"Craps Check" Question

So I just got back from playing a losing session of craps, where I was mostly playing the pass with odds, and also one "come" bet with odds per shooter. On one particular "come" bet, the shooter rolled three straight 3's and one 12 all back to back, so I lost $20 quick. The dealer yelled over at me, "You better start taking some craps check with those come bets!" What are your thoughts on doing a $1 craps check when betting some "come" bets? Is it a good strategy? Thanks.

Quote: IAchance5

So I just got back from playing a losing session of craps, where I was mostly playing the pass with odds, and also one "come" bet with odds per shooter. On one particular "come" bet, the shooter rolled three straight 3's and one 12 all back to back, so I lost $20 quick. The dealer yelled over at me, "You better start taking some craps check with those come bets!" What are your thoughts on doing a $1 craps check when betting some "come" bets? Is it a good strategy? Thanks.

It's a terrible idea. The craps bet has one of the highest house percentages on the table (11%). A $1 craps bet costs you MORE money than a $5 come bet!

The craps bet WOULD NOT HAVE prevented you from losing the come bet, in any case. It would be a separate bet with a separate outcome.

"thou shall not hedge"

To be accurate however, on a $5 pass line with a $1 craps check, you get the following:

=2*4/36+4*8/36+4*(6/36*3/9+8/36*4/10+10/36*5/11)-6*(6/36*6/9+8/36*6/10+10/36*6/11) = 3.636 percent HA. Of course that gets diluted when you take odds. Taking full odds at a 3 / 4 / 5 times table gets you from .374 percent to .914 percent HA on the full bet when you take the craps check.

Feel free to hedge thy bets but realize that it's a hedge and will overall dilute your action.

IAchance5

From my view-point, you are just making stupid bets with the come bet, but that’s my view-point! Here is why I say that, to start with I know that you probably read that come bets were good bets to make. Like most players starting out or even after playing twenty years, because we read it in a book we think that has to be right!

Quote:

On one particular "come" bet, the shooter rolled three straight 3's and one 12 all back to back, so I lost $20 quick. The dealer yelled over at me, "You better start taking some craps check with those come bets!"

You gave a good example as to why you shouldn’t make come bets, the craps points will kill you on new come out rolls, and the 7 will take down your come bets that you have already established. I know of course you get to keep your odds, when there was a new come out seven after the first point way made! So when there is a hot roll you stand to lose on every new come-out roll, and then you have to reestablish new come bets. Most players will have one or two come bet on the table at the same time you just lost $10 on a come out roll if the shooter made a 7 on the come-out roll if you had two come bets, plus the fact that you can also lose a new come bet to the crap numbers!

You have a bet that has to hit twice to win on and that adds to your dilemma! Why don’t you just place the point you want and you would be a lot better off, that way you can get paid on one hit of the dice, you are not going to lose your bets on the come out rolls, and you can take them down or turn them off at any time!
Here is a good example, you have a $5 come bet and now the shooter makes a 10, your bet goes to the 10 and you take $10 in odds. When the point hits again, that’s if it does you get paid $5 for you come bet and $20 for you odds grand total of $25 “great”

I place the 10 for $15 I got paid 27 if I didn’t have a buy on the point for the first 10 that was rolled, then I get paid $27 on the second point that was rolled, same $15 I made $54 what is wrong with this picture? You make up your own mind; at one time I played come bets because the dealers and the books said that they were the best kind of bets to make one pass-line bet with two come bets.

I am sure glad that I spent the $19 on that first book I bought on craps, it taught me one thing and that is to think for your self, and don’t believe every thing you read about the game of craps.
Now there are some books out that goes against the old way of thinking about the come bet being a good bet, you do the math and make-up your own mind. There are going to be a lot of guys that will tell you the come bet is a great bet to make.

By the way crap bets are just that crap! That is why the dealer is selling you that bet so he can keep his job, by winning money off of you! Any bet that the stickman is selling is a bad bet, just like the guys at the carnivals do they sell you a game that you can’t win on, how much did it cost you to win that $5 teddy bear for your girl friend, what a great job of salesmanship they did on you. You played a game that looks like it was so easy to win on but the truth was you didn’t stand a chance.

Here honey I won you this bear, it only cost me $50 to win it for you, you are the man!

Now all I got to do is sit back and brace my self for all the math guys that are gong to tell you I am wrong!

This is just my opinion, and I may be wrong, so take it for what it’s worth, and try it before you take it into a casino near you!

Lot’s of luck

Superrick

Quote: superrick

IAchance5

From my view-point, you are just making stupid bets with the come bet, but that’s my view-point! Here is why I say that, to start with I know that you probably read that come bets were good bets to make. Like most players starting out or even after playing twenty years, because we read it in a book we think that has to be right!

This sort of thinking is why the casinos make so much money. If everybody only made the best bets available, a crap table couldn't afford to pay its dealers.

The come bet is the third best bet available on the table. It is tied with the Pass Line bet, and VERY slightly behind the Don't Pass and Don't Come bets (tied for second). The best bet of all is the odds bet. The reason why these bets are better than the others available had nothing to do with whether some book says they are or are not--it's because they have the lowest house edge. A player will benefit from making those bets that have the lowest house edge.

And for what it's worth, the vast majority of the books out there that deal with the game of craps offer information that is absolutely correct. Most people don't have the ability to work out the mathematics of the game on their own--the books offer a shortcut. "Thinking for yourself" at the crap table is like not reading the directions for that new band saw you just bought.

Quote: superrick

IAchance5

You gave a good example as to why you shouldn’t make come bets, the craps points will kill you on new come out rolls, and the 7 will take down your come bets that you have already established. I know of course you get to keep your odds, when there was a new come out seven after the first point way made! So when there is a hot roll you stand to lose on every new come-out roll, and then you have to reestablish new come bets. Most players will have one or two come bet on the table at the same time you just lost $10 on a come out roll if the shooter made a 7 on the come-out roll if you had two come bets, plus the fact that you can also lose a new come bet to the crap numbers!

I started out with place bets and still go back to it at times when I simply want to bet less. Pass/Come + odds can get very expensive very quickly.

My reason for sticking with come bets though is I just feel the ramp up time is useful to me. You mention a situation were you actually hit a few 10s. Well, I've done that and there have been times were I never hit a 10 in 15 shooters. While I can still lose a ton on a streak like 4/5/6/8/9/10/7, I've decided I'd rather have this instead of getting point 7s and losing every place bet I just placed.

I'll take my 0.374% disadvantage in Vegas and enjoy the wild swings. :)

wow, here's turning a thread into the old come vs place betting arguments....you'll never convince these place betting dopes so why bother trying.

Despite the arguments, I still love the game, because everyone has a different method of betting. There is a tremendous variety of betters out there, from those who religiously stick to pass lines, to those who place numbers, to those who play the lottery in the center. There are come betters and don't betters. Some people throw the dice in the air attempting to get great control. Others just toss the dice without a care in the world. Everyone will argue that one system is better than the other.

But the math is absolutely right in the matter. Just because you are throwing dice makes everyone think that you can play a system that will beat dice and that the dice have magical memories or properties that will make some numbers appear more than others. What makes us believe this is our anthropological ability to learn to recognize patterns and to falsely believe that the past does affect the future. The fact is that the dice are random and all our betting patterns create are functions of expected value and variance that can be explained through statistics.

Playing the don't is better than playing the pass (marginally). Placing the 6 and 8 is very similar to pass line and odds when you consider the per roll HA. Buying a four or 10 when you pay commission on the win can be just as lucrative as the place 6 and 8s. That's why the game is so great. There are a variety of ways you can win, but the casino has rigged the games so that they take a cut of every single roll of the dice.

So is it mathematically better to just do place bets, rather than play the come bets w/ odds? I thought the come bets had a slightly less HA than the place bets?

Quote: IAchance5

So is it mathematically better to just do place bets, rather than play the come bets w/ odds? I thought the come bets had a slightly less HA than the place bets?

All place bets are worse than the Pass Line. The 6 and 8 are only moderately worse, but they're still about 15% more costly than an equivalent Pass Line bet. Place bets on 4/5/9/10 are horrible.

Some posters here have given the impression that come bets are somehow inherently undesirable. This is untrue; come bets are simply Pass Line bets with the next roll being the "comeout" for that particular bet. A come bet with full odds is a much better way to get your money into action than making a place bet. For instance, a $5 come bet costs you 1.4% HA, or 7 cents. If the bet moves to, say, the number 8, then you can take double odds for $10. You are now betting $15 to win $17. If you made an equivalent place bet, say $18 to win $21, you would be paying a house edge of 24 cents for functionally the same bet. (You get paid a little less on the come+odds combination, but you enjoy the advantageous comeout roll for the initial $5 bet.)

The above becomes even more true if the number the come bet moves to is 4/5/9/10. Then, the difference between come+odds and place bets is much larger.

Try to avoid thinking about "combinations" of bets and outcomes; the only consideration is making the bets that have the lowest house edge. Pass and come bets enable you to do that, both because those bets have a low house advantage, and because they allow you to make odds bets, which have NO house advantage.

It is not mathematically better to just do place bets. Come bets with free odds are identical (mathematically) to Pass line bets (1:1 in the front, and true odds in the back). However, like Pass line bets, they are contracts, so you cannot take them down like place bets. Place bets pay less than true odds for that option.

Quote: mkl654321

All place bets are worse than the Pass Line. The 6 and 8 are only moderately worse, but they're still about 15% more costly than an equivalent Pass Line bet. Place bets on 4/5/9/10 are horrible.

Try to avoid thinking about "combinations" of bets and outcomes; the only consideration is making the bets that have the lowest house edge. Pass and come bets enable you to do that, both because those bets have a low house advantage, and because they allow you to make odds bets, which have NO house advantage.

This seems to me to be a very narrow definition of "better" and "best". The HA should not be the only consideration. Of course, superrick has a different kind of blinders on - he tries to compare the place bets with the pass/come AFTER a point has been established, which is a logical error, of course.

Each bet in craps has various aspects, among them the ev (expected value - in craps an expected LOSS), a degree of variance, flexibility (or lack thereof), who controls the bet, etc. I don't believe any one of these, by itself, should be looked at in isolation.

The expected loss determines how much of a "handicap" the player needs to overcome in order not to lose money. The variance (standard deviation) determines how lucky the player has to be to overcome the "handicap", and how much the player wins with good luck and loses with bad luck. The rightside line bets are contract bets, which means that you cannot take them down after a point is established. You can take the "darkside" line bets down, but you are giving up the advantageous half in doing so. Place bets and odds bets have maximum flexibility, as you can call them off or take them down any time. OTOH, place bets are dealer-controlled, which bothers some people.

In general, the casino makes the player pay for higher variance. For example, most of the inside bets have very high variance compared to line bets, and have a correspondingly high HA. The reason for this is that the casino loses more money when the players are lucky. If there were no variance, then the casinos would always win, but there would be no incentive for players to play.

Let's examine a few bets.

$10 pass line: HA .01414, ev -$.14, SD $9.99, contract bet, player-controlled
$12 place six: HA .01515, ev -$.18, SD $12.95, dealer-controlled, can be called off or down any time
$5 boxcars: HA .1389, ev -$.69, SD $25.47, dealer controlled, one-roll bet
$5 Field (w/triple): HA .0278, ev -$.14, SD $5.71, player controlled, one-roll bet

Let's compare 60 pass bets with 50 place bets, 120 boxcars and Field bets, all adding up to $600 in bets.

$10 pass line, 60 bets: ev -$8.48, SD $77.45, ev/SD .1095, breakeven probability .4564
$12 place six, 50 bets: ev -$9.09, SD $91.54, ev/SD .0993, breakeven probability .4604
$5 boxcars, 120 bets: ev -$83.33, SD $279.03, ev/SD .2986, breakeven probability .3826
$5 Field, 120 bets: ev -$16.67, SD $62.56, ev/SD .2665, breakeven probability .3949

The ev/SD ratio tells you how lucky you have to be, in terms of fractions of a standard deviation better than expectation, to overcome the handicap of the expected loss. In this case, for the same bet handle, the place six gives you a very slightly better chance to break even or better than the pass line, due to the increased variance.

Let's examine how these bets compare with different degrees of luck.

bet	$10 pass	$12 place six	$5 Field	$5 12
-2 SD	-$163	-$192	-$142	-$641
-1 SD	-$86	-$101	-$79	-$362
+.5 SD	+$30	+$37	+$15	+$56
+1 SD	+$69	+$82	+$46	+$196
+1.5 SD	+$108	+$128	+$77	+$335
+2 SD	+$146	+$174	+$108	+$475

The probabilities associated with these standard-deviation outcomes are:

-2 SD or worse, or +2 SD or better: .0228
-1 SD or worse, or +1 SD or better: .1587
+.5 SD or better: .3085
+1.5 SD or better: .0668

As you can see, if you're very lucky that boxcar bet wins you more than the others, but if you are unlucky you get hammered. The bet on the 12 is risky for the casino, because they expect to win 5 units out of 36, but if the player wins one extra one out of 36, the casino loses 26 units. Also, if the player wins the 12 and parlays, the casino can lose 930 units, bang! That's why they charge so much for the bet.

The place six and pass are pretty close, and you can find the "crossover" point where one bet comes out better than the other. The difference in ev is very small, just $.61, and the difference in SD is $14.09. If you divide .61 by 14.09, you get .0433, which tells you that, at .0433 of a standard deviation better than expectation, the place six passes the pass and continues to pull away with better luck. To check:

.0433 * 91.54 = 3.963 - 9.09 = -5.127
.0433 * 77.45 = 3.354 - 8.48 = -5.126

Of course, the way to get more variance without increasing the expected loss is to take or lay odds behind the line bets, which maximizes the ev/SD ratio and the probability of breaking even or better. The standard deviation of a series of bets tells you, also, how unlucky you have to be to bust a given session bankroll. It's not all about HA, folks.
Cheers,
Alan Shank
Woodland, CA

Quote: goatcabin

Let's compare 60 pass bets with 50 place bets, 120 boxcars and Field bets, all adding up to $600 in bets.

$10 pass line, 60 bets: ev -$8.48, SD $77.45, ev/SD .1095, breakeven probability .4564
$12 place six, 50 bets: ev -$9.09, SD $91.54, ev/SD .0993, breakeven probability .4604
$5 boxcars, 120 bets: ev -$83.33, SD $279.03, ev/SD .2986, breakeven probability .3826
$5 Field, 120 bets: ev -$16.67, SD $62.56, ev/SD .2665, breakeven probability .3949

The ev/SD ratio tells you how lucky you have to be, in terms of fractions of a standard deviation better than expectation, to overcome the handicap of the expected loss. In this case, for the same bet handle, the place six gives you a very slightly better chance to break even or better than the pass line, due to the increased variance.

Let's examine how these bets compare with different degrees of luck.

bet	$10 pass	$12 place six	$5 Field	$5 12
-2 SD	-$163	-$192	-$142	-$641
-1 SD	-$86	-$101	-$79	-$362
+.5 SD	+$30	+$37	+$15	+$56
+1 SD	+$69	+$82	+$46	+$196
+1.5 SD	+$108	+$128	+$77	+$335
+2 SD	+$146	+$174	+$108	+$475

"In this case, for the same bet handle, the place six gives you a very slightly better chance to break even or better than the pass line, due to the increased variance."
THERE IT IS! Place 6 has a 46.04% chance of coming out ahead, vs 45.64% chance for the PL bet, for the same bet handle of $600.

If I read this right, I have a slightly better chance of coming out ahead by placing the 6/8, as opposed to making a PL bet with no odds. I assume this has to do with the variance, but I am still missing something I guess. I hope I have not taken this out of context.

So HE on the PL bet, with no odds is 1.414%, while the HE on the Place bet is 1.52%, yet a place 6 has a better chance of winning?

I'm not doubting you Alan, I am asking to make sure I have understood this correctly. My head feels ready to explode....... ;-)

Quote: RaleighCraps

THERE IT IS! Place 6 has a 46.04% chance of coming out ahead, vs 45.64% chance for the PL bet, for the same bet handle of $600.

If I read this right, I have a slightly better chance of coming out ahead by placing the 6/8, as opposed to making a PL bet with no odds. I assume this has to do with the variance, but I am still missing something I guess. I hope I have not taken this out of context.

So HE on the PL bet, with no odds is 1.414%, while the HE on the Place bet is 1.52%, yet a place 6 has a better chance of winning?

I'm not doubting you Alan, I am asking to make sure I have understood this correctly. My head feels ready to explode....... ;-)

By itself, "chance of winning" is meaningless. If you bet on the "draw a number between 1 and 100 game", and if you draw 1 through 99, you win fifty cents, but if you draw 100, you lose all your money, you are arrested, thrown in prison, raped, your house is burned to the ground, and you get the world's biggest wedgie from a guy named "Hammer", you indeed have a 99% chance of winning, but I doubt you would want to play.

Likewise, "coming out ahead" is meaningless unless the AMOUNTS you "come out ahead" are equal to the amounts you "come out behind". In the case of the place bets, the relative proportion of the wins and losses is what matters.

Confusions in terminology such as this are why mathematicians--and casino operators--talk primarily in terms of "expected value".

Quote: RaleighCraps

"In this case, for the same bet handle, the place six gives you a very slightly better chance to break even or better than the pass line, due to the increased variance."
THERE IT IS! Place 6 has a 46.04% chance of coming out ahead, vs 45.64% chance for the PL bet, for the same bet handle of $600.

Not quite; the place six has a slightly better chance of BREAKING EVEN or coming out ahead.

Quote: RaleighCraps

If I read this right, I have a slightly better chance of coming out ahead by placing the 6/8, as opposed to making a PL bet with no odds. I assume this has to do with the variance, but I am still missing something I guess. I hope I have not taken this out of context.

This is because of the higher variance on the place six. The expected loss is higher, because the HA is slightly worse, but the standard deviation is higher, too. This means you don't have to be quite as lucky to overcome the expected loss. These figures are for 60 passline vs. 50 place bets, which result in the same bet handles.

Quote: RaleighCraps

So HE on the PL bet, with no odds is 1.414%, while the HE on the Place bet is 1.52%, yet a place 6 has a better chance of winning?

Yes, but this also means that you will lose more on the place six if your luck is average or worse. Variance works both ways! Look at the numbers for -1 and -2 SD.

Cheers,
Alan Shank
Woodland, CA

Quote: mkl654321

By itself, "chance of winning" is meaningless. If you bet on the "draw a number between 1 and 100 game", and if you draw 1 through 99, you win fifty cents, but if you draw 100, you lose all your money, you are arrested, thrown in prison, raped, your house is burned to the ground, and you get the world's biggest wedgie from a guy named "Hammer", you indeed have a 99% chance of winning, but I doubt you would want to play.

Excuse me, but I did not give the chance of winning "by itself", but rather with a whole set of outcomes associated with different "degrees of luck" expressed as deviations from expectation.

Quote: mkl654321

Likewise, "coming out ahead" is meaningless unless the AMOUNTS you "come out ahead" are equal to the amounts you "come out behind". In the case of the place bets, the relative proportion of the wins and losses is what matters.

Here again, I did give specific amounts, and, of course, you always lose more for the same variance from the ev, than you win, by the amount of the expected loss. The numbers for all those bets are symmetrical around their respective ev's.

Quote: mkl654321

Confusions in terminology such as this are why mathematicians--and casino operators--talk primarily in terms of "expected value".

My terminology is not confused. Mathematicians do not ignore variance, and neither do casino operators. Once you start piling up odds multiples, variances becomes the dominant factor.
Cheers,
Alan Shank
Woodland, CA

Thanks mkl and goat. I think I have a better idea now of what these numbers really mean. Darn, I thought I had found something for nuttin'.

Quote: mkl654321

Confusions in terminology such as this are why mathematicians--and casino operators--talk primarily in terms of "expected value".

I think it's worth while to examine the interplay of expected value and variance (as expressed by the standard deviation, the square root of variance). The expected value of a bet (let's call it expected loss, since it's always negative in craps) increases linearly with the number of bets. So, if the ev of a $10 pass bet is -.1414, the ev of 100 of them is -$14.14, 1000 -$141.41, etc. etc. The standard deviation, however, increases with the square root of the number of bets, so that the ev, which begins much smaller in absolute magnitude, gradually catches up. At some number of bets, the ev will equal (or nearly so) the standard deviation. At that point, the probability of a player breaking even or better is about 16%. After that, the ev continues to "pull away" from the SD, so it becomes less and less likely for a player to break even or better.

A couple of examples:

$10 pass: ev -$.1414, SD $9.999
We want to solve for the number of bets where the ev will equal the SD in magnitude.

ev * x = SD * x^.5

.1414 * x = 9.999 * x^.5 divide both sides by the square root of x
.1414 * x^.5 = 9.999 divide both sides by .1414
x^.5 = 70.714 square both sides
x = 5000.51

So, on the 5001st bet, the ev will exceed the standard deviation is absolute magnitude, and the probability of a player breaking even or better will be about .16. At this point the ev is -$707, as is the SD. This means that the "symmetrical player" to the one that breaks even (-1 SD, rather than +1) is going to lose $1414. When the ev gets to be twice the SD, the probability of a player breaking even or better will be a little more than .02.

Let's try $10 pass with 3, 4, 5X odds.
ev -.1414, SD 49.153

This time, the same calculation yields a figure of 120,837 bets, at which point the player still has a .16 probability of being even or better. However, the ev for that many bets is -$17,086, so the other side of that coin is that a player has a .16 probability to have lost at least $34,173 over that same number of bets.

It takes a long, long time for the ev to become "dominant" when you take or lay high odds multiples, but never forget that variance magnifies both good and bad luck in terms of money outcomes.
Cheers,
Alan Shank
Woodland, CA