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The next table is for the Deuces Wild game in level 3. The wins in the column for "this level" have been pre-multiplied by 4, for the 4x multiplier in level 3. The "future levels" column shows the value of advancing to level 4. The "total value" shows the combined value, current and future. The lower right cell shows a return of 8.00 times the bet per level. So, if you can get to level 3, the rest of the game (levels 3 and 4 combined) is worth 8 times the bet per level.
The next table is for the All American game in level 2. The wins in the column for "this level" have been pre-multiplied by 2, for the 2x multiplier in level 2. The "future levels" column shows the value of advancing to level 3. The "total value" shows the combined value, current and future. The lower right cell shows a return of 5.63 times the bet per level. So, if you can get to level 2, the rest of the game (levels 2-4 combined) are worth 5.63 times the bet per level.
The final table is for the Jacks or Better game in level 1. The “future levels” column shows the value of advancing to level 2. The “total win” shows the combined value, current and future. The lower right cell shows a return of 3.60 times the bet per level.
So, this game is worth 3.603417 units, assuming optimal strategy. However, you have to bet 4 coins to play, making the return 90.1%.
This issue has bothered me for many years. In 1999, my father took me to Vegas for my 21st birthday. We were playing blackjack at the same table, I with roughly $25 in bets on the table, my father with about $40. The dealer had 20, but miscalculated and thought she busted. She paid us as if we won. Roughly 15 minutes later, three suits came down, put their hand on our shoulders, essentially appraised us of the situation, and mandated that we pay back the "winnings" or leave the casino. We decided to leave the casino, and gamble elsewhere that evening. Is that standard operating procedure or is this more the exception to the rule? — Mike from Buffalo Grove, IL
In my opinion the two most sacrosanct things in gambling are no cheating, and honoring a bet. No expiration dates, no excuses, a gentleman honors his gambling debts. You didn't say how many points you had. The right thing to do would be to return the winnings only if you had a 20, or the winnings plus the original wager if you had less than 20. If they were rude in the way they asked, I wouldn’t blame you for leaving, but I still would have paid. I’ve been asked this before, so I think that this is the standard operating procedure.
I know the commandment to not make side bets. However, I have seen a side bet in blackjack that pays 11 to 1, if the player has a pair in his first two cards. Would it be possible using a count system to gain an advantage? — Brian from Las Vegas
It sounds like you are talking about Lucky Pairs, a side bet that wins if the player’s first two cards are a pair. Many baccarat tables also offer this bet. As I show in my baccarat page, the house edge is 10.36%, assuming eight decks. In either game, you would pretty much need to eliminate all cards of at least one rank to have an advantage. To know that, you would need to keep 13 different counts. In baccarat, this could be done, since you are allowed to take notes while you play. However, based on some very extensive analysis, profitable opportunities don't happen often enough for this to be a practical use of time.
Hello, Wizard. I read your Texas Hold 'Em questions, and I noticed you calculated a 59.85% chance of seeing an ace or king on the board, while holding pocket Queens. How did you come up with that figure? — Jacob from Atwater, CA
There are combin(50,5)=2,118,760 combinations of five cards out of the remaining 50 in the deck. 42 of those cards are 2-Q. The number of combinations of 5 cards out of 42 is combin(42,5)=850,668. So, the probability of not getting a king or ace is 850,668/2,118,760 = 40.15%. Thus, the probability of getting at least one ace or king is 1-40.15% = 59.85%.
An alternative calculation is 1 - pr(first card in flop is not ace or king) × pr(second card in flop is not ace or king) × pr(third card in flop is not ace or king) × pr(fourth card in flop is not ace or king) × pr(fifth card in flop is not ace or king) = 1 - (42/50) × (41/49) × (40/48) × (39/47) × (38/46) = 59.85%.
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Ask the Wizard! (No. 222)