Ask The Wizard #122

Absolutely! The crucial aspect of this scenario is that Monty is always aware of where the car is and thus will inevitably open a door that has a goat behind it. This ensures that, no matter what the players initially select, he can always show a goat first. This situation is known as the 'Monty Hall Paradox.' The confusion often arises because it's not always emphasized in the question that the host knows the location of the car and consistently reveals a goat first. Some of the miscommunication can be attributed to Marilyn Vos Savant , who poorly framed the query in her article. Let’s assume the car is behind door 1. Now let’s analyze the outcomes if the player (the second contestant) chooses not to switch.

• Player picks door 1 --> player wins
• Player picks door 2 --> player loses
• Player picks door 3 --> player loses

Now let’s examine the results if the player opts to switch doors.

• If the player goes with door 1, the host reveals a goat behind door 2 or door 3, prompting the player to switch to the other door, which results in a loss.
• If the player selects door 2, the host then uncovers a goat behind door 3, leading the player to switch to door 1, resulting in a win.
• If the player picks door 3, the host displays the goat behind door 2, and the player switches to door 1, resulting in another win.

Thus, when the player refrains from switching, their chance of winning remains at 1/3. In contrast, opting to switch provides a doubled winning chance of 2/3, which clearly suggests that switching is the optimal choice.

For more insights into the Monty Hall paradox, you might find the article at. Wikipedia .

I believe casinos still profit from this game due to the errors players often make. Moreover, there are variations of video poker here in Las Vegas that can return over 100% with the optimal strategy. Nevertheless, casinos rely on player mistakes to keep returns under that threshold. With most games offering returns exceeding 100%, the margin is usually so minor that it wouldn’t justify relying on it as a primary income source. Yet, if you're going to play anyway, it's wise to choose the games with the best odds.

I appreciate your kind remarks. I'm familiar with the game you mentioned. Just to illustrate, let’s say your initial hand is JJQQK and you decide to keep both jacks. To figure out how many ways you could draw one jack and two other cards, you'd calculate 2 times combin(45,2), which equals 1980. The combinations for pulling two jacks during the draw would tally up to 45. In terms of achieving a three of a kind from the draw, there would be 10 times 4 plus 1, giving you 41. Therefore, the total ways to enhance your hand into a three of a kind or better would be 1980 plus 45 plus 41, totaling 2066. The overall possibilities of drawing any three cards from the 47 remaining ones is combin(47,3), which is 16215. Thus, the probability of advancing your hand to three of a kind or better would be 2066 divided by 16215, which is roughly 12.74%. Conversely, if you decided to keep both pairs, your chances of improving to a full house would be 4/47, approximately 8.51%. This leads me to agree that retaining just one of the higher pairs is indeed the smarter strategy.

And I want to thank you for your kind words as well. For those unfamiliar with Texas Hold’em, this inquiry is similar to questioning how likely a player holding an ace and a king would be to pair up by the river when drawing five more random cards from the remaining 50. Out of those 50 cards, 44 are neither kings nor aces. The combinations of drawing five cards from these 44 would be combin(44,5), amounting to 1,086,088. Meanwhile, the total combinations for drawing five cards from the full 50 is combin(50,5), which equals 2,118,760. Hence, the probability of not pairing an ace or king would result in 1086088 divided by 2118760, approximately 51.26%. Therefore, the probability of successfully pairing is 1 minus 51.26%, giving around 48.74%. This figure is very close to 1 in 2.

Often, with progressive machines, a portion of every bet contributes to funding the following jackpot. This ensures that when one hits the jackpot, the next potential payout doesn’t start at a tiny sum but rather at a significant amount that has already been accrued. The percentage allocated toward this secondary jackpot isn’t always fixed; it can increase as the primary jackpot grows. By the way, although you didn’t inquire, some games, such as those at 'Be the Dealer', feature different jackpots depending on the denomination, with each jackpot reflecting that particular denomination's contributions. My theory is they utilize what I refer to as a 'super meter', to which all denominations contribute. Each individual denomination then receives a share of the super meter proportional to its value against the total of all contributions. For instance, suppose they have a progressive video poker machine with denominations of 5 cents, 25 cents, $1, and $5, and a super meter currently shows $100,000, then the meter for the $1 game would equate to (1/6.75)*100,000, which equals approximately $14,814.81.

I haven’t encountered this theory before, but I’m quite certain that they do not add caffeine to their beverages.

That's intriguing. Essentially, it appears to function as a two-way counter to assist players in keeping track of the running count in blackjack. From what I've gathered, it lacks true count conversion or indexed aid. Still, having knowledge of the running count and adjusting bets accordingly is certainly more advantageous than not tracking at all. It's a clever ruse as well. However, remember that employing any device for calculating probabilities in Nevada casinos is illegal and carries penalties that are similar to those for bank robbery.