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Ask the Wizard #46

How many different ways are there of rolling 3 ones using 6 dice?

Jamie from Croydon, England

First there are combin(6,3)=20 ways you can choose three dice out of 6 for the three ones. Then each of the other three can be any of five numbers. So, the total ways are 20×53=2500. The total number of ways to throw all the dice are 66=46,656, so the probability of rolling exactly three ones is 2500/46656=0.0536. For help with the combin function see my probabilities in poker section.

At the Turning Stone Casino in NY, they offer $1 10/7 and 25-cent 9/6 jacks or better video poker. Being a low roller, I'm not prepared to bet $5 per hand. Am I better off playing 1 coin in the dollar game or 5 coins in the quarter game?

Stuart S. from Lake Katrine, U.S.

I assume by "10/7" you mean double bonus. As my video poker cheat sheet shows, the return for that game, with five coins bet, is 100.17%. However, if $5 is too rich for your blood, then you can use my video poker analyzer to get the return for a one-coin bet. Just put in 250 per coin bet for a royal flush. The calculator defaults to 4,000, so change it to 1,250. Press "analyze" and you'll see the return is 99.11%.

So, you're much better off playing five quarters in 9-6 Jacks or Better.

How would I modify the strategy you give for jacks or better video poker to a jacks or better game with the following payout schedule:
J’s or better:1
There is a bank of these at Harrah's in East Chicago Indiana, on a progressive jackpot. Any info would be appreciated.

Bruce from Mahomet, Illinois

You can get close to optimal strategy for just about any video poker game by using my video poker strategy maker.

Have you heard of the "Reverse Labouchere" method, described in detail in Norman Leigh's book "Thirteen Against the Bank"?

Greg from Ottawa, Canada

No, and I don't really care what it is either. All betting systems are equally worthless.

Do you have a strategy for the California "No bust" Blackjack, where the players can be the bank. It is played at several Casinos including Larry Flynt's Hustler Casino in Gardena CA.

Chris from Hermosa Beach, California

Since you wrote I analyzed one version of California no-bust blackjack. However, that set of rules has since disappeared. It seems every casino in the L.A. area has their own rules and it is hard to keep up. With the fee to play, it is a lousy bet anyway, unless you're banking. Sorry I can't be of more help.

When reading literature about the formula casinos use to determine comps, the only formula I ever see used as an example is a formula for blackjack. Assuming the casino determines your average bet by your spread, what formula is typically used by casinos to determine expected craps losses, which in turn, determines available comps.

Tim from San Antonio, Texas

I asked my friend Larry Drummond, a craps dealer and former webmaster of Next Shooter for help on this question. Larry can be a bit abrasive but is a good source of hard to find information on craps. Here is what he said, "Comps for craps vary from Casino to Casino and from Boxman to Boxman. A player should get to know the Boxman. The Boxman sets the players average bet and tracks the TIME that a player is at the table. It is easier for the Boxman to track action for COMPS. if the player is consistent in their wagering pattern. Now, I ask you ... if a player goes $52 or $54 across after a Point is established with a $5 flat bet on the Pass Line. Is that a $57 or $59 average? ... Or a $5 average with a whole bunch of other INDIVIDUAL bets? The answer is... IT DEPENDS ON HOW WELL YOU KNOW THE BOXMAN and HOW MANY TIMES TO YOU ATTEND THIS PARTICULAR CASINO."

Larry added in another e-mail the following, "In addition to the information I already sent to you ... ODDS on Pass Line and Come bets are often NOT included in the AVERAGE for comps. Same with LAYING ODDS on the Don’t Side... as in the long run this should be a WASH. But ... If a smart boxman wants someone who is spending big bucks on ANY 7, the worst bet on the table... he would probably average the ODDS and the LAYS to keep the sucker coming back to the casino ... you can re-word this to make it a little more palatable for your site ... In addition ... A good boxman will COMP to the MAX if he sees that the PLAYER is "betting for the boys."

If I know the variance on a game of video poker, how do I figure out the bankroll I would need to have a 90%-95% probability of avoiding ruin? Great site! Thanks in advance for your answer!

Dave from Mulvane, USA

I hope you're happy, I spent all day on this question. Please visit my new video poker appendix 1 for the answer. There is no easy way to get a risk of ruin figure with just the variance. It depends on exactly what the returns are for each hand and their probability.

In baccarat, you address the question of counting to zero the house edge. But what about the (true) count (for ’player’) necessary to make ’player’ the better bet than ’dealer’? According to your table in baccarat appendix 2, removing cards 5 through 9 increases the relative probability of a ’player’ win. In an 8-deck shoe, is there a point at which the house edge on ’player’ falls below that on ’dealer,’ even though the edge on both may remain positive? Where is that point?

Richard from Glendora, USA

Here are the values to assign each rank for counting the Player bet, from my blackjack appendix 2. The true count is the running count divided by the number of decks remaining.

Player Bet Count
Removed Player
0 -178
1 -448
2 -543
3 -672
4 -1195
5 841
6 1128
7 817
8 533
9 249

I show that if the true count exceeds 17,720 then the Player bet house edge is reduced to 1.06%, and becomes as good of a bet as the Banker. At true counts greater than 17,720, the Player is the better bet.

I can't help but say that you can just walk over to a blackjack table and have a much lower house edge with basic strategy.

I was sitting at first base in an 8-deck blackjack game. The dealer finished shuffling, and as she put the cards in the shoe she fumbled them slightly, flashing the first two cards: a jack and an ace. Since I knew the jack would be the burn card, I also knew I would get the ace as my first card. This is obviously to my advantage, but how much, percentage-wise? I ended up betting $50 instead of my normal $5 on the first hand from the shoe. I wish it had a happy ending, but I was dealt a soft 18, and ended up busting since the dealer was showing a ten. Thanks for your time, and for the great site!


Thanks for the compliment. I don't have numbers readily available for eight decks, but in a four deck game, where the dealer stands on soft 17, I get the player advantage given that the first card is an ace to be 51.66%. In Basic Blackjack by Stanford Wong he says the advantage in a six-deck game where the dealer stands on soft 17 to be 50.5%. Sometimes a Las Vegas fun book will have a coupon which can be used as an ace for the first card in blackjack. Wong also mentions the kind of situation that happened to you in his book.

Thank goodness I just discovered your great site. I have been trying to solve the following, and keep getting different answers. If I am dealt a pocket pair (in Hold'em) what are my chances of getting either three of a kind, or four of a kind on the flop (next three cards).

Elliot from Harwich, Massachusetts

For probability questions, I like to take the number of combinations the event you're interested can happen divided by the total number of combinations. First review the combin function in my probabilities in poker section. The number of ways to get a four of a kind is simply the number of singletons in the deck, or 48. The number of ways to get a three of a kind (not including a full house) is the product of the number of ways to get the third card, 2, and the number of ways to get two other singletons, 2*combin(12,2)*42 = 2,112. The total number of ways the cards can come up in the flop are combin(50,3)=19,600. So, the probability of a four of a kind is 48/19600=0.0024, and the probability of a three of a kind is 2,112/19,600=0.1078.