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Ask the Wizard #271

Ever since Black Friday (when three major poker sites were shut down by the U.S. government) the forums are loaded with people claiming to having regularly made six-figure incomes for many years. All of a sudden these people have got the whole country asking themselves, "Why not me?" There have to be SOME losers.


The newspapers have plenty of stories of professional online poker players lamenting a lack of a way to make a living too. Indeed, you would think everybody is making money from online poker, both players and operators. However, there have to be losers to pay for it all, but I have yet to hear anyone confess to losing.

So, let me be the first. I've played plenty of online poker, usually $1-$2 to $4-$8 structured games, and I don't need to keep track to know I'm down. I don't even know if I'm strong enough to beat the rake. In my opinion, many online poker sites have become infested with bots and pros, aided with player tracking software, making it tough for the recreational players, like me, to have a chance.

If the U.S. government ever does legalize online poker, and I strongly feel it should, I hope a solid regulatory agency will oversee it and make sure it is only human beings playing on a level playing field.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

In blackjack, is it better to face a dealer two or dealer seven?


A seven. The following table shows the expected value in an infinite-deck game according to the up card and whether the dealer hits or stands on a soft 17. The values for a ten and ace are after the dealer peeks for blackjack, and confirms he doesn't have one.

You can see the player can expect to win 14.40% of his wager against a seven but 9.07% or 9.10% against a two.

Expected Value by Dealer Up Card

Up Card Stand Soft 17 Hit Soft 17
2 9.07% 9.10%
3 12.35% 12.38%
4 15.88% 15.85%
5 19.67% 19.65%
6 23.69% 23.40%
7 14.40% 14.40%
8 5.82% 5.82%
9 -4.06% -4.06%
10 -17.36% -17.36%
A -36.92% -33.78%

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Imagine a slot machine bonus with a field of icons. Some are prizes and some are "Party Poopers," for lack of another term. The player keeps all prizes he finds until he accumulates so many Party Poopers. Is there a formula for the number of prizes the player can expect to get?

"Anonymous" .


p = Number of Party Poopers.
w = Number of wins.
e = Party Poopers needed to end bonus.

Imagine the w wins in a row, like a loaf of bread. Then place the p Party Poopers like raisins equaly along the bread, creating the same distance between consecutive raisins as well as from each end. For example, if the bread was 12" long, and you had 3 raisins, you would place the raisins at the 3", 6", and 9" points, creating 4 segments of 3" each.

The expected number of wins picked is the product of the length of each segment and the number of Party Poopers needed to end the bonus, or e × (w/(p+1)). Let's look at an example.

Suppose there is a field of 40 icons with 8 Party Poopers. It takes three Party Poopers to end the bonus. That would leave 32 wins. So, p = 8, w = 32, and e = 3. The expected number of successful picks is 3 × (32/(8+1)) = 32/3 = 10.67.

The Choctaw Casino in Durant Oklahoma has recently introduced a craps table using cards. They use 48 cards, ace to six, from eight decks. The shooter calls a number between one and six. Then the dealer counts off that number and the next two cards are used to establish the roll. The cards are shuffled after each "roll." How do these rules affect the odds?


The odds change a little bit compared to craps played with dice, due to the effect of card removal. Regardless what the first card is, there is less than a 1/6 chance that the next card will be the same value. As an example of the effect, the house edge on the pass bet is 1.41% with dice, but 1.34% in this game. I indicate the house edge for all the bets in a new table at the end of my page on Card Craps, for various numbers of decks, including eight.

There is a blackjack game online with the following rules:
  • 8 decks.
  • Cards shuffled after every hand.
  • Blackjack pays 2 to 1.
  • Dealer hits on soft 17.
  • No doubling down.
  • Split pairs once only.
  • No surrender.
Between your house edge calculator and effect of rule variations I would be able to figure out the house edge, except you don't indicate the effect of no doubling. Can you help fill in the missing piece of the puzzle?


Using my blackjack house edge calculator, I get a house edge of 0.82%, before factoring the 2-1 on blackjacks and no doubling. 2-1 on blackjacks is worth 2.26% to the player. No doubling is worth 1.37% to the dealer.

So I show the player edge is 0.82% -2.26% + 1.37% = 0.07%.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.