Ask The Wizard #5

In the initial 20 rounds of blackjack, the chance of hitting exactly x victories can be calculated as (1/2), (20,x). Let me provide you with the probabilities for drawing 0 to 3 blackjacks among those first 20 rounds.²⁰ * combin Therefore, the chances of securing three or fewer blackjacks are 1 in 776. This statistic falls short of indicating any cheating behavior and may simply reflect unfortunate luck.

Pr(0) = 0.0000010
Pr(1) = 0.0000191
Pr(2) = 0.0001812
Pr(3) = 0.0010872
Pr(3 or less) = 0.0012884

Personally, I tend to become cautious only when I observe four standard deviations below the norm (which translates to a probability of 1 in 31,574), and I wouldn't formally accuse anyone unless I reach five standard deviations (with a probability of 1 in 3.5 million).

According to the blackjack software you referred to, both systems provide players with a minor advantage. If that's accurate, how do these casinos remain profitable when using such systems? Is it because a relatively low number of players apply optimal strategies and good gambling management?

Online casinos The dealer benefits since the player must act first. If both the player and dealer exceed 21, the round doesn't end in a tie; rather, the player loses.

Typically, single-deck blackjack rules are quite strict. Cards are dealt face down, and doubling is generally limited to 9 to 11 or 10 to 11. The dealer usually hits on a soft 17, and doubling after a split is often prohibited. Additionally, ensure you avoid games where blackjack pays less than 3 to 2, as this is common with single-deck tables.

If you roll six standard six-sided dice, what are the chances of achieving six of a kind?

Imagine a bingo session with 75 random cards. If 12 random numbers are drawn based on standard bingo guidelines, would the probability of someone getting a bingo be calculated as 75 * 0.00199521? (I derived the 0.00199521 from your standards on bingo probabilities for an occurrence within 12 drawn numbers.) If that's incorrect, what would the actual probability of achieving a bingo be? Your website is very informative. Wizard of Vegas site.

The answer is 6*(1/6)⁶ = 6/46,656 = 1/7,776 = 0.0001286 .

However, in the context of bingo, we cannot apply the above method, as all cards compete against the same set of drawn balls. It's a complex explanation, yet due to the configuration of the cards in five columns of 15 potential numbers each, the expected number of drawn balls is indeed interconnected. Conducting a random simulation would be necessary to accurately resolve your inquiry. Without such a simulation, I estimate the probability of a bingo happening to be approximately 13.9%. probabilities in bingo When I calculate the combinations of player and dealer hands for your specific scenario, I arrive at only 3,986,646,103,440 compared to your figure of 19, etc. There seems to be a discrepancy by exactly a factor of five. My calculation was based on (52,5)*combin(47,5). Could you help me identify my mistake? I appreciate your fantastic website.

Thanks for your kind words! You found a discrepancy due to the dealer being able to have any of the 5 cards facing up. This means the order matters in the dealer's hand since the first card is dealt face up. The correct way to calculate total combinations is combin(52,5)*47*combin(46,4), which equals 19,933,230,517,200.ⁿI was fortunate enough to hit four of a kind at a local casino, which led to an invitation to participate in a tournament with approximately 300 competitors vying for significant cash prizes. My query concerns what the best strategy would be in this context. Each participant is allocated $5,000 in play chips, with a minimum bet of $25 per hand. The tournament comprises different heats, starting with all but 100 players eliminated in the first round, then 25 in the second, six in the third, and finally determining the champion.⁷⁵= 1 - .9980048⁷⁵= 1 -.8608886 = .1391114.

Developing a strategy for table game tournaments can be quite intricate. In summary, I would focus on playing conservatively in the initial rounds. Often, opponents might exhaust their chips, allowing you to progress without much effort. As the rounds progress, particularly with about five hands remaining, you'll need to seize opportunities to challenge those ahead of you. This is the moment to push for a lead or risk busting while trying. It’s also strategic to plan your larger bets for when you are acting after the main competitors.

Next, we need to consider the combinations for scenario (2). There are 2 suits remaining to complement your existing pair. The number of ways to form a pair from the 3 ranks with 3 remaining cards is (3,2), and for the ranks with 4 remaining cards, it's combin(4,2). Hence, the total combinations for case (2) would be 2*(3*combin(3,2) + 9*combin(4,2)) = 2*(3*3 + 9*6) = 126. The overall number of ways to achieve a full house is the sum of the two cases: 39 + 126 = 165. Since there are combin(47,3) = 16,215 possible ways to arrange the three cards drawn on the second attempt, the probability of drawing a full house is calculated as the number of full house arrangements divided by the total combinations, giving us 165/16,215, or approximately 0.0101758, which is around 1 in 98.

Generally speaking, most communication at the table, including tipping, should be done with hand signals and chip placements. Most of the time, players will make bets on behalf of the dealer. To do this, simply position the tip on the edge of the betting area next to your own wager in the middle. The tip does not contribute to the table minimum as it is considered part of your bet, designated for the dealer. If you double or split your own bet, ensure you apply the same practice for the dealer's tip. When you win, the dealer will payout your bet and the tip separately. It's vital not to interfere with the tip or its winnings; allow the dealer to collect them. Once, I mistakenly attempted to take my tip and winnings back, forgetting that the tip was meant for the dealer, leading to a rather embarrassing moment when the dealer pointed it out.

Let's rephrase this inquiry. If the lottery encompasses 10 million combinations, and all players select their numbers randomly (acknowledging potential duplicates), how many tickets must be sold for the probability of at least one winner to reach 90%? Let p denote the winning probability, and n signify the number of tickets sold. The likelihood of one person losing is represented as 1-p, while the probability of everyone losing would be (1-p)^n. Therefore, the probability of having at least one winner is expressed as 1 - (1-p)^n and we need to equate this to 0.9, solving for n.

Thus, to ensure a 90% probability of at least one winner, the lottery must sell approximately 23,025,850 tickets. In case you are curious, had the lottery sold exactly 10 million tickets, the probability of at least one winner would approximate to 63.2%, calculated as approximately 1-(1/e).

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When analyzing the initial 20 blackjacks, the chance of getting exactly x blackjacks is (1/2),ⁿ(20,x). Allow me to save you some time. Here are the statistics for receiving exactly 0 to 3 blackjacks during the first 20 hands:ⁿConsequently, the likelihood of landing 3 or fewer blackjacks is 1 in 776. This figure isn’t alarming enough to accuse anyone of dishonest play. It might just be a case of unfortunate luck.

.9 = 1 - (1-p)ⁿ
.1 = (1-p)ⁿ
ln(.1) = ln((1-p)ⁿ)
ln(.1) = n*ln(1-p)
n = ln(.1)/ln(1-p)
n = ln(.1)/ln(.9999999)
n = 23,025,850.

Personally, I prefer to see outcomes that are four standard deviations lower than what is expected (which equates to a probability of 1 in 31,574) before becoming concerned. I would need to witness five standard deviations (with a probability of 1 in 3.5 million) before I would formally allege any cheating.