Come bets are a 3% hedge against the 7?

3/4/5x odds, $10min

Passline/Continuous come + max odds.
I view the Come bet as a Hedge against the 7. That's why I do Continuous Come.

if I have every # (4,5,6,8,9,10) then I have $300 on the table.
$10 on the Come when a 7 rolls.
10/300 = 3.33%

so can a Come bet be viewed as a 3% hedge against the 7?
or does losing to 2,3 and 12 neutralize it?
(but the 2,3,12 doesn't kill your other #s)

I don't know.... but they are different bets and if you happened to make them at different tables you wouldn't really think of it as a hedge. It is a way to give you "action" instead of waiting around for the next time the stickman croons something about "coming out". It has a different marker but gets settled by the same "7". Just eats thru the bankroll faster than sticking around with only ONE passline bet awaiting its determination.

I like the Come bet, but for reasons other than as a hedge. The only time it's a hedge is on first roll after point. I don't take odds on my PL, so if 7 hits they cancel out. I bet only 1 Come bet. Your method is mistaken because you are ignoring that even if you win your $10 on the Come when 7 hits, you're losing the $300 you have down. You know this, I'm sure, & you're only interested in the 3% figure, which may be true. But, it's immaterial.

Does not really matter what you do at craps table, yes, you can use come bets as hedge against 7out, but how are you going to win by hedging ? Negative EV says it all.

Quote: Dicenor33

Negative EV says it all.

So where ever is there going to be positive EV??? Its ALL negative.

Cont'd . Some men prefer blondes, others brunettes, but once you get married, your Negative EV shoots through the roof.

That's a very expensive "hedge." Win $10 but lose $300??? Really???

Quote: AlanMendelson

That's a very expensive "hedge." Win $10 but lose $300??? Really???

Yeah I agree with Alan here. Why would you ever want to hedge 3% against? (Even if it was 3% whcih it cannot be because you haven't figured in the losers on the come)

If you want to hedge you have to reasonably consider getting back at least 50% of your outlay. The only way to do that is by having $37.5 on the big red and at 16.67% you aint doing youself any favors ;)

I would consider yours a strategy not a hedge.

I fear many dealers learn about craps in the same manner as you mention players do.

Oh, ... by the way, a Come Bet is indeed a great bet to make, though you are perfectly free to post some reason for it being a bet made only by the unwashed, uncivil hoi poloi who learn craps without reference to a book which is probably the manner in which ninety percent of craps players in history learned the game!

Quote: 100xOdds

3/4/5x odds, $10min

Passline/Continuous come + max odds.
I view the Come bet as a Hedge against the 7. That's why I do Continuous Come.

if I have every # (4,5,6,8,9,10) then I have $300 on the table.
$10 on the Come when a 7 rolls.
10/300 = 3.33%

so can a Come bet be viewed as a 3% hedge against the 7?
or does losing to 2,3 and 12 neutralize it?
(but the 2,3,12 doesn't kill your other #s)

You have a pass bet on a point, and you make a come bet in like amount.
If you roll a seven the bet results cancel out. This is sort of like the infamous "doey-don't", but only if you roll that seven.
If you establish a point on the come bet, then you have twice as much at risk. This increase in variance is much greater than the decrease in variance when you have point, come, seven.
Keep in mind that come bet and pass bets are really the same thing and are independent. Hedging is to reduce variance; you minimize variance by betting just the pass line, which is a far cry from what you are describing. The reason to do "continuous come" is to maximize your action, so that if you hit a number, it's off & back on. Of course, with max odds, that is a Heeeeeeeeeeeeeeeeeeeeeel of a lot of variance.
Cheers,
Alan Shank

Quote: goatcabin

You have a pass bet on a point, and you make a come bet in like amount.
If you roll a seven the bet results cancel out. This is sort of like the infamous "doey-don't", but only if you roll that seven.
If you establish a point on the come bet, then you have twice as much at risk. This increase in variance is much greater than the decrease in variance when you have point, come, seven.
Keep in mind that come bet and pass bets are really the same thing and are independent. Hedging is to reduce variance; you minimize variance by betting just the pass line, which is a far cry from what you are describing. The reason to do "continuous come" is to maximize your action, so that if you hit a number, it's off & back on. Of course, with max odds, that is a Heeeeeeeeeeeeeeeeeeeeeel of a lot of variance.
Cheers,
Alan Shank

Alan, I would appreciate it if you could explain to me what "variance" means. I understand variance in terms of whether or not a particular roll progression is close to the pyramid (low variance) or away from it (high variance). IOW, a low variance table means a cold table, because the seven is showing close to the pyramid. Vise-versa if other numbers are showing more than the seven.

I've seen others use "variance" in different contexts, so it must not mean just what I understand. Could you explain to me what you mean by it in the context you used it? Please keep in mind I'm not a mathematician, so if you could speak in general terms, I'd appreciate that, also.

Quote: Sonny44

Alan, I would appreciate it if you could explain to me what "variance" means. I understand variance in terms of whether or not a particular roll progression is close to the pyramid (low variance) or away from it (high variance). IOW, a low variance table means a cold table, because the seven is showing close to the pyramid. Vise-versa if other numbers are showing more than the seven.

Variance can go either way,"hot" or "cold". Low variance would not necessarily mean "cold"; it just means that no numbers are occurring with unusual frequency. Table watchers might say the table is "choppy". High variance could apply to the either more or fewer sevens, or any other number. BTW, I really hate these terms, because they imply that there is some reason to expect an observed (from past results) condition to continue. You will hear people say, "Don't bet the pass at a cold table," whereas past rolls have no effect whatsoever on future rolls. Don't expect a trend to continue, or to change, for that matter.

Quote: Sonny44

I've seen others use "variance" in different contexts, so it must not mean just what I understand. Could you explain to me what you mean by it in the context you used it? Please keep in mind I'm not a mathematician, so if you could speak in general terms, I'd appreciate that, also.

Strictly speaking, variance is the sum of the weighted squared differences from the mean, and the standard deviation is the square root of the variance.
For a bet in craps, here is how it is calculated:
Take a bet on the 12, paid off at 30-to-1, but expected to win only once in 35 trials.
1 win pays 30
35 losses lose 35
net loss of 5 units per 36 bets = -.1389 This is the "expected value" of the bet (the mean result).
To figure the variance:
result - mean difference squared weighted
-1 - (-.1389) -.8611 .74149 x 35 25.95
+30 - (-.1389) 31.1389 908.35 x 1 908.35
---------
934.30 / 36 = 25.9528
This is the variance. The standard deviation is the square root of the variance, or 5.094.
So, the standard deviation of the bet on the 12 is 5.094 times the amount of the bet.
OTOH, the SD of a passline bet is almost exactly the amount of the bet, because it's paid at even money and the probabilities of winning and losing are very close.

These are measures of "dispersal" around a mean.

Take these sets of numbers:
6 6 6 6 7
1 9 3 9 15
They both have the same mean, 6.2, but obviously the second has more variability, so its variance and standard deviation will be higher.

The significance of this for a craps player is that bets or betting systems with higher variance are expected to experience wider swings in their outcomes. If you bet one unit on the pass line, you can either win one unit or lose one. Unless you get really unlucky, your bankroll will not change a great deal with constant bet amounts. Your net change will be the same as your win-loss record. OTOH, if you are betting the 12, your bankroll can change quite quickly. Same thing with taking or laying odds on pass or don't pass bets.

I often use the term "variance" in a general way to mean "variability" or "volatility", rather than in its strict definition.
The poster here is talking about making a lot of bets and taking odds on them. Suppose he is standing next to a bettor who is just making a passline bet and not taking odds. They are experiencing the same sequence of rolls. While the one bettor is going +1 -1 -1 +1, etc., etc., the other could be winning or losing large amounts of money.

Hedging reduces variance, because bets offsetting each other reduce the magnitude of the change in bankroll. If you have a passline bet on a point and you make a come bet in the same amount, if the seven rolls you have a net outcome of zero.

Hope this helps, probably too detailed.
Cheers,
Alan Shank

Yes, it does help, Alan. I now understand "variance" in craps (perhaps in other things, also) is used to measure different things. It can be used in the sense I've understood it, but it can also mean what to expect in making various bets. I do have (thanks to Wikipedia) some idea of SD and variance. I just thought it applied only to the pyramid's working out on the table.

BTW, I've noticed scientists & mathematicians frequently square amounts or take the square root of them. Photography, once my hobby, gets into these things, esp. working w/ light. Shows how proportional nature is & how they have figured that out. Another BTW: I'm assuming taking the sq. root of something is a way of reducing a larger number to a smaller, more workable number.

I bet you if Craps were to have just been invented today they would not have named it the Come Bet.



ZCore13

There is a reason variance and variation are pretty close in meaning. If you know how often each number should be rolled on average, yet see a distribution that varies from it, call it variation, call it variance.

Variance in statistical math has come to have a particular meaning perhaps, is all.

To take the Sq root of it and call it standard deviation, well it turns out you can do something with that, make a bell curve and get some meaning.

Makes sense to me now, but it wasn't so long ago that I pondered it and got irritated with the circular explanations that variance was standard deviation squared; so what was standard deviation? the sq. root of variance. I'm not kidding.

To do the math can be challenging for someone like myself, but a basic sense of what we are talking about is actually not so hard.

What would you call it, Zcore13? It's a second, third, etc., Come-Out. It gives a player the opportunity to have more than one PL bet w/ odds. Kind of like the Place bet, but w/ lower HA & more money when the numbers hit. Sometimes a player might want to let the dice pick the numbers, instead of doing it, himself. It's all in what a player wants. There's advantages/disadvantages w/ both bets.

Quote: Sonny44

Yes, it does help, Alan. I now understand "variance" in craps (perhaps in other things, also) is used to measure different things. It can be used in the sense I've understood it, but it can also mean what to expect in making various bets. I do have (thanks to Wikipedia) some idea of SD and variance. I just thought it applied only to the pyramid's working out on the table.

To put it more simply, variance refers to the range of possible outcomes. Pass bet: win one unit or lose one unit. So after 60 bets, the possible range is from -60 to +60. If you take odds, the possible range of outcomes increases.

Quote: Sonny44

BTW, I've noticed scientists & mathematicians frequently square amounts or take the square root of them. Photography, once my hobby, gets into these things, esp. working w/ light. Shows how proportional nature is & how they have figured that out. Another BTW: I'm assuming taking the sq. root of something is a way of reducing a larger number to a smaller, more workable number.

Here's why they square and then take square root:

The differences from the mean can be positive or negative, but we are interested in the MAGNITUDE of the differences, i.e. the absolute value. So, when you square a negative number you get a positive one. Then, you take the square root to get back to the original magnitude.
Cheers,
Alan Shank

I think I understand what you might mean by hedging. If you place $10 on the Pass then you'll either win or lose depending on whether the shooter does. However if you place $5 on the Pass then $5 on Come you

(i) Lose out if the initial roll was a 7 or 11 (-$5: you only win $5 rather than $10).
(ii) Gain if the initial roll was a Craps (+$5: you only lose $5 rather than $10).

(a) break even on an immediate 7-out (+$10: this sounds like the main reason for doing it this way)
(b) lose out if the second throw is Craps (-$5) or (c) win if the second throw is Eleven (+$5) (Craps is twice as frequent as 11)
(d) lose out if the second throw is the first point (as you now have to get the point again to win both bets).
(e) land up waiting on two points before the 7 (i.e. less likely to win both, more likely to come out even, but might lose both).

The nice thing is breaking even on an immediate 7-out, but this is at a cost that you've less chance of winning both bets. So it's a valid method to reduce the risk (or in mathematical terms decrease the variance).

If you were playing a fair coin-tossing game (assume you always call Heads) the one-bet method would either
(i) H: win +$10 (50%) or
(ii) T: lose -$10 (50%).
The two-bet method would
(i) HH: win both +$10 (25%)
(ii) HT or TH: break-even (50%)
(iii) TT: lose both -$10 (25%)
Thus with one-bet you always win or lose $10, whereas using two-bets you only win or lose $10 half the time.

The ev and the SD are the primary determinants of how a bet in craps works out.
As I mentioned above, the SD for a passline bet is just about equal to the amount of the bet. If you take single odds, the ev is the same, but the SD is higher.

$5 pass bet: ev -$.0707, SD $4.995, average bet $5
$5 pass bet, single odds: ev -$.0707, SD $9.79, average bet $8.56
The more odds you take, the higher the SD and average bet, but the ev is always the same.

Those SD numbers are not too helpful, though. If you make one bet, you are either going to win $5 or lose $5. With odds, it's more complicated, but there are still few possible outcomes.

However, suppose you are going to the table to play for a couple of hours. Let's say 60 passline decisions. What can this math tell you about the possible results? A lot.

Since the ev is the same for every bet, the ev for 60 bets is just 60 x -.0707 = -$4.24. So the mean result is expected to be losing four or five dollars. But the SD does not increase like that, but with the square root of the number of bets. This is because (or maybe why) the more bets you make, the more likely you are to be somewhere near the mean. You can get a feel for this by examining the possible results of one, two and three $5 passline bets:

1 bet
+5
-5

2 bets
+10 WW
0 WL
0 LW
-10 LL

3 bets
+15 WWW
5 WWL
5 WLW
5 LWW
-5 LLW
-5 LWL
-5 WLL
-15 LL
You can see that the more extreme outcomes are becoming less probable.
Anyway, the SD increases with the square root of the number of bets, so more slowly than the ev.

So, take a look at these figures, for 60 resolved bets:

$5 pass: ev -$4.24, SD $38.73
w/single: ev -$4.24, SD $75.84
w/double: ev same, SD $110.67
w/3, 4, 5x: ev same, SD $190.37
w/5 x: ev same, SD $228.15
w/10x: ev same, SD $418.63
w/100x: ev same, SD $3903.72 !!!!!

If a very large number of people play 60 $5 pass bets and we plot their outcomes, we are going to see a bell-shaped graph, centered on -$4 and spreading out from there in both directions, the height of the graph getting smaller and smaller. In the extremely unlikely event that someone won or lost all 60 bets, the end points would be at +- $300. In a distribution like this, about 68% of the outcomes will fall between one standard deviation to either side of the mean (ev). About 95% of the outcomes will fall between two standard deviations. So, we expect to find 68% of the outcomes between about -$43 and +$34. IOW, a range of two standard deviations (one positive, one negative) on either side of the ev. Note that we expect some of these people to come out even or better. Falling one standard deviation above and below the mean are equally likely, so we expect roughly the same number of people to lose $43 as to win $34. So, how lucky do you have to be to break even or better in this scenario? If you divide the ev by the SD, you get .109, so you have to experience "positive variance" of about 1/10 of an SD to break even. In "The Mathematics of Games and Gambling", Edward Packel has a chart that shows the probabilities of samples showing a particular percentage or multiple of a standard deviation. The probability for 1/10 of an SD is about .46

So, here are the +/- one SD figures:

-1 SD +1 SD
no odds -43 +34
single -80 +72
double -115 +106
3,4,5 -195 +186
5x -232 +224
10x -423 +415
100x -3908 +3900

As you put more and more money on the odds bets, the ev is dwarfed by the variance, so that it's almost a coin flip. If you take 3,4,5x odds, the ev/SD is just over .02, with a corresponding probability of just about .49.

BTW, all of these figures assume that all the players complete all 60 bets, i.e. no one goes broke.

So, the higher the variance, the less lucky the player has to be to break even or better. So, is high variance "good"? If you're lucky, you win more with high variance; if you're unlucky, you lose more, so bankroll considerations are going to come into play.

So, can you just put in a loss limit, and cut off the left tail, leaving the rest of the graph the same. NO, but that's for another day.
Cheers,
Alan Shank