Ask The Wizard #125

Certainly! My suggestion is to place your bet on the side facing up at the beginning of the toss. Based on research, Science News Online the chance of a coin landing on the same side it started is around 51%. This is due to the fact that when a coin is flipped, it doesn't rotate perfectly around its axis, which creates the illusion of continuous flipping. This theory holds true only if the coin is caught directly in a hand, preventing it from bouncing. Additionally, the data suggests that when a penny is spun, it will land on tails approximately 80% of the time, likely due to the heavier side (the head) favoring downward momentum. However, I'm a bit doubtful about this; I tested it 20 times and achieved 11 heads and 9 tails. The odds of getting 9 or fewer tails in 20 spins, given an 80% chance, is roughly 1 in 1775.

As with most betting systems the cancellation system This method typically results in successful sessions but can also lead to sporadic significant losses. When this cancellation method experiences losses, the consequences can be devastating. During the periods where it feels like you are losing continuously, the stakes accumulate rapidly, which can quickly drain your finances if luck isn’t on your side.

The table below illustrates the probabilities of completing a Yahtzee on the final roll, based on the number of extra dice required.

The next table outlines the chances of improvement. The left-hand column indicates how many dice you need before each roll, while the top row reveals how many are required after the roll. The body of the table presents the probability of achieving varying levels of improvement.

Following that is a table that shows the likelihood on the first roll when needing between 0 to 4 additional dice to form a Yahtzee.

This next table presents the chances of enhancement and ultimately succeeding, depending on how many dice you need after your initial roll. For instance, if a player requires 3 additional dice to achieve a Yahtzee, the likelihood of reducing this to needing 2 after the second roll and then making the Yahtzee on the third roll is 0.010288066.

To determine the final result, use the dot product from the count needed after the first roll from the previous table and the chances of eventual success from the final column of the preceding table. This results in 0.092593*0.012631 + 0.694444*0.029064 + 0.192901*0.093364 + 0.019290*0.305556 + 0.000772*1 = 4.6028643%. To verify, I simulated 100,000,000 games, achieving a simulated probability of 4.60562%.

There exists a 'lookup' table that connects different random numbers to specific stops on the reels. However, I was uncertain how they transition from that stage to actually determining the player’s payout. I decided to consult a former mathematician in the slot machine industry, who wished to remain anonymous. He explained, \"Your initial assumption is correct. The position on each reel is randomly picked through the RNG. The code then checks the symbols across each activated payline to uncover winning combinations. This method is also applicable for scatter payouts. All major video slot manufacturers utilize this approach. You can think of the algorithm as a large series of if-then statements, but actual implementation may be even more sophisticated.\" I hope this information is useful.

P.S. Following the publication of this column, I received another email about this topic. It's somewhat lengthy, and I am providing you with it. this link .

Thank you for the compliment! There is still some benefit, particularly in a full table context. However, under standard single-deck rules (like the dealer hitting on soft 17 and no doubling after splits), I don't believe the advantage is substantial enough to overcome the 0.19% house edge.

1/combin(52,4) = 1 in 270725.

I like to compute the house percentage as 1-(probability of winning * payout - probability of losing). In this case, it would be 1-((1/3)*1.8 - (2/3)) = 6.67%. Alternatively, if you're familiar with both the fair payout and the actual payout, a straightforward formula for calculating the house edge is (f-a)/(f+1), with f representing the fair payout and a denoting the actual payout. Therefore, in this instance, (2-1.8)/(2+1) = 0.2/3 = 6.67%.

Statistically, both options yield the same expected return. However, I'd prefer the 10-play version, as it tends to have lower variance and I find it more enjoyable.

You're welcome! I appreciate your gratitude. Remember, the dice do not have memory. After four throws, there’s no increased likelihood of rolling a seven. It's quite possible to throw 1000 non-sevens and still be just as close or far from achieving a seven as you were on the first roll. There isn't a 'perfect' number of come bets; just play as many as you find enjoyable.

The chance of losing any single spin is 1-(8/37), which stands at 78.38%. Consequently, the probability of losing all 15 spins is .7838.¹⁵= 2.59%.