# Ask the Wizard #257

Could you please, for your visually impaired gamblers out there, make an accessible blackjack strategy chart? Unfortunately, screen readers (programs that read text allowed as speech) don’t read the chart very well. Instead, could you write a step-by-step guide? An accessible chart will be greatly appreciated!

JordanN

Never let it be said I’m not a friend of the ~~blind~~ visually impaired. Here is my Wizard’s Simple Strategy in easy text form. This is not the standard basic strategy, which is more powerful, but would be lengthy to put into words.

Always:

- Hit hard 8 or less.
- Stand on hard 17 or more.
- Hit on soft 15 or less.
- Stand on soft 19 or more.
- With 10 or 11, double if you have more than the dealer’s up card (treating a dealer ace as 11 points), otherwise hit.
- Surrender 16 against 10.
- Split eights and aces.

If the player hand does not fit one of the above "always" rules, and the dealer has a 2 to 6 up, then play as follows:

- Double on 9.
- Stand on hard 12 to 16.
- Double soft 16 to 18.
- Split 2’s, 3’s, 6’s, 7’s, and 9’s.

If the player hand does not fit one of the above "always" rules, and the dealer has a 7 to A up, then **hit**.

For the full basic strategy in text form, please see my 4-deck to 8-deck basic strategy.

Two 54-card decks (including two jokers) are shuffled together. A player is given half of them. What is the probability that the player got all four red threes?

Doc

There are 4 red threes and 104 other cards. There is just one way to get all four red threes. There are combin(104,50)= 1.46691 × 10^{28} ways the player could get 50 of the other 104 cards. The total number of combinations is combin(108,54)= 2.48578 × 10^{30}. combin(104,50)/combin(108,54) = 0.059012.

If you don't like dealing with such large numbers, here is an alternative solution. Number the four red threes 1 to 4. The probability that the first red three is in the player's stack is 54/108. Now remove the first three. The probability that the player has the second red three is 53/107, because the player has 53 cards left, and there are 107 remaining cards. Likewise, the probability that the player has the third red three is 52/106, and the fourth red three is 51/105. (54/108) × (53/107) × (52/106) × (51/105) = 0.059012.

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

Which video poker game has the most variance?

rudeboyoi

My best guess is Royal Aces Bonus Poker. I’ve seen it only once in Mesquite years ago. It pays 800 for four aces, but compensates with a lowest paying hand of a pair of aces, as opposed to the usual jacks. Here is the return table.

### Royal Aces Bonus Poker

Hand | Pays | Combinations | Probability | Return |
---|---|---|---|---|

Royal flush | 800 | 490,090,668 | 0.000025 | 0.019669 |

Straight flush | 100 | 2,417,714,292 | 0.000121 | 0.012129 |

Four aces | 800 | 4,936,967,256 | 0.000248 | 0.198140 |

Four 2-4 | 80 | 10,579,511,880 | 0.000531 | 0.042460 |

Four 5-K | 50 | 31,662,193,440 | 0.001588 | 0.079421 |

Full house | 10 | 213,464,864,880 | 0.010709 | 0.107090 |

Flush | 5 | 280,594,323,000 | 0.014077 | 0.070384 |

Straight | 4 | 276,071,121,072 | 0.013850 | 0.055399 |

Three of a kind | 3 | 1,470,711,394,284 | 0.073782 | 0.221346 |

Two pair | 1 | 2,398,705,865,028 | 0.120337 | 0.120337 |

Pair of aces | 1 | 1,307,753,371,584 | 0.065607 | 0.065607 |

Nothing | 0 | 13,935,843,099,816 | 0.699126 | 0.000000 |

Total | 19,933,230,517,200 | 1.000000 | 0.991982 |

The standard deviation is 13.58! That is over three times as high as 9-6 Jacks or Better at 4.42.

However, if you limit me to games that are easy to find, my nomination is Triple Double Bonus, with a standard deviation of 9.91. Here is that pay table.

### Triple Double Bonus Poker

Hand | Pays | Combinations | Probability | Return |
---|---|---|---|---|

Royal flush | 800 | 439,463,508 | 0.000022 | 0.017637 |

Straight flush | 50 | 2,348,724,720 | 0.000118 | 0.005891 |

4 aces + 2-4 | 800 | 1,402,364,496 | 0.000070 | 0.056282 |

4 2-4 + A-4 | 400 | 3,440,009,028 | 0.000173 | 0.069031 |

4 aces + 5-K | 160 | 2,952,442,272 | 0.000148 | 0.023699 |

4 2-4 + 5-K | 80 | 6,376,626,780 | 0.000320 | 0.025592 |

4 5-K | 50 | 31,673,324,076 | 0.001589 | 0.079449 |

Full house | 9 | 206,321,656,284 | 0.010351 | 0.093156 |

Flush | 7 | 311,320,443,672 | 0.015618 | 0.109327 |

Straight | 4 | 252,218,322,636 | 0.012653 | 0.050613 |

3 of a kind | 2 | 1,468,173,074,448 | 0.073655 | 0.147309 |

Two pair | 1 | 2,390,581,734,264 | 0.119929 | 0.119929 |

Jacks or better | 1 | 3,944,045,609,748 | 0.197863 | 0.197863 |

Nothing | 0 | 11,311,936,721,268 | 0.567491 | 0.000000 |

Total | 19,933,230,517,200 | 1.000000 | 0.995778 |

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

Five sailors survive a shipwreck. The first thing they do is gather coconuts and put them into a big community pile. They meant to divide them up equally afterward, but after the hard work gathering the coconuts, they are too tired. So they go to sleep for the night, intending to divide up the pile in the morning.

However, the sailors don’t trust one another. At midnight one of them wakes up to take his fair share. He divides up the pile into five equal shares, with one coconut left over. He buries his share, combines the other four piles into a new community pile, and gives the remaining coconut to a monkey.

At 1:00 AM, 2:00 AM, 3:00 AM, and 4:00 AM each of the other four sailors does the exact same thing.

In the morning, nobody confesses what he did, and they proceed with the original plan to divide up the pile equally. Again, there is one coconut left over, which they give to the monkey.

What is the smallest possible number of coconuts in the original pile?

David Filmer from MA (Cantab)

"Scroll down 100 lines for the answer.

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There were **15,621** coconuts in the original pile. Scroll down another 100 lines for my solution.

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Let c be the number of coconuts in the original pile and f be the final share for each sailor after the last division.

After sailor 1 takes his share and gives the monkey his coconut there will be (4/5)×(c-1) = (4c-1)/5 left.

After sailor 2 takes his share and gives the monkey his coconut there will be (4/5)×(((4c-1)/5)-1) = (16c-36)/25 left.

After sailor 3 takes his share and gives the monkey his coconut there will be (4/5)×(((16c-36)/25)-1) = (64c-244)/125 left.

After sailor 4 takes his share and gives the monkey his coconut there will be (4/5)×(((64c-244)/125)-1) = (256c-1476)/625 left.

After sailor 5 takes his share and gives the monkey his coconut there will be (4/5)×(((256c-1476)/625)-1) = (1024c-8404)/3125 left.

In the morning each sailor’s share of the remaining pile will be f = (1/5)×(((1024c-8404)/3125)-1) = (1024c-11529)/15625 left.

So, the question is what is the smallest value of c such that f=(1024×c-11529)/15625 is an integer. Let’s express c in terms of f.

(1024×c-11529)/15625 = f

1024c - 11529 = 15625×f

1024c = 15625f+11529

c = (15625f+11529)/1024

c = 11+((15625×f+265)/1024)

c = 11+15×f+(265×(f+1))/1024

So, what is the smallest f such that 265×(f+1)/1024 is an integer? 265 and 1024 do not have any common factors, so f+1 by itself is going to have to be divisible by 1024. The smallest possible value for f+1 is 1024, so f=1023.

Thus, c = (15625×1023+11529)/1024 = 15,621.

Here is how many coconuts each person, and monkey, received:

### Coconut Problem

Sailor | Coconuts |

1 | 4147 |

2 | 3522 |

3 | 3022 |

4 | 2622 |

5 | 2302 |

Monkey | 6 |

Total | 15621 |

David Filmer, the one who challenged me the with question, already knew the answer. Actually, he asked me the formula for the general case of s sailors, but I had enough trouble with the specific case of 5 sailors. David notes the answer for the general case is c = s^{s+1} - s + 1.

I’ll leave that proof to the reader.

Here are some links to alternate solutions to the problem: