Craps Question for the Wizard and other craps players

I am very conservative and only play 2 games routlett(EUROPEAN--red-black or odd even) and CRAPS. But I do play a lot and often. My Craps question is:

I play "wrong" and it seems to me that taking odds when playing wrong is a disadvantage. I think(but please check me out on this) that the probablity of winning once the point is set is 60% if playing wrong.(I suppose this assumes a normal distribution of 4 5 6 8 9 10 and 7. Wouldn't taking odds merely combine a 50% chance of winning with a 60% chance therefore watering down the 60%?

I have taken odds a few times when a 4 or 10 comes up but only if I am close to my loss limit. When I do that I increase the probablity to 75% by taking insurance on the 4 or 7.

Please check me out on this because it's my own idea and I have never seen it in any of the the books on craps. I have done well playing this way.
Zippedup.

Quote: zippedup

I have done well playing this way.

Well, if its enjoyable for you as well as being generally profitable for you, I would suggest continuing doing it.

Now, as to some other matters that you raised:

When you are on the Don'ts, you are not taking the odds, you are laying the odds.

Whenever you "insure" any bet, you are diluting your potential win. Any sort of hedging is an expense and if it turns out that you would have won without the hedge, all you've done is lower the amount that you won for some false sense of protection against that loss. So I would not try to protect any bets. You've taken a chance... find out what will happen.

Hi Flea
Thanks for your thoughts. I realize that over time the insurance bet is a loser over the long haul and don't use it very often. It pays 8/1 and a few times I get real nervous about the amount I have bet like 500. Day in day out I never use it either.
zippedup

Quote: FleaStiff

When you are on the Don'ts, you are not taking the odds, you are laying the odds.

I concur withFleastiff on all this, plus you have to figure taking [laying] free odds is never detrimental. You can look at these as separate bets, not combined bets.

I am an inveterate Don't bettor too, as my handle implies.

The way to think about laying odds is to consider the TOTAL amount bet (flat bet + odds). For example, $35 on a $5 table with 3/4/5 odds ($5 flat + $30 odds). The house edge is lower if you bet the flat bet plus odds than if you bet the entire $35 on Don't Pass. When you lay odds you're only exposing $5 to the come-out roll, which is more likely to lose (eight ways) than win (three ways) on a single roll. Once you get past this hurdle then the $30 odds bet is break-even, and the $5 flat bet has a player advantage of roughly 3:2 (point numbers overall have about a 40% chance of winning).

Oultine---I don't need information on the craps odds on the combination of rolls one and two. I know them by heart.

The question remains Once the point is established when playing wrong the probablity of a winning 7 is 60% over the long haul. Is that 60% diluted if you start LAYING odds which yield 50% over the long haul. The intial bet that you made is history and has little impact on post point bet I would think.

Someone made an excellent point about its the single number that comes up rather than the average of all the numbers that counts. In other words in the paragraph above the 50 vs 60 is correct but only if you LAYED odds on every number that came up all the time.

I guess where I am coming out is that if you have a 4/10 with a 67% probability of winning you should bet it. In addition if you want to take the 67% to 75% by reverse betting double twos or or double 5's you can but it's costly but can help protect you if you are investing a large amount of money on your number

Wizard Can you help us out here???

My point (pun intended) is that the flat bet already "won" the come-out roll. The 60% chance of the point not passing is contingent upon establishing a point in the first place. If you are confident you are more likely to get past the come-out roll than chance dictates, you should bet your entire amount as a flat bet and lay no odds.

Odds yield 0%, not 50%. Or put another way, 100% of the original wager.

What's a similar scenario... in blackjack, you always split 8s and aces, but you have to be dealt such a hand in the first place.

Laying odds does not dilute the advantage your don't pass bet has earned after surviving the come-out roll. In terms of $ of expected value, your +EV is exactly the same with or without odds.

In fact, when considering your action at the table, you are increasing the house edge by not laying odds. That is, if you want to get a certain amount of action, you should be betting less on the don't pass and always laying maximum odds. Your expected loss in $ will be less than if you bet the full amount on the don't pass.

On a side note, dice rolls do not have a normal distribution...

Thanks 7 outline and dwheat

I have been betting exactly the way you indicate 7outline.

Dwheat has the exact opposite thinking

Where is the wizard when we need him lol.

I'll ask him to help us out in a separate post. I appreciate your taking the time to respond
zippeup

D Wheatly--You get the gold star and win the prize!!

I found a book at Borders book store yesterday that almost quoted your statement.
Said as follows:
If betting wrong---minimize your bet on the come out roll but bet enough so you don't have to bet above the amount of odds the casino is providing on the point rolls. In other words set a win goal first then figure your come out roll bet based on how much in odds the casino is willing to provide and how much you are willing to risk.

Then bet fully up to the extent of the odds provided for 10/4 especially.

Thanks all!! Guess I will change my process. Will be interesting to see if we get a response from the Wizard and what he says.
Sorry for all the posts and thanks for everyones thoughts
ippedup

Quote: zippedup

My Craps question is:

I play "wrong" and it seems to me that taking odds when playing wrong is a disadvantage. I think(but please check me out on this) that the probablity of winning once the point is set is 60% if playing wrong.(I suppose this assumes a normal distribution of 4 5 6 8 9 10 and 7. Wouldn't taking odds merely combine a 50% chance of winning with a 60% chance therefore watering down the 60%?
Zippedup.

I suppose the betting is up to you. Say you are a flat 5x table and you are betting $5 on the don't pass. 2/3rds of the come out roll a point is established. On average, you will win 59.39% of the bets. When you came out the odds were 13.889% against you, now the odds are 18.79% for you.

If you take the $25 in odds and distribute it anywhere else on the table, your return on that $25 (minimum house advantage = 1.52% for the 6 / 8) would be less than then odds on the DP (with a HA of 0%). All you are doing is insuring your don't pass and getting action on the other numbers.

Quote: zippedup

... taking odds when playing wrong is a disadvantage[?]...

I really thought this was a settled question. Free Odds are free of house advantage, they do not confer advantage or disadvantage under any circumstance, other than the variance that comes from wagering more.

Quote: zippedup

... taking odds when playing wrong is a disadvantage[?]...

It only seems like a disadvantage, because, the wrong flat bets pay even money, while the wrong free odds pay less than even money.

On the surface, yeah, it seems like a disadvantage.

It's only when you dig below the surface that you realize that the even money flat bet pays less than true odds that the disadvantage evaporates.

I thought about this too at one point. It does not dilute at all. If you are at a 60% advantage and do an odds bet, you are still have a 60% advantage. Think of any true odds payments as diluting for future come outs to lessen the house advantage. It just seems like diluting the 60% because you have an advantage at that point, and don't want to bet more to win less for another bet. But think of that other bet as a separate entity. It's a free odds bet period, and you are allowed it because you exposed your come out bet when in risk. The true odds bet will help offset for the future.

Quote: playpianoking

I thought about this too at one point. It does not dilute at all. If you are at a 60% advantage and do an odds bet, you are still have a 60% advantage. Think of any true odds payments as diluting for future come outs to lessen the house advantage. It just seems like diluting the 60% because you have an advantage at that point, and don't want to bet more to win less for another bet. But think of that other bet as a separate entity. It's a free odds bet period, and you are allowed it because you exposed your come out bet when in risk. The true odds bet will help offset for the future.

Yes, I'm quoting myself. I thought about this more recently. Maybe you have a point. If the pass line odds is supposed to dilute the 1.41% down to .37% with 3x-4x-5x odds, would the same not be true on the don't pass? Ok, so what is the come out on the don't pass side, 1.29% or something? - so I guess you would also say max odds dilutes that down to be better for the player, but after the point is established odds immediately change. Here is an example: Think in poker about AK vs 66 preflop. Essentially a race at that point, but if the flop comes K 8 2 rainbow, the odds immediately shift strongly in favor of AK.

So while pass line bettors shift favor back to the house after a point, they dilute it down to their favor by being given a free odds bet. But don't pass betters after the fact are actually sitting better than a free 50/50 bet.

I'm beginning to confuse myself...

Quote: playpianoking

I'm beginning to confuse myself...

Have you seen the wealth of craps info at Wizard of Odds?

Be sure to check out the appendices at the bottom of that page. Hopefully the details on the calculation of the HA for don't pass will shed some light.

The HA for don't pass without odds is 1.364%. The house advantage is defined as the expected value divided by the expected value bet.

When you add odds, the HA decreases because the payout is fair. You get paid 5:6 on 6 and 8, 2:3 on 5 and 9 and 1:2 on 4 and 10. The EXPECTED value does not change.

So on a $5 table, you can bet $30 to win 15, 20, or 25 based on the point.

The house advantage is now .273% but the expected loss is still 1.364% on the original bet. But because your average bet is $25 instead of $5 because of the odds, you divide the house advantage by 5.

All odds do is increase volatility on both sides. For right side betters betting odds, you will win alot more money when points hit and lose less when points don't. For wrong betters betting odds, you will lose alot more money when points hit and win less when points don't.

Let's take an example where you have an 11-sided die and 5 of the results are 8 and 6 of the results are a 7. One player plays $5 with no odds and the other player plays $5 with $500 odds. There are 100 trials where the point is 8.

In the first case, the expected value for the first player is 100 x ((5/11) * 5 - (6/11) * 5) = -$45.454 for a house advantage of 9.09% and an expected value of -$.4545

In the second case, the expected value is 100 x ((5/11) * 605 - (6/11) * 505) = -$45.454 for a house advantage of .0909% (since the bet is now $505) and an expected value of -.4545.