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Straight pass line bets (without odds) have a house advantage of about 1.4% - i.e. over the long run, you will win 48.6% of your bets and the house will win 51.4%.

I recommend that first two additional bets you should read about are odds on the pass line bet and place bets on the 6 and 8. Those three bets (pass line, odds and 6/8 place) are three of the best bets on the table. If you are willing to make bets of $20 or more, the fourth good bet is a buy on the 4 or 10. Once you fully understand those four bets, you can ignore all other bets on the table and have a reasonable chance to win (still always less that 50% long term).

Besides understanding the available bets, the next most important thing is managing your money. That's a longer discussion, but lots of books on craps suggest how best to do that.

Quote:Mission146It barely matters and no, in that order. Don't Pass is the better bet by one one-hundreth of one percent in House Edge terms. Short-term results are meaningless.

For those in the "pushes count as action" crowd the difference is five hundredths of one percent. But one often overlooked difference is the fact that most casinos these days allow players to lay greater amounts of odds on the Don't than they're allowed to take on the Do. We all know this is because they allow you to lay what the passline odds would pay.

For example, at a 2x table a $10 Passline bet can take $20 odds on point 4 whereas a Don't Pass bet can lay $40. This means that a Don't player can get more action with a 0% advantage than a Passline player can for a given amount of flat action.

Combined EV on flat plus 2x odds:

Passline 0.61%

Don't Pass 0.46%

So the real difference is about fifteen hundredths of one percent which at these low percentages makes the DP w/2xodds combo about 25 percent better! ( 0.15 / 0.61 )

Steen

Quote:Gamesetmatch4I played craps for the first time yesterday. As a newbie I put 5 on pass line each time and lost like 5 times out of 6. Should I do not pass from now on?

It's up to you. Don't Pass is very slightly better from a purely numerical perspective. The house edge is 0.05% less, meaning if you bet $5 on Pass, it costs you an extra 0.25 cents per resolved bet. Let's estimate there are about 32 resolutions per hour. This means it costs you 8 cents more per hour to play $5 on Pass.

Personally, I think most of the appeal of craps is having fun with everyone on the same "team." It is certainly worth 8 cents an hour to me to join in the fun with the rest of the table. If it isn't fun, why even play?

Quote:Gamesetmatch4Also I noticed most people playing the numbers after come out roll. Are my odds netter doing this instead?

They may have been making odds bets, which have zero house edge. Odds bets just increase your variance. So again, it's up to you. It won't cost you anything in expectation to make them. You'll have more chance of busting out quickly but also more chance of a bigger win overall. Just depends on what is comfortable for you. Don't let anyone, whether it be players or dealers, pressure you into making or not making them.

There are also place bets, hardway bets, and all sorts of other goodies (or maybe baddies). They are all negative expectation, but some of them aren't too punishing. The stuff in the middle of the table is generally high house edge. Placing 6 and 8 are only a small disadvantage. I personally don't do much of that, my style is more conservative, but if you hit the numbers you can take off like a rocket. The problem is hitting the numbers. :)

I'd say check out the Wizard's craps page and the free craps game on his site if you want to practice.

Quote:SteenCombined EV on flat plus 2x odds:

Passline 0.61%

Don't Pass 0.46%

you mean HE, not EV

Quote:MrGoldenSunI'd say check out the Wizard's craps page and the free craps game on his site if you want to practice.

agree, but the best page is the old page, and that is hard to find now

here is a link

https://wizardofodds.com/games/craps/basics/

Yahoo!Quote:Gamesetmatch4I played craps for the first time yesterday.

you were a virginQuote:Gamesetmatch4As a newbie

no big deal.Quote:Gamesetmatch4I put 5 on pass line each time and lost like 5 times out of 6.

happens about 10% of the time or one would expect out of 100 other craps players around

about 10 would be just like you and even one could be worse

of course

most say to win on the pass is only due to how good the shooter is

at least you could have bet on other bets too

sureQuote:Gamesetmatch4Should I do not pass from now on?

the story can easily be the same with similar results

it is just a game

yes, those are the major sucker betsQuote:Gamesetmatch4Also I noticed most people playing the numbers after come out roll.

and for many that play that way

suckers are sweet

odds are better to lose more real cash the longer you do thatQuote:Gamesetmatch4Are my odds netter doing this instead?

but most that do that have more fun losing that way

craps could not be offered if all stopped making those number bets

really

play to lose and have fun!

Quote:odiousgambityou mean HE, not EV

How so?

Quote:odiousgambityou mean HE, not EV

Quote:SteenHow so?

I think it is clear to me now - a surprise - that a lot of people, including knowledgeable people, mix these terms up. A respected member told me in another thread, paraphrasing, "when most people say EV they expect you to know they mean HE". I guess it is true.

I am not trying to flame up this thread, and I think anyone you'd like to quiz can confirm I am no Troll. It bugs me though, so maybe I'll be a one-man crusader on the matter.

So here is the definition, my definition, but also I think everyone's definition before things get sloppy:

EV meaning expected value and HE meaning house edge, of course.

EV = Bet * HE

or

HE = EV/Bet

By these definitions formed by the formula, the EV is typically shown as a monetary value and the HE as a percentage.

Steen, I think it should be pointed out that in fact the EV does not change on an x-amount line bet, darkside or rightside either one, when a player adds his free odds bet. The HE does change, something I have pointed out many times only to be told it doesn't matter, to look at the EV only. This has created many threads here with two sides both quite sure the other is wrong .

I am certainly NOT the math guy around here, and I might have some terms mixed up occasionally. But odiousgambit, I think you must have left something out of those equations. (I would present my own versions, but I would likely screw it up even more.)Quote:odiousgambitEV = Bet * HE

or

HE = EV/Bet

For my $xx wager, I would like to have a very high expected value and a very low house edge, but you have described them as directly proportional. Can that be???

Edit: OK, I will go ahead and expose myself to additional ridicule by presenting my own proposed equations.

EV = Bet * (1 - HE) or HE = 1 - (EV/Bet)

Quote:DocI am certainly NOT the math guy around here, and I might have some terms mixed up occasionally. But odiousgambit, I think you must have left something out of those equations. (I would present my own versions, but I would likely screw it up even more.)

For my $xx wager, I would like to have a very high expected value and a very low house edge, but you have described them as directly proportional. Can that be???

Not sure what you mean. Just fill in the blanks. Say $10 bet.

say the edge is known

EV = 10 * -1.4% = 10 * -0. 014 = -0.14 dollars

say the EV is known

HE = -0.14/10 =- 0.014 =-1.4%

the HE can change with free odds added to the bet, the EV does not. Say $20 in free odds

EV = [10 *- 0.014] + [0 * 20] = -14 cents

HE = -0.014/20 = -0.007 = -0.7% [this is slightly the wrong answer I think because if you factor in how often you are allowed to make the free odds bet; Wizard says -0.00606] https://wizardofodds.com/games/craps/basics/

PS: just saw your edit and will have to ponder that

I did not use negative values like I should, does that explain it? going back and fixing that

Quote:DocMust be a different definition of what expected value you are talking about. I mean the expected monetary value of what the wagered amount will turn into. If I wager $10 on the pass line, either I will lose and have $0 or win and have $20. The EV will be slightly less than $10.

I should have used negative values for EV etc; is that it? If so perhaps we can say the formula works for absolute values

Quote:odiousgambitI should have used negative values for EV etc; is that it?

You should have, but that's not what Doc means. Doc is saying that in his or her mind, the EV is $9.86 on a $10 wager, because your expectation is to lose 14 cents and thus will have $9.86 remaining.

Your definition of EV is correct, though. The EV of a $10 passline bet is -14 cents.

Quote:MrGoldenSunYou should have, but that's not what Doc means. Doc is saying that in his or her mind, the EV is $9.86 on a $10 wager, because your expectation is to lose 14 cents and thus will have $9.86 remaining.

Your definition of EV is correct, though. The EV of a $10 passline bet is -14 cents.

good

gamesetmatch4, if he ever came back, must be ready to shoot himself LOL

Quote:DocI think it's back to definitions: Does "EV" mean the "expected final value of the money left on the table after the bet is resolved" or "expected dollar amount of the loss" or something else?

It's relative to what you had before you placed the bet. It's equal to the sum of the probability of each outcome times the gain/loss of that outcome. In the case of passline, your result will either be -10 or 10. The -10 happens just slightly more often so you end up with a small-magnitude negative number.

Note that saying a bet has zero house edge means the EV is zero, and vice versa. You will not, in expecation, gain or lose in the long-term by making zero-EV bets. Odds bets are an example.

Also note that if you make a lot of bets, the EV of all of them together can be calculated by knowing the house edge. If I plan to play 50 spins of roulette at $15 each with a -5.26% house edge, I can calculate my overall EV as

(50)*(15)*(-0.0526) = -$39.45

I lose $39.45. If I start with $300, I will expect to have 300-39.45 = $260.55 at the end.

Quote:MrGoldenSun... It's equal to the sum of the probability of each outcome times the gain/loss of that outcome.

I'm more accustomed to expected value being something like "the sum of the probability of each outcome times the value of that outcome."

A lost wager no longer has any value and a winning wager becomes worth X times the amount wagered.

I think we're just looking at different expected values.

EV = EV(money placed at risk) or EV = EV(gain/loss)

I use the first definition. I'm willing (for entertainment purposes) to place a wager for which the expected value is a little less than what it costs me. I would not be willing to put my money up for something that had zero expected value.

Quote:odiousgambitI think it is clear to me now - a surprise - that a lot of people, including knowledgeable people, mix these terms up. A respected member told me in another thread, paraphrasing, "when most people say EV they expect you to know they mean HE". I guess it is true.

I am not trying to flame up this thread, and I think anyone you'd like to quiz can confirm I am no Troll. It bugs me though, so maybe I'll be a one-man crusader on the matter.

So here is the definition, my definition, but also I think everyone's definition before things get sloppy:

EV meaning expected value and HE meaning house edge, of course.

EV = Bet * HE

or

HE = EV/Bet

By these definitions formed by the formula, the EV is typically shown as a monetary value and the HE as a percentage.

In fairness, I've seen you post on this subject before so I'm aware of your belief but I wanted you to restate it before I gave you my thoughts.

I respectfully disagree with your contention that a lot of people and I are wrong on this.

Arguments depend on premises, so let's begin with definitions of EV and HE.

I think you'll agree that EV (expected value) is the calculated average outcome of something (like a wager). It's calculated by summing all possible outcomes each weighted by their probability of occurring. Is it possible to express such an average more than one way? Can I say that the EV of a $1 Passline bet is -$0.0141 or -1.41 cents? Of course. Can I also say that it's -1.41% ? Of course I can! However, it's important to indicate which of these answers is being given by writing or stating the symbols used such as $, cents, %, etc.

How can I calculate EV as a percentage? One way would be to use the outcomes of a $1 wager, then multiply the result by 100. Another would be to divide the EV$ by the Bet$ and multiply by 100:

EV% = 100 * EV$/Bet$

Now what about the HE (house edge)? It's the house's (casino's) average expected profit from a wager. And how is that calculated? Well, you start by finding the EV, right? Right! Can it also be expressed different ways? Of course it can! Now I'll agree that it's most often expressed as a percentage, but that in no way prevents it from being expressed as a monetary value. In fact I've seen many people express it as such. Can I say that the HE on a $21 Buy bet is $1 ? Of course I can! But just as with EV, it's important to indicate which answer is being given by using the proper symbols.

You've defined EV and HE as:

HE = EV/Bet

Then you stated that EV was typically expressed in monetary units and HE as a percentage. I'm glad to see you wrote "typically" because it shows that you admit the terms can be expressed different ways. Nevertheless, let's add the symbols and write your formula as:

HE% = EV$/Bet$

Is your formula correct? No. To express HE as a percentage, it should be:

HE% = 100 * EV$/Bet$

Notice however, that this expression is exactly the same as the EV%:

EV% = 100 * EV$/Bet$

So, HE% is in fact EV%, just as HE$ is EV$.

Well, sort of. It depends on another wrinkle in defining EV and HE. "Typically" neither the EV nor the HE is expressed in terms of the house's wager (although we certainly could do so if we wanted). Instead, they're expressed in terms of the player's wager. When we calculate the player's EV we usually find that the results are negative, IOW the player has a disadvantage. But since a disadvantage for the player is an advantage for the house, we can swap terms just by switching signs. So for example, the player's disadvantage of -1.52% on Place6 becomes the house's advantage (edge) of 1.52%.

So really, your correct formula should be:

HE% = 100 * -EV$/Bet$

I'll admit that I sometimes omit the sign when I write the EV if I feel that it's a given. So if you want to ding me for that, I accept. But then ding yourself as well for also forgetting the sign and not multiplying by 100.

HE% = -EV%

HE$ = -EV$

Quote:Steen, I think it should be pointed out that in fact the EV does not change on an x-amount line bet, darkside or rightside either one, when a player adds his free odds bet. The HE does change, something I have pointed out many times only to be told it doesn't matter, to look at the EV only. This has created many threads here with two sides both quite sure the other is wrong .

You seem to be advocating that EV applies only to individual bets whereas HE applies to any combination. This is wrong. EV is an average outcome. A combined EV is an entirely valid calculation for any number of wagers. However, just because you can compute a combined EV doesn't mean that the EV's of the individual wagers have changed. The same is true for HE.

Steen

I'd even argue it's backwards -- the combined expected percentage loss is less interesting than the combined expected loss in dollars. If you make $5 pass + 100x odds, your expected loss is about 7c. The percentage of total wager is miniscule, and it would continue to be miniscule even if you added in a $5 horn high 2 bet. But I wouldn't ever suggest that $5 pass + 100x odds + $5 horn high 2 bet is a better combination than $10 pass with no odds. You need to look at the loss in dollars, not as a percentage of action, to evaluate the cost of gambling.Quote:SteenYou seem to be advocating that EV applies only to individual bets whereas HE applies to any combination. This is wrong. EV is an average outcome. A combined EV is an entirely valid calculation for any number of wagers. However, just because you can compute a combined EV doesn't mean that the EV's of the individual wagers has changed. The same is true for HE.

As to the lexical debate between EV and HE (or HA), I use the terms interchangeably. It's clear from my context what I mean (including the sign), and it's incumbent upon everyone to understand the difference between a percentage and a quantity. When people don't understand that distinction is when they get misled by things like "the house edge on 100x odds is so tiny it's the best bet in Vegas."