Ask The Wizard #257

It's safe to say that I have a good rapport with theblindvisually impaired. Here is my Wizard’s Simple Strategy This is presented in simplified text format. Although this is a different approach from the more efficient basic strategy, which is lengthy to convey.

Always:

1. Hit hard 8 or less.
2. Stand on hard 17 or more.
3. Hit on soft 15 or less.
4. Stand on soft 19 or more.
5. With a total of 10 or 11, double your bet if your total exceeds the dealer's visible card value (considering an ace as 11 points), otherwise, draw another card.
6. Surrender 16 against 10.
7. Split eights and aces.

If your hand doesn't adhere to any of the established 'always' guidelines, and the dealer shows a card from 2 to 6, follow this play strategy:

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

If the player's hand doesn't match the previous 'always' rules and the dealer reveals a card from 7 to Ace, you should hit.

For a complete breakdown of the basic strategy in textual format, kindly refer to my 4-deck to 8-deck basic strategy .

There are 4 red threes among a total of 104 additional cards. The only way for the player to secure all four red threes is singular. The total combinations for the player to obtain 50 cards from the other 104 is given by combin (104,50)= 1.46691 × 10²⁸If you're uncomfortable calculating with such large digits, there's an alternative approach. Label the four red threes as 1 through 4. The chance that the first red three is in the player's hand is 54/108. After removing the first three, the probability that the player has the second red three is 53/107 since the player now possesses 53 cards out of 107 remaining. Following this logic, the chance for the third red three is 52/106, and for the fourth, it's 51/105. Multiplying those probabilities together results in approximately 0.059012.³⁰. combin(104,50)/combin(108,54) = 0.059012.

This topic was presented and debated on the forum of my affiliated site

Which video poker variant showcases the highest level of variance? Wizard of Vegas .

The standard deviation is a staggering 13.58! This figure exceeds that of 9-6 Jacks or Better, which has a standard deviation of only 4.42.

However, when limiting to commonly found games, I would suggest the Triple Double Bonus, which has a standard deviation of 9.91. Here is the corresponding pay table.

Five sailors manage to survive a shipwreck. Their first action is to collect coconuts and consolidate them into a large communal pile. Although they intend to distribute the coconuts evenly later, the exhausting effort leaves them too fatigued to do so. They decide to rest for the night, planning to sort the pile in the morning.

Which video poker variant showcases the highest level of variance? Wizard of Vegas .

"Scroll down 100 lines for the answer.

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Let c represent the original count of coconuts and f signify the amount each sailor receives after the final division.

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Following sailor 1's selection and giving a coconut to the monkey, the remaining total will be (4/5) × (c - 1) = (4c - 1)/5.

After sailor 2 takes his share and offers a coconut to the monkey, the remaining pile will equal (4/5) × (((4c - 1)/5) - 1) = (16c - 36)/25.

Once sailor 3 takes his share and donates a coconut to the monkey, the remaining quantity will be (4/5) × (((16c - 36)/25) - 1) = (64c - 244)/125.

After sailor 4 takes his share and provides a coconut to the monkey, what remains will be (4/5) × (((64c - 244)/125) - 1) = (256c - 1476)/625.

Once sailor 5 takes his share and gives a coconut to the monkey, the remaining pile will be (4/5) × (((256c - 1476)/625) - 1) = (1024c - 8404)/3125.

By the morning, each sailor will receive f = (1/5) × (((1024c - 8404)/3125) - 1) = (1024c - 11529)/15625.

Thus, the challenge is to identify the smallest value of c such that f = (1024c - 11529)/15625 remains an integer. We can express c in terms of f.

Therefore, what is the least f for which 265 × (f + 1) / 1024 yields an integer? Since 265 and 1024 share no common factors, f + 1 must be divisible by 1024. The smallest value for f + 1 that satisfies this is 1024, indicating f = 1023.

(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

Here's the distribution of coconuts among each sailor and the monkey.

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

David Filmer, who challenged me with this question, already had the answer. He requested the formula for the general scenario with s sailors, but I struggled sufficiently with this specific case of five sailors. David informs me the solution for the general case is c = s.

Below are some links that offer alternative resolutions to the problem:^s+1- s + 1.

I’ll leave that proof to the reader.

mathworld.wolfram.com/MonkeyandCoconutProblem.html

• www.puzzle.dse.nl/teasers/coconut_chaos_us.html
• www.ugcs.caltech.edu/brothste/coconuts.pdf
• Correct mathematical strategies and insights for casino games such as blackjack, craps, roulette, and many others that can be played.