place the hard ways The Orleans

Just got back from the Orleans and the craps tables have a hard way place bet that pays 5 to 1. It is on the table in front of the 4 6 8 and 10. Dealer said it only pays when number comes up hard 2/2 3/3 4/4 or 5/5 and you don't lose it when the number comes soft 1/3 etc.... It must be played for table min. ($5 this table). It was a full table and no one played it. Is it a sucker bet or something I should look into?

Well, there's only one way to win (on any of the hardways) and six ways to lose (on a seven-out), so the odds are 6-to-1 and the bet pays 5-to-1.

There are actually 4 ways to win.

I'm not any more confused than usual, I think. So, I'm ok admitting I'd have to consult a math maven. However, the following has become clear during a long career of gaming. Anything new that shows up in a game, sports book, or other casino enterprise is intended to increase the house edge. Why else innovate? I am absolutely willing to eat these words, however. Good Luck!

As I said I saw NOBODY playing it and when I go back for Super Bowl I wouldn't be surprised that it was gone. Plus it won't stop me from playing those $1 hard ways they are such fun to hit

Something has to be missing here. Either you DO lose if any of the hard ways come up soft as well as a 7 (which gives the house something like a 9% advantage on the bet) OR you can only bet one of the hardways like the usual hardway bet only losing on the 7.

If the bet was you win on any hardways (4 different ways to win) and only lose on a 7 (6 ways) the true odds are 3 to 2 yet they are paying 5 to 1? This would be an enormous player advantage.

There are 4 ways to win if you bet on all of the hardways. There is only one way to win if you bet on only one of the hard numbers. Either way, the odds are still 6-to-1; the odds of each number coming out before a seven. The odds don't apply in the aggregate. It would be the same as a place bet or bets on any number. Let's say you make a place bet on the six; the true odds are 6-to-5 on that bet. If you bet both the six and the eight, the true odds are still 6-to-5 on each bet. Making more than one bet does not change the true odds for each bet.

Thats where I am confused jestes. The original post implies a box where you place a bet and if ANY come up hardway you win and if 7 you lose. Thats what I am trying to clarify. If you still bet each number separate, the way you do on a standard hardways layout on on craps table with the only differences being only a 7 loses and the odds are changed then it makes perfect sense and still a lousy bet, the way most propositions on the layout are.

There is a bet, apparently, on some layouts where you do that, you place a single bet in a box and if the 4,6, 8 or 10 comes up hard you win, if the 7 or ANY soft hits you lose. That has been discussed on other forums and as I stated the odds are slightly over 9 percent in the houses favor.

Just trying to figure out exactly what the layout and bet is.

Sorry the box is in front of each number 4 6 8 and 10 and a separate bet must be made on each. If the number comes soft you don't lose the bet it just stays out there. It only loses on a 7. I believe that it is a house advantage I was just wondering how much. I think the play is to add $1 and place the 6/8 as you normally would and once it hits throw the extra $1 you win on the hard number in hopes of getting the bonus bucks as they call them.

That was the way that I interpreted your original post, that each hardway would be a separate bet and that you would have to make four bets of $5 each to cover all of the hardways, for a total of $20. Again, there is only one way to make each of the hard numbers and six ways to make a seven, so the odds are 6-to-1 and the bet pays 5-to-1, so the difference is the house advantage.

Marcr: to my knowledge, there is only one bet on a craps table where one bet covers multiple numbers and that is the field bet. I guess you could consider the pass/come and don't pass/don't come as covering multiple numbers, since they are affected by the 2, 3, 7, 11, & 12 but every other bet on the table covers only one number. Before someone says, "what about the horn bet?" that is actually four, separate bets (on the 2, 3, 11 & 12), which is why it must be bet in multiples of four ($4, $8, $12, etc). A "world" bet is simply a horn bet plus an "any seven" which is why it must be bet in multiples of 5 (one unit on each of the numbers).