Ask the Wizard #212
Can you explain the rules to the "guts" game, as played in the movie Lucky You?
Matthew from Fort Wayne, IN
I hope you're happy; I watched this scene over and over for at least an hour, trying to make sense of the rules. I’ve played guts lots of times, over many years and locations, and have never seen it played as was done in that movie. Let’s call the first player to act Player 1, and the second player to act (the dealer) Player 2. Here is my understanding of how they played.
- Both players ante (or re-ante).
- Each player gets two cards.
- Player 1 must declare “in” or “check.” If he checks, go to rule 4. If he goes in, go to rule 7.
- Player 2 must declare in or check. If he checks, go to rule 5. If he goes in, go to rule 6.
- Although two checks never happened in the movie, I assume both players would start over from step 1.
- The action goes back to player 1, who must declare in or fold. If he goes in go to rule 8. If he folds, go to rule 9.
- Player 2 must declare “in” or “fold.” If he goes in go to rule 8. If he folds, go to rule 9.
- The two hands are compared; and the higher hand wins. The winner collects the pot, and the loser must match it, creating a new pot. This is equivalent to the loser just paying the winner the amount of the pot. Although there was never a tie in the movie, I assume no money would move. Next, go to rule 10.
- When a player folds, the other player collects the pot. Then repeat with a new hand from step 1.
- An additional card is given to each player, to add to his existing 2-card hand, making a 3-card hand. The third card is dealt face down, on top of the face-up two card hand. I do not know whether straights or flushes counted at the 3-card stage. I prefer to play where they do count (but not at the 2-card stage).
- Steps 3 to 9 repeat. If both playes go "in," then go to rule 12.
- An additional two cards are given to each player, to add to his existing 3-card hand, making a 5-card hand. The fourth and fifth cards are dealt face down, on top of the face-up three card hand.
- Steps 3 to 9 repeat. Then start over at step 1.
If you watch the movie carefully, Huck should have lost $11,000 in total, when he had $10,000 to begin with. I watched the scene lots of times to try to find this missing $1,000. My best guess is that when he went in on the last two-card hand, he should have matched the $4,000 pot, but had only $3,000 left. I assume that, much as in regular poker, he could only stand to win what he was risking. In the last hand, Huck folded. I’m not sure if this was because his three-card hand couldn’t beat his father’s two-card hand on the table, or if he was forced to fold, because he didn't have the money to match the pot if he lost.
If my understanding of the rules or analysis of the scene is in error, I welcome correction.
Most of our casinos collect the commission on buy bets on a win only. They only charge $1 for wager of $20–$39. Does this actually make buying the 4 or 10 for $33–$39 a better bet than placing the 6 or 8?
Ron from St. Louis
Let’s find the breakeven point. The expected value of placing the 6 or 8 is [(5/11)*7 + (6/11)*-6]/6 = -(1/11)/6 = -1.52%.
Let b be the buy bet. The expected value is [(1/3)*(2b-1) + (2/3)*-b] / b = (-1/3)/b
Equating the two bets:
-1/66 = (-1/3)/b
3b = 66
b = 22
So, at a bet of $22 the odds are the same. The odds are better on the buy bet for bets of $23 to $39.
Great Site!! If I have pocket Queens, what is the chance that an Ace or King will come by the river? A simple fundamental question, but one that will help me tremendously.
Ed Miller from Banning CA
Thanks. There are 50 cards left in the deck, and 42 of them are not aces or kings. The probability of not seeing any aces or kings in five community cards is combin(42,5)/combin(50,5) = 850,668/2,118,760=40.15%. So, the probability of seeing at least one ace or king is 100% - 40.15% = 59.85%.
I’m a novice counter, using the hi-lo system. Knowing the count, do my odds increase significantly enough to make a side bet of "over 13, under 13" worthwhile?
Jared from Minneapolis
Yes! That side bet is extremely vulnerable to card counters. As long as the minimum is not too low, you should be using another strategy to exploit it, one that treats aces as a low card. Arnold Synder presents such a strategy in The Big Book of Blackjack. Otherwise, if you are using a standard hi-lo count, Synder says to only make the Over bet in very high counts.
In your Ultimate Texas Hold ’Em return table why is a large raise recommended for two-card hands listed in the table that have a negative expected return? For example, suited K/2.
Charlie Masterson from Quincy, MA
According to my two-player Texas Hold ’Em probabilities, the following are the possible outcomes with suited K/2:
My table on Ultimate Texas Hold ’Em shows that the player has the advantage on the Play bet, but a disadvantage on the Ante and Blind bets. In this case, the player is stuck with bad odds on the Ante and Blind. However, his odds are favorable on the Play. So, by making the maximum raise he is getting the most value out of his better than 50% chance of winning. The bad odds on the other two bets bring the overall value under 50%. That value would be even less with a smaller raise.
At the Borgata casino in Atlantic City, I noticed a new Pair Plus pay table. 100-1 Mini Royal, 50-1 Straight Flush, 40-1 Three of a kind, 6-1 Straight, 3-1 flush, 1-1 pair. What is the house edge for this pay table?
Pete Braff from Long Beach
The house edge under that pay table is a comparatively low 1.85%. Kudos to the Borgata, assuming your information is correct.
Based on viewer feedback, the Borgata lowered the win on a three of a kind to 30 to 1 sometime during 2008.
I was killing time in the Harrah’s in New Orleans the other day, and sat at a Let It Ride table for the first time in years. I noticed that the disclaimer stated that the aggregate payout one round was $25,000. This was generally at $75,000 when the game was first introduced. Since this was a $10 minimum table, this meant that in the unlikely event of a Royal Flush, even a minimum bet would not be fully paid out. How can they set it so that a minimum bet can’t be paid off? To me, this is like to a slot machine having a big sign saying "$1,000,000 jackpot" with fine print saying "Payout limited to $100,000". I understand that the aggregate payout amount is pretty much what the traffic will bear, but are there any guidelines that require a certain minimum level for the aggregate limit?
Rick from New Orleans
For the benefit of other readers, in Let it Ride the player starts with three bets, and may pull back two of them if his cards don’t look good. If the minimum were $10, he would start with $30 in bets. If the player has a possible royal flush, proper strategy says to always stay in the game. A royal flush pays 1000 to 1. With a royal flush, the player would win 1000 to 1 on three bets of $10, or a total of $30,000 on bets of $10. However, the maximum aggregate payout is $25,000, so the 1000 to 1 is impossible to achieve, unless the player deviates from proper strategy, and doesn’t raise with hopes of royal.
I completely agree with your point. In my opinion it is false advertising to offer a win that is impossible to get under proper strategy. So, to Harrah’s I say “shame on you.” They can afford to pay a $25,000 jackpot.
Here in Nevada, an aggregate payout rule must be in plain view, and it cannot apply to wins less than 50 to 1 (Nevada Revised Statute 5.190). So, unless there is another statute I don’t know about, this would be legal here too. However, I am not aware of the same kind of impossible jackpot here. The maximum payout is also usually $25,000, but some of the classier casinos have higher maximum payouts. For example, the Wynn is at $75,000. The minimum bet here is usually $5, so as long as you stay at bets of $8 or less, the win for a royal will stay under $25,000. With a $1 side bet, the win would be exactly $25,000, so they would be allowed to deduct any wins of other players against you. My advice is to never bet so much that the aggregate payout rule might apply, on principle alone.