WOO logo

Wizard Recommends

Ask The Wizard #423

I heard that the outcome in electronic roulette games is predetermined by a random number generator. Then the game somehow gets the ball to land in the number it is supposed to go in. Perhaps it adjusts the wheel speed, uses magnets or blows air on the ball. What is really going on Wiz?

anonymous

I file this one in my very large file of casino conspiracy theories. There is absolutely no reason for any maker of electronic table games to do this. A skeptic might say the motive is to land the ball where players would lose the most. However, this would be illegal in most jurisdictions and easy for someone like me to test for. Double-zero roulette already has a 5.26% house advantage, which is plenty.

This topic comes up on my forum at Wizard of Vegas once in a while. Never once have I seen any data to back up this allegation, just anecdotal evidence at best.

Let's say there is a heterosexual dating event attended by m men and w women. Each person writes c names on a card of people they are interested to meet. After the event, the host will match up any pair of people who picked each other. What are the odds I get at least one match? You may assume that picks are random.

anonymous

Assuming you're male, the probability of any one woman picking you is c/m. So, the chances of any one woman not picking you is (m-c)/m. The chances that c women don't pick you is ((m-1)/m)c. The chances of at least one match is 1 less than this probability or 1-((m-1)/m)c. Note that the number of women playing doesn't matter. Just the ones you put on your card.

Let's look at an example. Suppose there are 50 men and five names per card. The probability of at least one match is 1-((50-5)/50)5 =~ 40.95%. By the way, this can be closely approximated as 1-e-(c2/m).

I saw this picture on Twitter. What are the odds of the dealer dealing himself eight aces?

 

DJTeddyBear

Let's assume a fresh six-deck shoe. There are combin(32,8)= 10,518,300 ways to choose eight aces out of the 32 in the shoe. There are 312 total cards in the shoe. There are combin(312,8) = 2,034,346,802,568,790 ways to choose any eight of them. Thus, the probability is 10,518,300 / 2,034,346,802,568,790 =~ 1/193,410,228.

Another way to solve it pr(first card is an ace) × pr(first card is an ace) × ... &times pr(eighth card is an ace) = (32/312) × (31/311) × ... × (25/305).

You receive an average of one Email every six minutes. The probability of receiving an Email any given moment is constant and independent of the time since the last Email was received. Every minute you roll a six-sided die, starting one minute from now. What is the probability you receive an Email before rolling a six?

Ace2

(1-e-1/6)/(1-(5/6)e-1/6) =~ 0.52111006

 

Here is my solution (PDF).

This question was asked and discussed in my forum at Wizard of Vegas.