Ask The Wizard #14

I'll begin by calculating the likelihood of a king high hand and will quickly outline the formulas for the other two hands. The probability is determined by taking the total number of king high hands and dividing it by the overall number of possible hands. There are 11 ranks below the king. To form a king high hand, we must select two different ranks from these. The number of arrangements possible for this selection is combin (11,2) = 55. Out of these combinations, one is the king-queen-jack, which forms a straight, so we eliminate that from our count, leaving us with 54 valid options that do not create a straight. For each chosen rank, there are four different suits available, leading to 64 total suit combinations. However, four of these result in a flush, which leaves us with 60 combinations that aren't flushes. Therefore, the total number of king high combinations is calculated as 54 times 60, resulting in 3240. When considering the total ways to arrange 3 cards from a deck of 52, we find there are combin(52,3) = 22,100. Hence, the likelihood of forming a king high hand is 3,240 divided by 22,100, which equals approximately 0.1466063. The calculation for an ace high hand is: (combin(12,2)-2)*(4³-4)/combin(52,3)=0.1737557. The reasoning behind the -2 instead of -1 accounts for both ace-2-3 and queen-king-ace straights.³The chance of achieving a queen high hand is determined by: (combin(10,2)-1)*(4', "Hello! I'm really enjoying your website! If I'm dealt a blackjack, and I see the dealer has an ace, I'm faced with the choice of taking even money or carrying on with the hand. Which option should I choose?

This is akin to evaluating the insurance decision in blackjack. By taking even money, the expected return is clearly 1.0 times the amount wagered. For simplicity, let’s operate under the assumption of an infinite card deck. The likelihood that the dealer has a blackjack is 4 out of 13, while the probability of them not having one is 9 out of 13. If the dealer does indeed get a blackjack, you push, meaning you neither win nor lose. If they don't get a blackjack, you win 1.5 times your bet. The expected value of opting out of insurance is calculated as (4/13)*0 + (9/13)*(1.5) = 13.5/13, which simplifies to about 1.0384615, exceeding 1.0. Therefore, opting out of insurance or even money and choosing to play your hand is the more advantageous option. In a real game with a limited number of decks, the odds are even more favorable because there’s one less ten in play (in your own hand), reducing the dealer's chances of hitting a blackjack.³-4)/combin(52,3)=0.119457.

Hello, I'm currently researching Bingo and I’m curious about how to calculate the chances of winning at Bingo. I want to know the probabilities of achieving horizontal, vertical, and diagonal lines, filling the entire card, and obtaining the four corners. I've seen your probability table and I'm interested in the specific formula you used.

Calculating the probability of achieving bingo (which typically means getting five consecutive marks) is quite complex, especially because of the existence of the free space in the card. I utilized a computer program to assist with this. However, calculating the four corners is more straightforward. The probability of marking the four corners on a card, given that x marks have been made, is given by

I've heard suggestions that if you have one come bet in a game of craps, you should collect the odds on the initial roll, but if you possess multiple, you ought to keep these odds in play. The reasoning is that with two or more, the chances of rolling one of those specific points improves over rolling a total of seven. However, with just one bet, the likelihood of winning is generally higher than losing. combin Players should consistently keep their odds active, irrespective of how many come bets they've placed. When weighing the available options, one must consider more than just the chances of winning. It's true that with a single come bet, the risk of losing that bet is higher than the chance of winning, but the potential reward is also notably larger than the possible loss. The main reason to maintain the odds is that this bet has a zero house edge. If you turn off the odds, you inadvertently increase the disadvantage of your game, making it more favorable to the house edge bets and thus raising your expected losses compared to your total wagers.

I remember reading about a strategy that involves placing two bets on the same roll in craps, which could help lower the house edge to its bare minimum. While I don't expect to bank a huge profit, I'm also not looking to take substantial losses. It might sound unexciting, but I prefer a steady, cautious approach. My plan is for my wife and I to sit separately at the table, effectively balancing each other out—in one case, one could win big, while the other may end up losing considerably. If we bring along enough funds, we could extend our gaming session for a few hours. I believe the bets I was referencing are Pass & Don't Pass.

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