rudeboy99 - 06 March 2012 02:58 AM
No. Okay, for example, the components of a $6 place bet are in reality $1 “flat” which pays even money and a $5 portion which pays true odds of $6 to $5 for a payoff total of $7. Another example. Say you have a $10 place on the 4. The “flat” is $2 and the other $8 pays off true odds of 2 to 1 for $16 + $2 “flat” for a total of $18. That’s how place bets were configured when they were invented and became an integral part of the game.
That’s how Place bets were configured when they were invented, huh? I don’t think I’ve ever heard that, but I guess it’s at least possible that it’s true. Interesting that the supposed “flat” portion is always equal to the number of units.
rudeboy99 - 06 March 2012 02:58 AM
When I broke in, many moons ago, we were trained to handle place bets in this fashion.
IMHO, that seems like a really inefficient way to get dealers to learn payouts. $20 Place bet on the 5… ok, that’s four units, so $4 flat… $20 - $4 = $16… $16 should pay $24… $24 + $4 = $28. “$28, same bet sir?... What was the call again?” I prefer the method of figuring out the number of units, multiplying that number by two, then adding it to the bet.
rudeboy99 - 06 March 2012 02:58 AM
I’m sure you’ve seen dealers offset a portion of the place 6 & 8. That’s the “flat” part of the wager and all the dealer needs to do is multiply the"flat” times 7 to figure out the payoff. Additionally, if the player makes a full press when hitting a bet, the dealer simply hands off the “flat” to the customer and incorporates the entire payoff atop the remainder to come up with place bet that’s exactly double of the previous wager. Sabe?
Again, I believe it’s much simpler to add the cap to the bet to calculate the payout for the 6/8. But most importantly, neither calculating the payoffs nor doing full press moves work with improper caps such as $54 and $102 Place bets on the 6/8.