Why do don't pass betters get shortchanged on laying odds versus pass line players?
So if the point is 4, a pass line better with a $10 flat bet can bet $30 in odds while a don't pass better with a $10 flat bet can lay $60. If the point is made, the pass line better will win $70 ($10 + $60) while if there is a 7out, the don't pass better will win only $40 ($10 + $30).
Why aren't don't pass players allowed to lay a sufficient amount that they would win the same as a pass line player? In my example, why don't they allow a don't pass player lay $120 against a $10 flat which would yield a win of $70 ($10 + $60)?
Quote: FleaswatterWhile playing at a 3x 4x 5x odds craps table, the maximum amount of odds that can be layed against the don't pass line bet is 6x the bet. Why is the limit 6x?
So if the point is 4, a pass line better with a $10 flat bet can bet $30 in odds while a don't pass better with a $10 flat bet can lay $60. If the point is made, the pass line better will win $70 ($10 + $60) while if there is a 7out, the don't pass better will win only $40 ($10 + $30).
Why aren't don't pass players allowed to lay a sufficient amount that they would win the same as a pass line player? In my example, why don't they allow a don't pass player lay $120 against a $10 flat which would yield a win of $70 ($10 + $60)?
I suspect it's because the lay odds bets are expected to win more often than lose, the opposite of the pass bets.
On the pass line, the probability a point will be made is .406, times $70 is $28.42. (with 3, 4, 5X odds all the payoffs are $70, including the flat part)
On the DP, here are the breakdowns
4/10 .16666 of points, .6667 probability of winning $40 = $4.44
5/9 .33333 of points, .6 probability of winning $50 = $10
6/8 .41666 of points, .5454 probability of winning $60 = $13.63
They add up to $28.07.
On the right side, all the payouts are the same.
On the "wrong" side, all the lay odds bets are the same, different winnings.
Cheers,
Alan Shank
Quote: FleaswatterWhile playing at a 3x 4x 5x odds craps table, the maximum amount of odds that can be layed against the don't pass line bet is 6x the bet. Why is the limit 6x?
Ever wonder how they came up with the 3x/4x/5x limit? Regardless of what the point is, if you bet the maximum odds with pass, you are putting up 3/4/5x against the casino's 6x (3x @ 2-1, 4x @ 3-2, 5x @ 6-5). Presumably, the casino wants to put the same limit on the don't pass odds bettors - in this case, they're putting up the 6x against the casino's 3x/4x/5x.
Quote: FleaswatterWhile playing at a 3x 4x 5x odds craps table, the maximum amount of odds that can be layed against the don't pass line bet is 6x the bet. Why is the limit 6x?
Because 6x happens to be the inverse odds for all of 3x/4x/5x. If you're at a straight 10x odds table, the lay limits are 20x, 15x, and 12x. The key is to realize the rules are take X odds to win Y, lay Y odds to win X. The odds limits are stated in terms of X. Y is implied based on the true odds payouts.
And a big part is tradition. If a don't bettor could go to a 3/4/5x table and, on a $5 point of 4, lay $60 to win $30, then everyone on the pass side would be yelling about how they can't make a $30 odds bet.
At a 3-4-5 table, pass bettors:
* risk $30 to win $60 on the 4 and 10.
You fade that bet, risking $60 to win $30. So you win the maximum amount the right-way bettor can risk. Makes perfect sense to me.
* risk $40 to win $60 on the 5 and 9.
You put up the $60 and if the point fails, you take the $40.
* risk $50 to win $60 on the 6 and 8.
Again, you are putting up the $60 to win the $50.
That's just how don't odds work.
Quote: ThatDonGuyEver wonder how they came up with the 3x/4x/5x limit?
The tables game manager at Caesars palace, Jimmy Wike, is credited with being the one who "standardized" the 3, 4, 5 odds system because he wanted to make it simpler for the dealers to pay out max odds at the tables. Each bet pays the same.
Quote: sodawaterThink of it this way. When you play the don't odds, you are taking the other side of the pass odds.
At a 3-4-5 table, pass bettors:
* risk $30 to win $60 on the 4 and 10.
You fade that bet, risking $60 to win $30. So you win the maximum amount the right-way bettor can risk. Makes perfect sense to me.
That's just how don't odds work.
+1
