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What are the chances of 30 to 1?

With the standard 30-1 you'd get at most craps tables, bets on 2, 12 or hard hop bets give the house a 13.89 percent edge. If the payoff is 31-1, the edge drops only to 11.11 percent. True odds are 35-1, so there's a long way to go from 30-1 before these become viable bets.

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Gaming Odds Explained: 30 To 1 vs. 31 For 1

By John Grochowski on Thursday March 22, 2018

A craps player wrote to me recently to say he’d found an automated game that looked like it had better odds than he found at table games. “Instead of 30-1 on 2 or 12, it pays 31-1 and instead of 15-1 on 3 or 11 it pays 16-1,” he wrote. “Hardways pay 10-1 and 8-1 instead of 9-1 and 7-1.” I told him he needed to check the machine again. It was likely the machine was paying odds-for-1 instead of odds-to-1, and that the payoffs he was seeing were just normal returns expressed in a different way. It’s common for machine games to pay odds-for-1 because the game has already taken your bet before the roll takes place. On a table that pays 30-1, if you win your bet on 12, you keep your 1-unit bet and get 30 units in winnings, for a total of 31. On a machine that pays 31-for-1, the machine keeps your bet when you make it, but pays 31 units on a win — still a total of 31 on your side after a win. So 31-for-1 is the same as 30-to-1, 16-for-1 is the same as 15-to-1, 10-for-1 is the same as 9-to-1, and so on.

What if the game really was paying 31-1 on 12, and not 31-for-1?

With the standard 30-1 you’d get at most craps tables, bets on 2, 12 or hard hop bets give the house a 13.89 percent edge. If the payoff is 31-1, the edge drops only to 11.11 percent. True odds are 35-1, so there’s a long way to go from 30-1 before these become viable bets.

What about the hardway payoffs the reader described?

When your payoff is 9-1 on hard 6 or 8, the house edge is 9.09 percent, and when you’re paid 7-1 on hard 4 or 10, the house edge is 11.1 percent. If the pays were increased to 10-1 and 8-1. they’d match the true odds of winning these bets. The house edge would be zero. That reinforces the likelihood the game is paying in odds-for-1. Any time the odds paid on a machine version of a table game look higher than the table version, check to see if the payoffs are for-1.

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What are the odds of 2 people picking the same number out of 100?

The answer is 1/100. If both people pick a number, the number that person 1(or person 2, it doesn't matter) chooses does not matter. When that number is selected, person 2 has a 1/100 chance of picking the same number, no matter what number person 1 picked.

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I am doing an introductory course in statistics and I'm struggling with one question in particular. The question is: Suppose two people each have to select a number from 00 to 99 (therefore 100 possible choices).

(a) The probability that they both pick the number 13 is:

2/100 1/100 1/200 1/10 000 2/10 000

(b) The probability that both persons pick the same number is:

2/100 1/100 1/200 1/10 000 2/10 000

Ok, I understand the (a) part of the question. The probability for the first person to pick 13 is 1/100 and the probability for the second person to pick 13 is also 1/100. Since these are two independent events I use the rule P(A and B) = P(A) * P(B) = 1/10 000. The (b) part is where I am stuck at the moment. At first, I thought the probability is the same, but instead of only looking for the combination of both people picking 13 we now have to consider the probability of every combination of the same number that the two people can pick.

How do I approach this?

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