34 $314.62 $369.83
+sessions: 19389 18312 20437 14569 24671
0sessions: 127 104 104 93 86
-sessions: 20484 21584 19459 25338 15243
max win : $477 $477 $124 $10917 $890
max loss : $445 $114 $445 $529 $12629
busts : N/A 13964 N/A N/A N/A
win goals: N/A N/A 13124 N/A N/A
First, notice the effects of taking odds on all the "guys": about the same mean outcome, but lots higher SD, more winning sessions, larger wins and larger losses. No surprises. Even doubling the stop-loss and win goal, more shortened sessions for Guys #2 and #3. Of the 40,000 sessions, Guy #4 had 392 wins of over $1000, 18 of over $5000. Guy #5 had 36 sessions losing at least $5000, 503 losing at least $1000. Of course, those are still very low percentages of all the sessions.
Comparing the different strategies in the double-odds scenario, we see pretty much the same relationships as we saw without odds, somewhat less variance for Guys #2 and #3, lots more for #4 and #5, more winning sessions for #3 and #5, more losing sessions for #2 and #4.
(Notice that the stop-loss and win-goal can be "overshot" by a bit when you're taking odds.)
Now, for $5 passline (base bet), taking 3, 4, 5X odds, stop-loss and win goal each $150
60 bets: ev -$4.24, SD $190.37
player Guy 1 Guy 2 Guy 3 Guy 4 Guy 5
mean net : -$3.49 -$2.63 -$2.79 -$10.83 -$2.33
std. dev.: $189.88 $167.66 $169.89 $547.42 $634.56
+sessions: 19475 17641 21312 15045 24724
0sessions: 357 284 269 50 42
-sessions: 20168 22075 18419 24905 15234
max win : $835 $835 $180 $19994 $1706
max loss : $725 $175 $725 $1488 $21271
busts : N/A 16318 N/A N/A N/A
win goals: N/A N/A 15793 N/A N/A
Kicking the odds up for Guys #2 and #3, even with $150 as the cutoffs, still results in more shortened sessions. Guy #4 has 15 sessions winning over $10,000, 59 winning over $5000 and 1064 winning over $1000. Still, the odds are about 37-to-1 against winning $1000 or more. Guy #5 has 1261 sessions losing at least $1000, 83 losing at least $5000 and 26 losing $10,000.
If I had kept the magnitudes of the stop-loss and win-goal constant, the number of shortened sessions would have been much greater, of course.
Next, I will combine some of these strategies, giving Guys #4 and #5 a stop-loss, a win goal, and maybe both.
More about variance
We've seen that variance is minimized as the probability of winning gets closer to .5 and the payout gets closer to even money. This implies that, if a player wants to be able to play for a certain time on a short bankroll, playing the minimum on the passline (or don't pass) without odds gives him/her the best chance to do this. This is because the expected loss is very low and in order to bust the player has to have an extremely unfortunate session, losing many more bets than winning. OTOH, the low variance means it takes more luck to overcome the edge, small as it is.
The crucial factor in determining how lucky a player has to be to be ahead after a given number of bets is the ratio of expected value (ev) to standard deviation (SD). If you know the ev and SD for a given bet and the amount of the bet does not change, then the ev/SD for any given number of bets is given by:
(ev per bet / SD per bet) * sqrt(number of bets)
For example, suppose we take a $5 pass bet, with an ev per bet of -.0707 and a standard deviation of $4.9995. For 60 bets, we would get:
-.0707 / 4.9995 * sqrt(60) = -.1095
To check our work, let's do it the longer way:
-.0707 * 60 = -$4.242
sqrt(60) * 4.9995 = 38.726
-4.242 / 38.726 = -.1095
The ev/SD answers the question, "How many standard deviations (or fractions thereof) better than expectation does a player have to 'achieve' in order to overcome the expected loss?" Once you know this figure, you can look it up in a statistical table to determine the probability of that occurring. The probability of having a session where the result is one SD (or more) BETTER than the ev is about .16, or odds of about 5.3 to 1 against this happening. Likewise, the probability of having a session where the result is one SD WORSE than the ev is the same. This also means that the probability of a session result falling within one SD on either side of the ev is about .68. The probability of a session result falling within two SD on either side of the ev is about .95. That seems very high, but it also means that we expect one out of every 20 session, on average, to fall outside two SD, and we fervently hope it's on the "good" side.
"Luck" is symmetrical
A player is equally likely to come out one SD BETTER than the ev or WORSE. So, you can see that, for a given ev, the more variance you build in the better your chances of coming out ahead, but also the better your chances of losing more. That is, the payoff for one SD of good luck is higher, but the penalty for one SD of bad luck is also higher. This symmetry, of course, is on either side of the ev, not the zero point, so for a given "degree of luck", you lose more than you win. For example, let's look at plus-minus one SD if we played the passline for 500 bets:
ev = -.0707 * 500 = -$35.35
SD = 4.9995 * sqrt(500) = $111.79
ev/SD = .3162
So, if we had a session result one SD BETTER than the ev, we would win about $75, while we would lose about $145 with an equal degree of bad luck. We would need to come out a little more than three tenths of a SD BETTER than the ev in order to break even or better, the probability of which is about .38.
So, if you want to win and are willing to pay the price for high variance, minimizing the expected loss by sticking to low-edge bets (especially "free odds") and at the same time boosting variance is the way to give yourself the best shot. The odds bets are the only way to get variance without increasing the expected loss.
For a given bet handle, making fewer bets maximizes variance
Huh? What does that mean?
Suppose you want to double your bankroll, and you're willing to invest all of it to do so. Simply bet your entire bankroll on one of the line bets. You have a .4929 or .4930 probability of winning. (Of course, you cannot exceed the table maximum, unless maybe you're at Binion's.)
Another factor that helps determine variance is the number of different outcomes there are. In the above case, there are only two possible outcomes. OTOH, if you bet half your bankroll twice, there are four possible outcomes, WW, WL, LW, LL. The ev is the same, since you are betting the same total amount, but the SD is now only .707 of the bankroll. Keep splitting your bankroll up into smaller bets and you keep decreasing the variance while keeping the same ev.
$5 pass, 300 bets, ev = -$21.21, SD = $ 86.59, ev/SD = .245
$10 pass, 150 bets, ev = -$21.21, SD = $122.46, ev/SD = .173
Same bet handle, same ev, more variance with fewer, bigger bets, better chance to come out ahead (or lose more). In the first case, the probability of breaking even or better is about .40, in the second case .43.
This principle can illuminate the issue of making come bets, and how many. For any given amount of money you are going to expose to the HA, you get more variance if you bet it on fewer bets. So, you get more variance by betting $100 pass line than by betting pass, come, come, etc., totaling $100. Of course, you may not want more variance.
I ran a little experiment using WinCraps. The first "Guy" bet $20 passline, no odds, no come bets, for 200-roll sessions. The second "Guy" bet $8 pass and up to two $8 come bets, no odds. I set the bankrolls to $3000, so nobody would bust, and ran 10,000 sessions. The average bet handles were pretty close, $1199 for Guy #1 and $1158 for Guy #2, but the SDs of the session net outcomes were $156 for Guy #1 and $94 for Guy #2. Guy #1 broke even or better 47.5% of the time, Guy #2 44.2%. However, overall Guy #1 lost a bit more, because it's always edge * action, isn't it?
I hope this information helps somebody to decide how much variance to build into his/her strategy.
Exploding the Gambler's Fallacy
On the Wizard of Odds' Website there's a "Gambler's Fallacy FAQ" where there's an exchange between the Wizard and a guy who keeps asserting the Gambler's Fallacy, i.e. that because there is a cumulative-probability calculation that says the probability of n events(x) in a row is (a small) p, that after n-1 events(x) the probability of another (x) is very, very small and the probability of non-X is very high.
The basis for this argument is that the probability of the series of events does not change as the events in the series take place, rather than the probability of each event remaining constant. There's a very easy way to explode this argument. Suppose we flip a fair coin three times. Here are the possible outcomes:
There are only eight possible arrangements of heads and tails for three throws, and they are equally likely. So the probability of each is .125. By definition, impossibility is zero and certainty is one, with all other outcomes somewhere between.
So, before we throw a coin, the probability of HHH is .125. The Fallacy would argue, then, that after two heads have been thrown, the probability of the next flip being a head is just .125, while the probability of a tail must be .875, since that is the only other possibility. IOW, the individual-event probabilities change, while the cumulative probabilities stay the same.
Here, again, are the eight possibilities and their associated probabilities:
Now, we can combine some of these if we ignore the order and just concentrate on the number of each outcome:
3 heads .125
two heads .375
two tails .375
3 tails .125
So, let's step through the series and examine what happens, keeping in mind that the total of the probabilities of possible outcomes must always equal 1.000.
We flip the coin! Result, heads!
Now, one of the three events is now determined; since the first flip was heads, all of the originally-possible outcomes starting with a tail are now eliminated, right? If we started out with a .125 probability of three tails, what is the probability of TTT now? It's zero, isn't it? In fact, all four of the possible outcomes starting with a tail are now impossible, with probabilities of zero, right? Now, since the total of the probabilities of the possible outcomes must always equal 1.0, what happened to the 4 * .125 for those four starting-with-a-tail outcomes? THEY HAVE TO BE RE-DISTRIBUTED among the still-possible outcomes. We are left with:
which are all equally likely, so each has a probability of .250, summing to 1.000.
So, we see very clearly and unambiguously that the probabilities associated with a series of events MUST CHANGE as the events unfold. TTT started out at .125, but it is now 0.000.
This is the end of the argument for the Gambler's Fallacy. It clearly fails. In fact, we only need to consider two events:
First flip is heads. Now TT and TH can no longer happen, right? End of story. Game over. p(TT) and p(TH) are now 0.00 and HH and HT are each .5.
It's really such a simple concept, and yet some quite intelligent people persist in believing that, if the probability of six sevens in a row is just .000021433, then after five sevens in a row the probability of another seven is very, very small. Or, they believe that, if the probability of six dice rolls without a seven is just .3349 (.8333333^6), that after five rolls without a seven, the probability of the next roll being a seven is .6651, not .1667.
The odd thing is that all these calculations of cumulative probability are based on a single probability for the event in question. Duh! So these people recognize a constant probability of a single event, then turn around and deny it. Well, it's anti-intuitive, isn't it?
All other things being equal, the more variance in a bet or system of betting, the less lucky you need to be to overcome the "handicap" of the house advantage. The flip side of that, of course, is that for any given degree of bad luck, i.e. any fraction of or number of standard deviations WORSE than expectation, the more you lose. It's just as likely to come out one standard deviation better than expectation or one worse; due to the house advantage, you lose a little more with -1 SD than you win with +1 SD.
So, is more variance "good"? No one can decide that for anyone except him/herself, because no one else knows what your goal(s) is (are) in playing. If you're well-heeled and willing to risk a lot for a chance to win big, more variance gives you a better chance for that. If you have a very limited bankroll and want to play without much risk of losing a lot, low variance tends to "damp the swings".
Has anyone heard of the "doey-don't"? It is a system that damps out almost all variance on the comeout roll, in order to get past its disadvantage for the Don't Pass and lay odds on points, where the player is more likely to win and the casino pays true odds. This is how it works:
comeout: bet equal amounts on pass and don't pass
these will offset each other exactly, except for the 12
point established: the flat portions will always offset each other after a point is established; the player can then either take or lay odds in whatever multiples he/she wants, up to the maximum the casino allows
I believe the original idea was to lay odds on the DP.
The basic problem with this system, in my view, is that you are subject to the HA on BOTH bets. This is true of any hedge using partially-offsetting bets. I describe the doey-don't here only to illustrate a point about variance.
Let's look at only the flat-bet part of the doey-don't; it is the ultimate in low variance. In fact, you can NEVER win, only lose the pass on the 12. The expected value is about -1.4% of THE SUM of the two bets, and the standard deviation is 0.08 of the sum of both bets. The only variation comes from the random appearance of the 12. Any session of playing this way would have a range from zero (if the twelve never showed) to a loss equal to the bet unit times the number of twelves that showed on the comeout.
Of course, presumably no one would ever play this way; I'm just using it as an illustration of why variance is necessary in order for the player to have any chance to break even or win.
So, the house advantage is like a "cover charge" to buy variance from the casino. High-variance bets have higher HA; you're buying more variance. Look at these figures:
|Field (triple 12)
|Field (double 12)
We see a general trend toward higher HA as SD goes up, although for Big Red and World you pay the highest HA and don't even get that much variance for it. Note, also, that the Field without the triple 12 not only has a higher HA, but lower variance, as well.
However, there is a way to get more variance without a higher HA; in fact, there is a way to get more variance without paying any additional HA at all: the odds bet behind the pass, don't pass, come and don't come. You can't make these bets by themselves, of course; you have to bet at least the table minimum on the flat part and the bet has to go to a point before you can take or lay odds, but the essential point is that the odds bet does not carry any HA itself. Why not?
There are two ways the casino builds the HA into bets:
1) charging a "vig" to make the bet, usually 5% of the bet amount, or the amount to be won, whichever is less.
2) paying less than the true odds of winning the bet
For example, when you place the six or eight, the odds against winning are 6 to 5, because there are six ways to roll a seven, against five to roll a six or eight; so, the payoff should be even money plus 20% (6/5 = 1.2) However, the payoff is 7 to 6, which is even money plus 16.7%; even if you win, you lose, so to speak, because the casino "short-changes" you. That's OK; they have to meet their expenses and make a profit, too.
On the odds bets, however, the casino actually pays you 6 to 5 on the 6/8, 3 to 2 on the 5/9 and 2 to 1 on the 4/10; they don't short you on the payoff.
Any of us who have read books about craps have seen the list of HAs for various multiples of odds taken, like:
pass + single odds: 0.83%
pass + double odds: 0.61%
pass + 5X odds: 0.33%
pass + 10X odds: 0.18%
These are based on taking the expected loss and dividing it by the total of the flat and odds bets. OK, that's one way of looking at it, but I think it's misleading. The fact is that the expected loss is ALWAYS .014 times the FLAT BET. No matter how much you take or lay in odds, the expected loss does not change, because the expected value of the odds bet is always zero.
It's more illuminating to show the expected loss, a money amount or number of units, as:
100 unit pass: 1.4 units
100 unit pass, single odds: 1.4 units
100 unit pass, double odds: 1.4 units
100 unit pass, 5X odds: 1.4 units
100 unit pass, 10X odds: 1.4 units
When you take or lay odds, you add variance, or call it volatility, not expected loss. Since the "handicap" remains the same, the higher variance (SD) means to don't have to be as lucky to overcome the handicap. However, you don't have to be as unlucky to get clobbered, either, since volatility works both ways.
Remember the two million people playing pass for two hours? Suppose those people had taken double odds on every point, still assuming they had enough bankroll to always last the two hours? The graph of all their results would still have the same peak, at about -1 unit, but it would have been a lot more spread out, with a lot wider range between the best and worst outcomes. Also, a lot more people would have come out ahead, about 48% instead of less than 43%, but those who lost would have lost more, because variance is symmetrical absent bankroll considerations.
Obviously, when you take or lay odds, you are risking more money, which means if the dice don't cooperate you will lose money faster and perhaps bust your session bankroll or reach your loss tolerance. Many people advocate progressive betting, where you start out at a minimum amount and increase bets after wins. Of course, the drawback is that when you increase a bet and lose, you throw away some past winnings. Progressions rely on streaks. A couple of members had a conversation about the relative merits of straight progression versus a regress-then-progress system that locks up an initial win. In my view, neither is "better", and it depends on the flow of the dice which comes out ahead in any given session. Consider, however, that progressing place bets always results in higher expected loss, because every dollar is subject to the HA; however, if you progress odds bets instead of place bets the expected loss does not increase, and if you have access to a casino that allows a high odds multiple, you can go quite a way without ever increasing your flat bet. Of course, you don't have as much flexibility, because you can only make the odds bets when there is a point, which happens roughly two-thirds of the time.
Consider this, assuming the casino allows 10X odds:
start at minimum pass bet
ignore comeout results
after a point win, take single odds on the next point
if you lose a point with odds, go back to the start
if you win another point, take double odds on the next point
You can be as aggressive as you like, or you can put in a regression somewhere. If you do reach max odds and win, add one unit to the flat bet, etc.
Adding Stopping Conditions
The symmetry of the bell curve recording the results of two million people playing the pass line for two hours each is the result of all of them playing for the full two hours.
Suppose those two million people started out with only ten times their unit bet? (There was a person posting here who plays with $100 and could only find a $10 table.) If that were the case, something over 400,000 of them would be expected to lose their 10 units before 200 rolls, some of them well before that. So there's going to be a big "spike" at -10 units, cutting off the left side of the graph there, while the right side, to the "good" of the expected loss, is going to look pretty much the same. The casinos are not going to win as much, because the unlucky players who lost 30 or more units in the first experiment could only lose 10 in this one. This means less total money was bet, and the less money bet the less the expected loss. We say this graph has "positive skew", because the right tail is longer than the left one. The other interesting thing is that the players who last the two hours are actually going to have an average win, because none of them lost as many as 10 units. It's sort of like life expectancy: as you grow older, your life expectancy increases, because everyone who has died by that age is no longer part of the "universe".
Now suppose we told those two million people with their 10-unit bankrolls to establish a 10-unit win goal. If they get 10 units ahead, they put 8 of them aside and play with 2, trying to win 5 more, but stopping if they lose the 2. They can keep going as long as they don't lose 20% of what they've won. Now we are going to "bump up against" peaks on both sides of the bell. We're still going to have over 20% who bust, but now we'll have some 12% who end up 8 units ahead, having gotten up 10 and then lost the 2. Others will not lose back the 2 and go on to win more. If you reach a win goal and decide to risk 20% of it, it's like starting play with a very short bankroll. If you lose it, you stop, but, if not, you keep playing until you reach the next win goal, again setting aside most of it and continuing to play with a couple of units. In rare cases, this can go on for several goals.
I ran a simulation of this, using a program I wrote. The first win goal is 10 units, so the bankroll would be 20. Upon reaching 20, set 18 aside, play with 2 and set the next goal at 25 (+15). Upon reaching 25, set 23 aside, play with 2 and set the next goal at 30, i.e. each time try for another half of the original win goal. Each time, it's like starting a new session with just two units and trying to win five units. It's not very likely, but the idea is that you don't want to lose back much of what you've won. If your luck continues, you keep going; else you quit.
In order to reach a win goal starting with a short bankroll, it helps to have a lot of variance. Of course, more variance means you're more likely to lose the short bankroll. I ran a simulation of 20,000 sessions like this:
start with $200
bet $5 pass, taking 3, 4, 5X odds (avg. bet will be about $19)
establish win goal of $100 (bankroll $300)
save $70 and play with $30, trying to win $50 more; repeat for each win goal reached
stop when the FIRST of these occurs:
bankroll is gone
you've lost back the $30 you played with after reaching one or more win goals
you've reached 200 rolls AND no bet is outstanding AND you've just lost a bet
The program keeps track of just about everything.
Here's how it came out:
reached a win goal: 55.4%
time limit: 19.2%
Over 5000 sessions busted, so there's a very large "spike" at -200 (net).
The are smaller spikes at +70 and +75 (2208 between them), because quite a few reached the $100 win goal (or just past it) and then lost back the $30 and quit. Another 1900 or so sessions came out between +$120 and +$150.
The biggest win was $620 in a session where nine successive win goals were reached! Six more "players" reached eight, ten reached seven and 50 reached six win goals.
Overall, there were 11,862 winning sessions, 80 broke even and 8058 lost. The mean net outcome was to lose $2.44, and the standard deviation was $145.
This illustrates the effect of a short bankroll (relative to the size of one's bets), a moderate win goal (half the bankroll) and pretty high volatility (from the 3, 4, 5X odds) on the shape of the graph of possible session outcomes. The method I simulated gives one a better-than-50% chance of increasing one's bankroll by 50%, or a lot more, along with about a 25% chance of losing it. "You pays your money and you takes your choice!" Whether that is desirable is up to each individual player.