# Ask the Wizard #88

I have one simple question. I know you are the Wizard of odds but I need help. I play Craps and Texas Hold’em in the casinos. I fell I could be an intimidating force (THUS RAISING MY ODDS OF WINNING) if I could just figure out how to shuffle my poker chips. I have practiced but just can’t get it. I was hoping you could point me in a direction to learn this. Thank you for your time.

Joseph

You asked the right person! I’m quite good at shuffling poker chips. Unfortunately I don’t often get to show it off because when I do play with chips it is usually counting cards or hole card reading, and in either case I don’t want to look like a pro. Anyway, start with a stack of 10 chips. Then cut them in half, putting two 5 stick chips side by side. Think of the two stacks as an 8. Put the 8 at about a 45 degree angle to your plane of symmetry. Put your thumb on the bottom of the 8, your index finger where the two stacks come together, and the other 3 fingers at the top of the 8. All five fingers should be barely touching the table. Then using your index finger gently lift both stacks while the other four fingers gently push the stacks together. After your index finger is about a quarter inch from the bottom of the table quickly move it away and keep pushing the stacks together with your other four fingers. It takes practice. I would recommend getting 10 nice clay chips and practicing at home or work. During commercials or whenever you have a minute to spare you can work on it. Before you know it you you’ll be riffling chips like a pro and casting fear in your fellow poker players.

Is there some way of telling when a progressive slot has reached a level where the house edge is zero? For example, what would the Major Millions jackpot have to be?

Anonymous

Although this is easy with video poker with slots there is no way to tell without knowing how the machine was programmed.

I rolled four hard 4’s without rolling a 7 or an easy 4. Any idea what the odds on doing that is? Can it be calculated?

Anonymous

The probability of winning the hard 4 bet is 1/9. So the probability of winning four times in a row is (1/9)^{4} = 1 in 6561.

Wizard- In a recent Ask the Wizard column there was a mention of Multiple Action BlackJack which I had played very successfully several years back at Foxwoods. I had a feeling there was a larger advantage to the player when the deck had excess A’s and 10’s since if the player hit blackjack he had it for all three hands while the dealer had to make a blackjack each time with his/her three hands. If this is true please do not post this on the site since it will be the end of this variation - it is no longer offered at Foxwoods. I find your site very informative and appreciate all your hard work.

Anonymous

Thank you for the kind words. Much like multi-play video poker the house edge is the same for multiple action blackjack as regular blackjack, assuming the same rules. It is true that if the player has blackjack he wins on all three hands. However if the player has a 16 he has to play it on all three hands. Overall everything balances out.

I know that mathematically anything is possible, but the other night at the casino I think that I witnessed something that would be a billion to 1 shot, not that those don?t ever happen. Here is what happened: In the course of 40 hands (40 single 3 card deals about 8 rounds with 5 players) at a let it ride table 3 four of a kinds were dealt. With a four of a kind being about 4100 to 1 what would be the odds of three of them being dealt within 40 deals? Please answer as this is killing me. Long time fan

Michael

For the sake of simplicity let’s assume each hand is dealt from a fresh deck. The probability of a four of a kind is 13*48/combin(52,5) = 624/2598960. The probability of exactly 3 out of 40 four of a kinds is combin(40,3)*p^{3}*(1-p)^{37} = 1 in 7378135, where p = 624/2598960. So that is more like a 1 in 7 million shot.

I was recently playing single deck blackjack with one other patron at the table and right after the shuffle I got 2 aces and split them and got 21 on both hands!(yea for me) the other guy got a nineteen 10-9 and the dealer had an ace up. the dealer had a soft 17 and drew a 10. My question is this: What should I do on the next hand considering that the deck just lost 3 aces and 4 tens and only 2 small cards? Should I go to the bathroom, leave the table (I had just sat down 2 hands prior.) or keep playing? Thanks for your great site. It is just as fun to learn about all the different aspects of the games as it is to use your advice and know I am playing the best game I can!

Anonymous

This would be a good time to either bet small, go to the bathroom, or just leave, depending on your style of play. Personally I hate bouncing up and down and would just bet small. There are lots of books that explain specifically how to count cards. However for the amateur if you see a disproportionate number of tens and aces leave the deck bet less. Likewise bet more if lots of small cards leave the deck, especially fives and sixes. Thanks for your kind words about my site.

I just witnessed a friend get four blackjacks in a row starting with the first hand of a newly shuffled single deck playing head to head against the dealer. I looked at the FAQ’s and saw the odds for getting one blackjack in single deck, but don’t know how to calculate them for getting four in a row off the top. Instead of a decimal probability, could you tell me the odds of this? It must be astronomical. Hope to hear from you.

Anonymous

I seem to get a variation of this question at least once a month. Let’s assume for now the deck is shuffled after every hand, to make the math easier. If the probability of something happening is p then the probability of it happening n times in a row is p^{n}. The probability of a blackjack in a single deck game is 4*16/combin(52,2) = 64/1326. So the probability of four in a row is (64/1326)^{4} = 16777216/ 3091534492176 = 1 in 184270. However the actual probability is much less, because as the player gets each blackjack the ratio of aces to cards left in the deck decreases. Without knowing what cards the dealer got I can’t tell you the exact answer.

First I wanted to tell you how much I look at and love your web site, and admire your math skills. I use 6 decks to deal blackjack, and added 3 jokers for reasons I won’t waste your time with but, what are the odds of dealing all 3 jokers to a player right in a row. Thank you very much.

Anonymous

You’re welcome, thanks for you compliments. The probability of being dealt 3 jokers in a row from a six deck shoe (plus the 3 jokers) is 1/combin(315,3) = 1 in 5,159,805. Another solution is (3/315)*(2/314)*(1/313).