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Ask the Wizard #180

A local casino had a promotion on their over/under 13 side bet in blackjack. If your first two cards are suited you receive a $5 "action chip." If your first two cards and the dealer’s up card are suited you receive a $10 action chip. The action chips are good for one bet only, the player keeps any winnings but always loses the action chips. The minimum bet is $10. Six decks of cards are used. What is the house edge on the over/under 13 bet?

John from Shakopee, NM

Before considering the bonuses, the house edge is lower on the over bet at 6.55%, as I show in my blackjack appendix 8. The probability of three suited cards is 4×combin(78,3)/combin(312,3) = 4×76076/5013320 = 0.060699. The probability the player’s two cards are suited, but the dealer’s card is not, is (4×combin(78,2)×234)/(combin(312,2)×310) = 2810808/15039960 = 0.186889. Let’s assume the action chips are worth 49.5% of face value. Then the bonuses are worth 0.495×(0.060699×$10 + 0.186889×$5) = $0.76301. The expected loss on the over bet is $10×0.0655 = $0.655. So each $10 over 13 bet is worth $0.76301 - $0.655 = 10.8 cents. The overall player advantage is 1.08% on a $10 over 13 bet.

Hi, Wizard. The casino here allows players to take over another player’s hand in Three Card Poker if he wishes to fold. The player taking over the hand must put up the play bet himself. Should I take over his hand if I know that one of the dealer’s card is 2 to J (low rank card)? What is the player edge? Thanks for your reply.

James from Genting, Malaysia

Yes, you should. If you see the dealer has a 2 to jack the odds favor raising on anything. Using this strategy does result in a player advantage. I get into the details in my book Gambling 102.

The regular baseball season is 162 games. If a team wins 92 games, it will probably make the playoffs. If the team has a 55% chance of winning each game, what are the odds that it will win exactly 92 games? What are the odds that it will win at least 92 games?

kemprolemslev from Los Angeles

The probability of winning exactly 92 games and losing 70 is 162!/(92!×70!)×0.5592×0.4570 = 0.056868. To get the exact probability of winning at least 92 you would need to sum this formula for all wins from 92 to 162. The answer for at least 92 wins is 0.353239.

With many online books offering halftime wagering, I’d love to know your opinion on "dutching". Sometimes it’s obvious, but at what point would you consider betting the opposite of your game bet at halftime? Also, what amount would you consider if you found a favorable situation?

Anonymous

I assume by "dutching" you mean hedging. The sixth of my ten commandments of gambling is "Thou shalt not hedge thy bets." The only time I would make an exception is when the hedge bet itself has a positive expected value, or life changing amounts of money are at stake.

I am interested in playing on-line, but the Bodog casino you recommend does not allow Canadian residents to play. Do you know why that is?

Allan from Toronto, Canada

Thanks for considering them. Bodog has their corporate headquarters in Vancouver. They feel it legally questionable whether they can take bets from their fellow Canadians, so they choose to be beyond reproach and not do so.

How many five-card combinations of a standard playing card deck have cards from exactly two suits?

Samantha from Belize

The two suits can be divided either 4 and 1 or 3 and 2. Let’s look at the 4/1 split first. There are 4 suits to choose from for the one with 4 cards, and 3 left for the one with 1 card. There are combin(13,4)=715 ways to choose 4 ranks out of 13. There are 13 ways to choose a single rank. So there are 4×3×715×13=111,540 ways to have a 4/1 split between the two suits. By similar logic there are 4×3×combin(13,3)×combin(13,2)=267,696 ways to have a 3/2 split. So the overall probability is (111540+267696)/combin(52,5) = 14.59%.

First off let me say unequivocally I understand and agree with your stance on betting systems. It’s quite simple: If you are at a disadvantage for an individual hand, the same holds for multiple hands, regardless of bet amount. End of story. I know the longer I play games in a casino, the higher my chances of leaving without money.

My question isn’t about winning long term with systems, as we know that’s impossible. But might systems have a usefulness in ’tailoring’ the losing experience? For example, player A prefers that each trip to the casino he will either win or lose a moderate amount of money (of course he’ll lose slightly more often than win). Player B prefers a chance to make a little money 4 out of 5 trips, and lose lots of money 1 in 5 trips.

Both will lose money in the long run, but is there a betting system that might help each accomplish his goal?

Anonymous

Yes. While betting systems can not change the house edge, they can be used to improve the probability of achieving trip objectives. Player A wants as little risk as possible. To minimize risk he should flat bet. Player B wants a high probability of a trip win. He should press his bets after a loss. Such a strategy carries the risk of a substantial loss. Although you didn’t ask, a player who wants to either lose a little or win big should press his bets after a win. This kind of strategy will usually lose, but sometimes will have a big win.

What is the probability that two bingo cards have no number in common? What is the probability they have every number in common?

Joe

The probability two bingo cards have no numbers in common is (combin(10,5)/combin(15,5))4×(combin(11,4)/combin(15,4)) = 1 in 83,414. The probability two bingo cards have all 24 numbers the same is (1/combin(15,5))4×(1/combin(15,4)) = 1 in 111,007,923,832,371,000.

What are the odds of being the dealt 2-3-4-5-7 unsuited? Thanks a lot, the site’s great!

Kevin from Massapequa

Thanks. (45-4)/combin(52,5) = 1020/2598960 = 1 in 2,548.

Dear Mr. Wizard, I have been recently trying to calculate the probability of getting a flush in Texas Hold 'Em if dealt two suited hole cards? My answer keeps on coming out to be 5.8% but this seems ulimately incorrect, your help would be much appreciated thanks

Nathan S. from New Plymouth

The probability of making a flush, with exactly three cards to the same suit as your hole cards, is combin(11,3)×combin(39,2)/combin(50,5) = 122265/2598960 = 0.057706. The probability of making a flush, with four more cards to the same suit as your hole cards, is combin(11,4)×combin(39,1)/combin(50,5) = 2145/2118760 = 0.001012. The probability of making a flush, with five more cards to the same suit as your hole cards, is combin(11,5)/combin(50,5) = 462/2118760 = 0.000218. The probability of making a flush on the board in another suit is 3×combin(13,5)/combin(50,5) = 3861/2118760 = 0.001822. Add this all up and you get 0.057706 + 0.001012 + 0.000218 + 0.001822 = 0.060759.

I have been "seeing" a guy for three years. He says he doesn’t want a relationship but always wants to be with me. Now we only sleep together. Will he ever want to be with me more or after this long can a guy just continue just to do one thing with me and pursue others? I don’t know what to say to him but feel sad all the time that I don’t get more.

Kristine from Tacoma

It seems he doesn’t want to buy the cow because the milk is free.