Use of matchplay coupons with blackjack

[FLASH1296]

This computation is wrong as you must compute what happens to BOTH bets - as they are placed simultaneously.

a $12 investment that fails when a 7 is rolled but is only 1/2 won when either a 6 or an 8 is rolled.

When either the 6 or the 8 is rolled [but not the other] the profit is ONE dollar (+$1).

When a 7 is rolled the loss is TWELVE dollars (-$12).

If both a 6 and an 8 are rolled - the profit is FOURTEEN dollars (+$14).

The shooter must avoid two sevens while rolling BOTH an 8 and a 6.
Rolling one of them results in a small profit, but a roll of 7 results in a total loss.
Rolling both is unlikely but provides a substantial profit.

 

[Mr. T]

Several of the area casinos where I live offer daily matchplay coupons to their customers. I visit these casinos on a daily basis and use these coupons at the blackjack table. A friend has told me that I should vary from basic strategy when using these coupons, mainly by standing on a few totals that I would normally hit (say 16 vs 10). His logic is that it becomes correct to risk busting less often when any win will pay you 2 to 1 (with the coupon in play). Is there a mathematical basis to his argument? If so, can someone suggest a modified basic strategy chart to be used when playing matchplays. Thanks.

16 vs 10 is a close call whichever way you play. Best play would be to Surrender if you are allowed to do so.

 

[ChefJJ]

This computation is wrong as you must compute what happens to BOTH bets - as they are placed simultaneously.

a $12 investment that fails when a 7 is rolled but is only 1/2 won when either a 6 or an 8 is rolled.

When either the 6 or the 8 is rolled [but not the other] the profit is ONE dollar (+$1).

When a 7 is rolled the loss is TWELVE dollars (-$12).

If both a 6 and an 8 are rolled - the profit is FOURTEEN dollars (+$14).

The shooter must avoid two sevens while rolling BOTH an 8 and a 6.
Rolling one of them results in a small profit, but a roll of 7 results in a total loss.
Rolling both is unlikely but provides a substantial profit.

I pretty much disagree here. Each roll of the dice is a separate probability calculation, therefore you cannot roll both the 6 and 8, and you are not trying to avoid two rolls of the 7. Impossible to happen in one roll. The numbers that matter to a player betting the Place 6 & 8 are: 6, 7, & 8. Only one of those can be rolled at one time.

Going back to my house edge calculation, your comments are addressed in it. Assuming $6 wagers on both the Place 6 and the Place 8, the win amount if either 6 or 8 is rolled is $7. That is represented in the calc. However, if the 7 is rolled, the whole $12 (both wagers) is lost. That is also represented in the calc.

The number of ways to win either of the bets is 10 (5 each for the 6 and 8), returning $7 to the player. The number of ways to lose the two bets is 6 (Big Red). Therefore, there are 16 combinations of the dice that will result in a win or loss. The other 20 are pushes as far as this bettor is concerned.

The assumption that a player must avoid 2 rolls of the 7 to win both bets is solely based on the philosophy that each roll of the dice is NOT a separate challenge of probability. Remember that whole thing of independent trials...it is primo in the game of craps.

Looking at craps wagers in a holistic manner like this is important. Just because there are many bets on the layout does not mean they are not connected when made by a single player. We could discuss this until the cows come home, especially for those contract bets that have two stages like the line and come bets...but I would be steering this thing WAY off topic.

So, for now, take a look back through the calculation with the above facts in mind. It does make sense. It may be counterintuitive to much that you have thought in the past, maybe? But it works.

good luck

 

[FLASH1296]

So ... let me see if I understand the case that you are trying to make.

You invest $12 and you declare it over and ready for a computation when

(frequently) 1/2 of your money invested remains at risk ?

Quelle Bizarre !

You are pretending that the situation is resolved when it is not.

Your figure of around 1% is obviously wrong.

 

[ChefJJ]

Each roll of the dice is a separate wager or set of wagers, as in the case of Placing 6 & 8. Nobody is making you keep either or both of those Place bets on the table from roll to roll. The Place bets are not contract wagers even though they are not "one-roll" bets.

It's the same formula and logic that you would use for making one, two, three, four, five, or six Place bets...or a Place bet with the Field bet...or a Place bet with a Lay bet...so on and so on.

So go back to your question:

You invest $12 and you declare it over and ready for a computation when

(frequently) 1/2 of your money invested remains at risk ?

And take into account that all of your money on the table is "at risk" for any roll of the dice. It doesn't matter whether a 6 or 8 came up on the previous roll and you collected or not. Why is the situation, as you say, "not resolved"?

Place bets either win, lose, or push on each and every roll of the dice. Sounds like a resolution to me. Look at your following options for each of the 11 number outcomes in the game of craps:

2, 3, 4, 5, 9, 10, 11, 12 - PUSH. Your choice to keep the Place 6 & 8 up, or take down one or both.

6 - WIN PLACE 6 BET: PAYS $7 for each $6 wagered. Your choice to keep the Place 6 & 8 up, or take down one or both.

7 - LOSE PLACE 6 & 8 BETS. Your choice to make another bet or not.

8 - WIN PLACE 8 BET: PAYS $7 for each $6 wagered. Your choice to keep the Place 6 & 8 up, or take down one or both.


One can see that for any number but the 7 rolled, it is up to the player to keep the bet(s) up or take down for the next roll. When the 7 is rolled, the choice is to make the bet(s) again or not. A choice must be made for each roll, even though most players keep the Place bet(s) active for each roll of the dice except the come-out roll, even though that is an option.

The bottom line for wagers in the game of craps that are not contract bets (i.e. Pass Line and Come bets that have traveled to a point number) is that each roll of the dice is akin to a new bet.

good luck

BTW - I have seen my rationale "disproven" if you will, but I still stand by it for better or worse.

 

[FLASH1296]

So ... If I place both the 6 and the 8 if I do not throw an immediate 7 (whether or not I roll a 6 or an 8) I can take the remaining place bet down as i see fit.

BUT I think that I have found your error.

When you place the 8 or the 6 you bet $6 to win $7.

The actual odds are 6 to 5 and you are paid $7 to $6.

You failed to plug in the proper wager of $6 to win $7.

Getting paid 7 to 6 instead of 6 to 5 is where the house earns its vig'

In 11 rolls of the dice you win 5 times and you lose 6 times.

When you win, you win $6 and when you lose you lose $5.

So 6 times you lose $6 for - 6 x 6 = $36. And 5 times you win $7 for + $35

You lose $1 on 11 x 6 = $66 for a net loss of 1.5% or precisely 1.51515152%

NOW ... in the scenario you described, lets see what happens:

You bet $12, A split place bet on the 6 and the 8.

If you roll a 7 you lose $12 That occurs with the six 7's.

So ... 16.7% of the time you are -$12

If you roll a 6 OR an 8 you are +$7.

So ... 27.8% of the time you win $6 and pull down the other place bet.

The other rolls do not result in a win or a loss so they are moot.

SO ...

in 100 rolls 27.8 times you win $6 27.8 x +6 = + $166.8

In 100 rolls 16.7 times you lose $12 16.7 x -12 = - $200.1

In 100 rolls 44.5 rolls are resolved and you LOSE $33.3

You total investment was 44.5 x $12 = $534

A loss of $33.3 on $534 equates to 6.2 % [or precisely 6.235955056179775 %]

Where am I wrong in the above ?

 

[ChefJJ]

You are wrong in that a winning $6 Place 6 or 8 bet is paid $6, or even that a $5 wager pays $6 (that's the Odds bet payout).

It is quite true that the casino's vig on Place bets is due to the discrepancy in probable outcomes for the Place bet's number vs. the 7's, and the mismatched payout. That's what makes the Odds bet a no-edge wager, but I digress.

Let's not get everything jumbled by the options that a player has after the roll happens (i.e. take down or leave up). This is a calculation for an impending roll, and I probably threw a curve ball by discussing that earlier.

Run through your calcs again with the aforementioned proper $7 payout for each $6 wagered. I read through your line-by-line breakdown twice and can't figure out where you are going, but we can probably rack that up to my day's worth of plumbing at the house :flame: But can't the math be any simpler?

good luck

 

[miplet]

post made with corrections in red

in 100 rolls 27.8 times you win $7 27.8 x +7 = + $194.6

In 100 rolls 16.7 times you lose $12 16.7 x -12 = - $200.1

In 100 rolls 44.5 rolls are resolved and you LOSE 5.50

You total investment was 44.5 x $12 = $534

A loss of $5.50 on $534 equates to 1.03 % [or precisely -0.0102996254%]

The only number I changed waas the win of $6 with a win of $7. I didn't check for any other errors.

 

[ChefJJ]

Miplet, way to weed through it all. Simple enough!

OK...let's put this one to rest. It was frying my brain :flame: and either way, Placing the 6 and/or 8 is a good bet in the game of craps.

good luck

 

[miplet]

http://wizardofodds.com/askthewizard/craps-otherbets.html
5th question down explains it.

 

[FLASH1296]

From: http://wizardofodds.com/askthewizard/craps-otherbets.html

What are your thoughts on this strategy and what would the true odds be, if you did take the bets down after one hit? - Michael Andrews

This is similar to a question I got last week. Yes, it is true that there are 10 ways to roll a 6 or 8 and 6 ways to roll a 7. However one must not look at the probabilities alone but weight them against the payoffs. The place bet on the 6 and 8 pays 7 to 6 odds when fair odds would pay 6 to 5. By making 6 unit place bets on the 6 and 8 and taking the other down if one wins the probability of winning 7 units is 62.5% and the probability of losing 12 units is 37.5%. If the player must cover both the 6 and 8 then the place bet is the way to go. This rate of return isn't bad but could be better. For the player who puts a priority on minimizing the overall house edge, the best strategy is to make combinations of pass, don't pass, come, and don't come bets, and always take the maximum allowable odds. June 18, 2000

 

[ChefJJ]

Nobody was talking about taking the other bet down after a win, FLASH...except for you. Miplet re-did your math, and the answer is approx. 1% for the edge on Place 6 & 8 wager.

Are we on the same page or is this one going to continue off course? :joker:

good luck

 

[FLASH1296]

So ... You guys are right and Michael Shakleford and the published experts on the math of gambling are wrong ?

A place bet on the 6 or 8 will has a House Edge of 1.5%
Worse than a pass line bet with a House Edge of 1.4%

You guys came up with wrong figures by simply betting on both the 6 and the 8
as if that could reduce the House Advantage. Obviously it cannot.

This should not become a pissing match.

We are here to learn as well as to share our limited expertise.

I would like a certified 'mathboy' to solve this debate.

 

[ChefJJ]

FLASH, it's not a pissing match. And I'm not really too concerned as to who is right or wrong.

You quoted Shackleford, but it didn't shed any light on our question.

Furthermore, you broke down the house edge calculation since you didn't care for mine, but it was slightly incorrect in that you had a $6 Place 6 or 8 bet paying $6 instead of $7. Miplet inserted the correct payout and we've got the same conclusion.

Why are we now diverting the subject away from the house edge of a combo Place 6 & 8 to a scenario that is looking beyond one roll...the impending roll?

good luck

 

[FLASH1296]

I just re-read the thread.

You said: " ... you had a $6 Place 6 or 8 bet paying $6 instead of $7"

Pulleeze show me where I stated that.

I do not see it.

 

[ChefJJ]

SO ...

in 100 rolls 27.8 times you win $6 27.8 x +6 = + $166.8

In 100 rolls 16.7 times you lose $12 16.7 x -12 = - $200.1

In 100 rolls 44.5 rolls are resolved and you LOSE $33.3

You total investment was 44.5 x $12 = $534

A loss of $33.3 on $534 equates to 6.2 % [or precisely 6.235955056179775 %]

Where am I wrong in the above ?

I made it orange and big above. Look at miplet's post at 8:52 yesterday to see how he corrected this error and determined that we are on the same page, for better or worse.

 

[ChefJJ]

FLASH, are you avoiding me :joker: :grin: ?

 

[FLASH1296]

I see your correction; but it was merely a typo that I made.
As you can see, the computations are unchanged.

 

[KenSmith]

If you pull down the other bet once you win $7, then it is true that you will on average lose 1.03% of the $12 you are risking.

However, you can't make two 1.52% house edge bets and magically reduce the vig to 1.03%. If you leave both bets up until both bets are decided, then not surprisingly, the house edge is still 1.52% on the $12 at risk.

Let's look at what happens. There are three possible outcomes:
If you hit a 7 before either 6 or 8, you lose $12 (p = 6/16).
If you hit a 6, then a 7, you win $1. (p = 5/16 * 6/11)
If you hit an 8, then a 7, you win $1. (p = 5/16 * 6/11)
If you hit both a 6 and an 8, you win $14. (p = 10/16 * 5/11)

Add it all up and you get an average loss of $0.18, which is of course 1.52% of the $12 risked.

Nobody was talking about taking the other bet down after a win, FLASH...except for you.

If you want that 1.03% to be correct, you'll need to pull down the other half your bet once you win.

Once you decide that you might not leave a bet up until it is decided, all sorts of strange things can happen. For example, if we just agree to place the 6 only for $6 and only leave it up for one roll, should we really gain any solace in the knowledge that we on average will lose only 0.463% of the amount risked? Hey, sounds like the best bet at the table when we look at it like that!

I'll stick with knowing that place bets on the 6 or 8 will cost me 1.52% of the amount risked, no matter what combination I play them in. To look at what happens if I pull the bets down before they are decided doesn't really tell me much.

 

[lucifer]

why are you idiots debating the house edge on place bets. its been known for 100 years. it hasnt changed. its a negative ev game. pass on it. period. Only in vegas can they advertise "give me 100 and ill give you 98 dollars" and people flock. "what a deal" flash sounds like a real rookie in the gaming biz. you should read a few more books before you pretend to be a counter or expert.