When a crap table states that the minimum bet is $25, is the player allowed to split his bet -- for example, $10 on the pass/come line and $15 on the don’t pass/come?
[Editor's Note: Andrew Uyal, ex-crap dealer and advantage player, current floor supervisor, and author of our upcoming book Blackjack Insiders, answers this question.]
This is a fairly common question and misconception about table limits, especially those games with multiple bets like craps or roulette.
If the minimum bet on the crap table is $25, that means $25 per bet, at least on the main game (I’ll get into that in a second.) If you want to play both the pass and don’t pass lines, it will cost you $50. The minimum will be the same for all the place bets, come bets, lay bets, etc.
The proposition bets, which consist of everything in the center of the table, have lower minimums, typically $5 on a $25 game. On a $10 or $15 game, the props will typically cost just $1 each.
The way to remember which bets cost more is as follows. Everything in front of the stick man, in the center of the table, is considered a prop bet and will have a lower minimum. Everything in front of the base dealer (the dealers on either side of the table) must meet the posted minimum requirement.
I’d like to add that playing both sides as stated in your question isn't advisable. It’s a similar situation to playing both red and black on roulette.
Take the come-out roll. The bets will hedge themselves on 2, 3, 7, and 11. But on a 12, you’ll lose the pass Line bet and push the don’t.
Roulette is a little more obvious. Betting both colors will result in a push or losing on the zero/double zero. Covering the zero also doesn’t help because something has to lose.
The house has the edge on both bets, so every time you bet both ways, the vig is grinding you down down down.
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Kevin Lewis
Nov-01-2018
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Pat Higgins
Nov-01-2018
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That Don Guy
Nov-01-2018
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Nov-01-2018
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Kevin Lewis
Nov-01-2018
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