# Ask the Wizard #277

Anonymous

First, it bears repeating that 16 vs. 10 is an extremely borderline hand between hit and stand. If you're allowed to surrender, that is much better than either hitting or standing for the basic strategy player. Otherwise, hitting is a tiny bit better, on average. It would take the removal of just one small card from an eight-deck shoe to sway the odds in favor of standing, because with one fewer small card there are more large cards left, making hitting more dangerous. That is why I say that if your 16 is composed of three or more cards you should stand, because a 3-card 16 has usually removed at least two small cards from the shoe.

Second, on the first hand after a shuffle, if the basic strategy and a card counting strategy differ on how to play the hand, then the basic strategy prevails. The basic strategy was carefully created to consider the exact deck composition based on the specific cards observed. A table of index values is a blunter instrument that is applicable throughout the shoe.

In this particular case a card counter could either hit or stand, depending on how he rounds the true count. If he rounds down, the true count will be -1, causing him to hit. If he rounds up, or to the nearest integer, the true count will be 0, causing him to stand. As long as I bring this up, according to Blackjack Attack by Don Schlesinger, the methodology of choice for rounding is "flooring," or rounding down, in this case to -1, causing the player to correctly hit.

Another similar situation is 15 vs. 10. 83% of the time (with a 10+5 or 8+7, but not 9+6), this results in a running count of -1 the first hand after a shuffle, and the index number to surrender is 0. Rounding down would cause the player to incorrectly hit, when surrendering is better.

The bottom line is that for the first decision after a shuffle, with no other cards known from other players, the card counter should use basic strategy. After that, resume using index numbers.

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

EvenBob

I don't know. What I think I can correctly say is the earliest casino game patent for a game played today is for Caribbean Stud Poker. There probably were other patents before it for games that didn't make it. The Caribbean Stud patent was filed on April 18, 1988 and issued on June 6, 1989. Patent number 4,836,553.

Not that you asked, but at that time casino game patents were valid for 17 years from date of issue, or 20 years from date of filing, whichever was more. In 1995 the term was extended to 20 years from date of filing. In the case of Caribbean Stud, the patent would have expired in 2008. However, I think it still has valid trademarks, meaning a casino could offer the game without paying royalties, but would have to think of another name that is not trademarked.

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

Krazycat

Yes! Bet on the side face up in the flipper's hand. The academic paper Dynamical Bias in the Coin Toss by Persi Diaconis, Susan Holmes, and Richard Montgomery concludes that the coin will land on the same side it started on 51% of the time.

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

Frank

For a near-exact answer to streak questions such as this we need to use matrix algebra. I answered a similar, yet easier, question in my June 4, 2010 column. If your matrix algebra is rusty I would look at that one first.

**Step 1**: Determine the probability of 0 to 6+ royals in the first 5,000 hands. Let's assume the probability of a royal is 1 in 40,000. The expected number in 5,000 hands is 5,000/40,000 = 0.125. Using the Poisson estimate, the probability of exactly r royals is e^{-0.125} × 0.125^{r}/r!. Here are those probabilities:

### Royals in 5,000 Hands

Royals | Probability |
---|---|

0 | 0.8824969026 |

1 | 0.1103121128 |

2 | 0.0068945071 |

3 | 0.0002872711 |

4 | 0.0000089772 |

5 | 0.0000002244 |

6+ | 0.0000000048 |

**Step 2**: Consider there to be seven states for the remaining 24,995,000 hands. For each one, the previous 5,000 hands can have 0, 1, 2, 3, 4, or 5 royals, or the player could have already achieved getting six royals in 5,000 hands, in which case success is achieved, and it can't be taken away. With each new hand, one of three things can happen to the player's state:

- Move down a level. This happens if the hand that was played 5,000 games ago was a royal, and is now dropping off, and the new hand was not a royal.
- Remain at the same level. This will usually happen if the hand played 5,000 games ago was not a royal, and the new hand is also not a royal. It can also happen if a hand 5,000 games ago was a royal, but the new hand is also a royal.
- Move up a level. This will happen if the hand played 5,000 games ago was not a royal, and the new hand is.

**Step 3**: Develop the transition matrix for the odds of each change of state for an additional game played.

The first row will correspond to level 0 before the new hand is played. The odds of advancing to level 1 in the next hand are simply 1 in 40,000. The probability of staying at level 0 is 39,999/40,000.

The second row will correspond to level 1 before the new hand is played. The odds of advancing to level 2 in the next hand are the product of the odds of not losing a royal on the hand dropping off and getting a royal on the new hand = (4999/5000)×(1/40000) = 0.0000250. The odds of going back to level 0 are the product of a royal dropping off and not getting a royal on the current game = (1/5000)×(39999/40000) = 0.0002000. The odds of staying the same is pr(no royal dropping off) × pr(no new royal) + pr(royal dropping off) × pr(new royal) = (4999/5000)×(39999/40000) + (1/5000)×(1/40000) = 0.9997750.

The probabilities for rows 2 to 6 will depend on how many royals are present in the history of the last 5,000 hands. The more there are, the greater the probability of one dropping off as a new hand is played. Let r be the number of royals in the last 5,000 hands and p be the probability of getting a new royal.

Pr(promote a level) = Pr(no royal dropping off) × Pr(new royal) = (1-(r/5000))× p.

Pr(remain at same level) = Pr(no royal dropping off) × Pr(no new royal) + Pr(royal dropping off) × Pr(new royal) = (1-(r/5000))× (1-p) + (r/5000)×p.

Pr(demote a level) = Pr(royal dropping off) × Pr(no new royal) = (r/5000)× (1-p).

Row 7 corresponds to having achieved the state of success for getting six royals in 5,000 hands. Once you achieve that accomplishment it can never be taken away, so the odds of staying in that state of success are 100%.

The rows in the transition matrix will correspond to the levels before the new hand, starting with level 0 in the top row. The columns will correspond to the levels after the new hand, starting with level 0 in the left column. The body of numbers in the matrix will correspond to the probabilities of moving from each old state to each new state in one game. Let's call this T1 =

0.999975 | 0.000025 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

0.000200 | 0.999775 | 0.000025 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

0.000000 | 0.000400 | 0.999575 | 0.000025 | 0.000000 | 0.000000 | 0.000000 |

0.000000 | 0.000000 | 0.000600 | 0.999375 | 0.000025 | 0.000000 | 0.000000 |

0.000000 | 0.000000 | 0.000000 | 0.000800 | 0.999175 | 0.000025 | 0.000000 |

0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.001000 | 0.998975 | 0.000025 |

0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 1.000000 |

If we multiply this transition matrix by itself we get the probabilities of each change of state in two consecutive games. Let's call this T2, for the transition matrix over two games:

0.999950 | 0.000050 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

0.000400 | 0.999550 | 0.000050 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

0.000000 | 0.000800 | 0.999150 | 0.000050 | 0.000000 | 0.000000 | 0.000000 |

0.000000 | 0.000000 | 0.001199 | 0.998750 | 0.000050 | 0.000000 | 0.000000 |

0.000000 | 0.000000 | 0.000000 | 0.001599 | 0.998351 | 0.000050 | 0.000000 |

0.000000 | 0.000000 | 0.000000 | 0.000001 | 0.001998 | 0.997951 | 0.000050 |

0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 1.000000 |

By the way, in Excel to multiply two matrices of equal size first select the region where you want the new matrix to go. Then use this formula =MMULT(range of matrix 1, range of matrix 2). Then do ctrl-shift-enter.

If we multiply T2 by itself we get the probabilities of each change of state in four consecutive games, or T4:

0.999900 | 0.000100 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

0.000800 | 0.999100 | 0.000100 | 0.000000 | 0.000000 | 0.000000 | 0.000000 |

0.000000 | 0.001598 | 0.998301 | 0.000100 | 0.000000 | 0.000000 | 0.000000 |

0.000000 | 0.000001 | 0.002396 | 0.997503 | 0.000100 | 0.000000 | 0.000000 |

0.000000 | 0.000000 | 0.000003 | 0.003193 | 0.996705 | 0.000100 | 0.000000 |

0.000000 | 0.000000 | 0.000000 | 0.000005 | 0.003989 | 0.995907 | 0.000100 |

0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 1.000000 |

So keep repeating this doubling process 24 times until we get to T-16,777,216:

0.882415 | 0.110305 | 0.006893 | 0.000287 | 0.000009 | 0.000000 | 0.000091 |

0.882415 | 0.110305 | 0.006893 | 0.000287 | 0.000009 | 0.000000 | 0.000092 |

0.882413 | 0.110304 | 0.006893 | 0.000287 | 0.000009 | 0.000000 | 0.000094 |

0.882385 | 0.110301 | 0.006893 | 0.000287 | 0.000009 | 0.000000 | 0.000125 |

0.881714 | 0.110217 | 0.006887 | 0.000287 | 0.000009 | 0.000000 | 0.000885 |

0.860229 | 0.107531 | 0.006720 | 0.000280 | 0.000009 | 0.000000 | 0.025231 |

0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 1.000000 |

If we doubled again we would overshoot our goal of T-24,995,500. So now we need to carefully multiply by smaller transition matrices, which we would have already calculated. You can arrive at any number using powers of two (the joys of binary arithmetic!). In this case T-24,995,500 = T-16,777,216 × T-2^{22} × T-2^{21} × T-2^{20} × T-2^{19} × T-2^{18} × T-2^{16} × T-2^{14} × T-2^{13} × T-2^{10} × T-2^{7} × T-2^{5} × T-2^{4} × T-2^{3} =

0.882375 | 0.110300 | 0.006893 | 0.000287 | 0.000009 | 0.000000 | 0.000136 |

0.882375 | 0.110300 | 0.006893 | 0.000287 | 0.000009 | 0.000000 | 0.000136 |

0.882373 | 0.110299 | 0.006892 | 0.000287 | 0.000009 | 0.000000 | 0.000138 |

0.882345 | 0.110296 | 0.006892 | 0.000287 | 0.000009 | 0.000000 | 0.000170 |

0.881675 | 0.110212 | 0.006887 | 0.000287 | 0.000009 | 0.000000 | 0.000930 |

0.860191 | 0.107527 | 0.006719 | 0.000280 | 0.000009 | 0.000000 | 0.025275 |

0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 1.000000 |

To be honest, in the interest of simplicity and saving time, you don't really need to bother with those last four multiplications. These correspond to the last 56 hands only, and the odds that those 56 will make a difference in the final outcome are negligible. I'm sure my many perfectionist readers would take me to the woodshed for saying that, if they could.

**Step 4:** Multiply the initial state after 5,000 hands by T-24,995,500. Let S-0, from step 1, be as follows:

0.8824969026 | 0.1103121128 | 0.0068945071 | 0.0002872711 | 0.0000089772 | 0.0000002244 | 0.0000000048 |

So S-0 × T-24,995,500 =

0.88237528 |

0.11029964 |

0.00689251 |

0.00028707 |

0.00000896 |

0.00000022 |

0.00013632 |

The number in the bottom cell is the probability of having achieved six royals within 5,000 hands at least once during the 25,000,000 hands. So a 1 in 7,336 chance.

My thanks to CrystalMath for his help with this question.