Posted: 2/11/2015 1:43:19 PM EDT
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These threads about the Powerball Lottery odds got me wondering. I went to Vegas and was eating lunch 1 day and decided to play Keno while I was eating. There are 80 numbers in Keno. I played a $5 ticket where you pick 20 numbers. There are different payouts according to how many numbers you select that match. It pays $500 if you don't get any that match. I played 3 games in a row where my 20 numbers failed to match any of the numbers drawn. What are the odds of that happening in 1 game much less 3 in a row? |
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If you played 20 numbers out of 80, then you had a 1 in 4 chance of getting at least 1 number. The probability is the same each time you play, regardless of how many times you play. So you have a 3/4 chance of losing every time.
Eta I Idk how many numbers are drawn in keno, but for one number drawn, the above would bE correct. |
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If you played 20 numbers out of 80, then you had a 1 in 4 chance of getting at least 1 number. The probability is the same each time you play, regardless of how many times you play. So you have a 3/4 chance of losing every time. Eta I Idk how many numbers are drawn in keno, but for one number drawn, the above would bE correct. They draw 20 of the 80 numbers each game I picked 20 They drew 0 of mine out of 80 It paid 100 to 1 (I bet $5 and won $500) but Vegas dosent pay true odds on Keno This happened 3 times in a row |
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They draw 20 of the 80 numbers each game I picked 20 They drew 0 of mine out of 80 It paid 100 to 1 (I bet $5 and won $500) but Vegas dosent pay true odds on Keno This happened 3 times in a row Quoted:
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If you played 20 numbers out of 80, then you had a 1 in 4 chance of getting at least 1 number. The probability is the same each time you play, regardless of how many times you play. So you have a 3/4 chance of losing every time. Eta I Idk how many numbers are drawn in keno, but for one number drawn, the above would bE correct. They draw 20 of the 80 numbers each game I picked 20 They drew 0 of mine out of 80 It paid 100 to 1 (I bet $5 and won $500) but Vegas dosent pay true odds on Keno This happened 3 times in a row First number: there are 60 numbers out of 80 possible that will not be on your list Second number: there are 59 numbers out of 79 possible that will not be on your list Third number: there are 58 numbers out of 78 that will not be on your list (60/80)*(59/79)*(58/78) .... (41/61) = the probability of that happening once 3 times in a row? I don't know, they didn't cover that any of the six times I took statistics 
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The odds of winning should be posted on the ticket. Quoted:
The odds of winning should be posted on the ticket. This. But for sake of teaching, let's work through how to solve this. There are 80 numbers in Keno.
I played a $5 ticket where you pick 20 numbers. There are different payouts according to how many numbers you select that match. It pays $500 if you don't get any that match. I played 3 games in a row where my 20 numbers failed to match any of the numbers drawn. What are the odds of that happening in 1 game much less 3 in a row? You have a set of 80 discrete numbers. You are required to pick 20 different numbers from that set. Order doesn't matter, and no duplicates can be selected. There are 60 possible "good" choices, out of 80 total numbers. On your first pick, the odds of selecting a number that is not one of your 20 is 60/80. Since we picked a "good" choice last roll, there's only 59 "good" choices left, out of 79 numbers left. On your second pick, the odds of selecting a number that is not on of your 20 is 59/79. Do this 20 times. In stats, when you're trying to determine the odds of multiple events happening, you either use Addition or Multiplication. If you trying to figure out the odds of A happening OR B happening, you use Addition. If trying to figure out the odds of A AND B, you use Multiplication. OR = + AND = * Based on this, we'll be multiplying all of our 20 results together. I'll spare you the math and tell you that the answer is 0.0011857057013244957, or 0.1186%. That's how we figure out the odds of winning 1 game. The odds of winning all 3 games can be phrased like this: What are the odds that I win game 1 AND game 2 AND game 3, or 0.001186 * 0.001186 * 0.001186 = 0.000000001668222856. Or 0.0000001668222856%. To convert that into something more understandable, take the inverse of the number (not the percent). That's 1/0.000000001668222856 or 1 in 599,440,294. For some perspective, the odds of winning the big Powerball prize is 1 in 175,223,510. Sounds like you played the wrong game. 
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This. But for sake of teaching, let's work through how to solve this. You have a set of 80 discrete numbers. You are required to pick 20 different numbers from that set. Order doesn't matter, and no duplicates can be selected. There are 60 possible "good" choices, out of 80 total numbers. On your first pick, the odds of selecting a number that is not one of your 20 is 60/80. Since we picked a "good" choice last roll, there's only 59 "good" choices left, out of 79 numbers left. On your second pick, the odds of selecting a number that is not on of your 20 is 59/79. Do this 20 times. In stats, when you're trying to determine the odds of multiple events happening, you either use Addition or Multiplication. If you trying to figure out the odds of A happening OR B happening, you use Addition. If trying to figure out the odds of A AND B, you use Multiplication. OR = + AND = * Based on this, we'll be multiplying all of our 20 results together. I'll spare you the math and tell you that the answer is 0.0011857057013244957, or 0.1186%. That's how we figure out the odds of winning 1 game. The odds of winning all 3 games can be phrased like this: What are the odds that I win game 1 AND game 2 AND game 3, or 0.001186 * 0.001186 * 0.001186 = 0.000000001668222856. Or 0.0000001668222856%. To convert that into something more understandable, take the inverse of the number (not the percent). That's 1/0.000000001668222856 or 1 in 599,440,294. For some perspective, the odds of winning the big Powerball prize is 1 in 175,223,510. Sounds like you played the wrong game. Quoted:
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The odds of winning should be posted on the ticket. This. But for sake of teaching, let's work through how to solve this. There are 80 numbers in Keno.
I played a $5 ticket where you pick 20 numbers. There are different payouts according to how many numbers you select that match. It pays $500 if you don't get any that match. I played 3 games in a row where my 20 numbers failed to match any of the numbers drawn. What are the odds of that happening in 1 game much less 3 in a row? You have a set of 80 discrete numbers. You are required to pick 20 different numbers from that set. Order doesn't matter, and no duplicates can be selected. There are 60 possible "good" choices, out of 80 total numbers. On your first pick, the odds of selecting a number that is not one of your 20 is 60/80. Since we picked a "good" choice last roll, there's only 59 "good" choices left, out of 79 numbers left. On your second pick, the odds of selecting a number that is not on of your 20 is 59/79. Do this 20 times. In stats, when you're trying to determine the odds of multiple events happening, you either use Addition or Multiplication. If you trying to figure out the odds of A happening OR B happening, you use Addition. If trying to figure out the odds of A AND B, you use Multiplication. OR = + AND = * Based on this, we'll be multiplying all of our 20 results together. I'll spare you the math and tell you that the answer is 0.0011857057013244957, or 0.1186%. That's how we figure out the odds of winning 1 game. The odds of winning all 3 games can be phrased like this: What are the odds that I win game 1 AND game 2 AND game 3, or 0.001186 * 0.001186 * 0.001186 = 0.000000001668222856. Or 0.0000001668222856%. To convert that into something more understandable, take the inverse of the number (not the percent). That's 1/0.000000001668222856 or 1 in 599,440,294. For some perspective, the odds of winning the big Powerball prize is 1 in 175,223,510. Sounds like you played the wrong game. Dang dude that's some impressive calculations. Thanks. I always figured the odds were phenomenally high but would never had expected 1 in 600 million. I used up enough luck in an hour to win the Powerball 3 times.  
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Dang dude that's some impressive calculations. Thanks. I always figured the odds were phenomenally high but would never had expected 1 in 600 million. I used up enough luck in an hour to win the Powerball 3 times. Quoted:
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The odds of winning should be posted on the ticket. This. But for sake of teaching, let's work through how to solve this. <snip math> Dang dude that's some impressive calculations. Thanks. I always figured the odds were phenomenally high but would never had expected 1 in 600 million. I used up enough luck in an hour to win the Powerball 3 times. I use to play tabletop wargames, specifically 40k. Eventually you figure out how to calculate the odds that you'll hit, wound, and kill the unit you're shooting at, and whether that's good enough to risk it.
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| Keno is probably rigged - since all of your numbers have to be input BEFORE the game begins, I'm guessing that the computer system calculates the lowest possible payout among those playing. Every now and then, it throws out a big winner to act as selective reinforcement. |
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Keno is probably rigged - since all of your numbers have to be input BEFORE the game begins, I'm guessing that the computer system calculates the lowest possible payout among those playing. Every now and then, it throws out a big winner to act as selective reinforcement. The odds of someone winning are prefigured on all the games in Vegas and favor the house. There is absolutely no reason for them to cheat. |
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The odds of someone winning are prefigured on all the games in Vegas and favor the house. There is absolutely no reason for them to cheat. Quoted:
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Keno is probably rigged - since all of your numbers have to be input BEFORE the game begins, I'm guessing that the computer system calculates the lowest possible payout among those playing. Every now and then, it throws out a big winner to act as selective reinforcement. The odds of someone winning are prefigured on all the games in Vegas and favor the house. There is absolutely no reason for them to cheat. It all depends on the legal definition of "cheat". It has been a while since I read the laws here in Maryland concerning slot machines, but I believe the only requirement is that they pay out occasionally. Other areas probably differ, and the odds will always favor the house. |
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I had to take a probability class to get my undergrad EE degree. The professor loved this little mental test:
Imagine that you are on Let's Make a Deal and Monty Hall tells you that you can pick either door #1, #2, or #3 and that behind one of them is a new car and behind the other two are goats. You choose door #2 but before he opens it, he tells you that he is going to show you where one of the goats is and opens door #1 -- revealing a goat. Do you have a higher, lower, or the same probability of picking the door with the car behind it if you change your guess from door #2 to door #3? Show your work. |
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I had to take a probability class to get my undergrad EE degree. The professor loved this little mental test: Imagine that you are on Let's Make a Deal and Monty Hall tells you that you can pick either door #1, #2, or #3 and that behind one of them is a new car and behind the other two are goats. You choose door #2 but before he opens it, he tells you that he is going to show you where one of the goats is and opens door #1 -- revealing a goat. Do you have a higher, lower, or the same probability of picking the door with the car behind it if you change your guess from door #2 to door #3? Show your work. Oh lord. Plane on a treadmill time.
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Oh lord. Plane on a treadmill time. Quoted:
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I had to take a probability class to get my undergrad EE degree. The professor loved this little mental test: Imagine that you are on Let's Make a Deal and Monty Hall tells you that you can pick either door #1, #2, or #3 and that behind one of them is a new car and behind the other two are goats. You choose door #2 but before he opens it, he tells you that he is going to show you where one of the goats is and opens door #1 -- revealing a goat. Do you have a higher, lower, or the same probability of picking the door with the car behind it if you change your guess from door #2 to door #3? Show your work. Oh lord. Plane on a treadmill time. There is a real answer here --- provable with math and everything! |
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I had to take a probability class to get my undergrad EE degree. The professor loved this little mental test: Imagine that you are on Let's Make a Deal and Monty Hall tells you that you can pick either door #1, #2, or #3 and that behind one of them is a new car and behind the other two are goats. You choose door #2 but before he opens it, he tells you that he is going to show you where one of the goats is and opens door #1 -- revealing a goat. Do you have a higher, lower, or the same probability of picking the door with the car behind it if you change your guess from door #2 to door #3? Show your work. You have been watching the movie 21 haven't you? |
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You have been watching the movie 21 haven't you? Quoted:
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I had to take a probability class to get my undergrad EE degree. The professor loved this little mental test: Imagine that you are on Let's Make a Deal and Monty Hall tells you that you can pick either door #1, #2, or #3 and that behind one of them is a new car and behind the other two are goats. You choose door #2 but before he opens it, he tells you that he is going to show you where one of the goats is and opens door #1 -- revealing a goat. Do you have a higher, lower, or the same probability of picking the door with the car behind it if you change your guess from door #2 to door #3? Show your work. You have been watching the movie 21 haven't you? Never heard of it. Like I said, I took a class. Did 21 come out before I took the probability class in 2001? |
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Never heard of it. Like I said, I took a class. Did 21 come out before I took the probability class in 2001? Quoted:
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I had to take a probability class to get my undergrad EE degree. The professor loved this little mental test: Imagine that you are on Let's Make a Deal and Monty Hall tells you that you can pick either door #1, #2, or #3 and that behind one of them is a new car and behind the other two are goats. You choose door #2 but before he opens it, he tells you that he is going to show you where one of the goats is and opens door #1 -- revealing a goat. Do you have a higher, lower, or the same probability of picking the door with the car behind it if you change your guess from door #2 to door #3? Show your work. You have been watching the movie 21 haven't you? Never heard of it. Like I said, I took a class. Did 21 come out before I took the probability class in 2001? Movie came out in 2008. In the movie Kevin Spacey asks the same question to his students in a class at MIT. The movie is about the MIT blackjack team. Good flick. And the answer is not what I would have deduced. Did you get it right? |
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Movie came out in 2008. In the movie Kevin Spacey asks the same question to his students in a class at MIT. The movie is about the MIT blackjack team. Good flick. And the answer is not what I would have deduced. Did you get it right? Quoted:
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I had to take a probability class to get my undergrad EE degree. The professor loved this little mental test: Imagine that you are on Let's Make a Deal and Monty Hall tells you that you can pick either door #1, #2, or #3 and that behind one of them is a new car and behind the other two are goats. You choose door #2 but before he opens it, he tells you that he is going to show you where one of the goats is and opens door #1 -- revealing a goat. Do you have a higher, lower, or the same probability of picking the door with the car behind it if you change your guess from door #2 to door #3? Show your work. You have been watching the movie 21 haven't you? Never heard of it. Like I said, I took a class. Did 21 come out before I took the probability class in 2001? Movie came out in 2008. In the movie Kevin Spacey asks the same question to his students in a class at MIT. The movie is about the MIT blackjack team. Good flick. And the answer is not what I would have deduced. Did you get it right? Oh - I think then that the film was based on the book "Bringing Down the House" which I did read. I never imagined that he came up with the problem himself so I am not surprised if it is out there and gets around in "geeky" circles. To answer your question -- I got it wrong. He asked on the first day and I had no idea about the math behind the problem. |
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Movie came out in 2008. In the movie Kevin Spacey asks the same question to his students in a class at MIT. The movie is about the MIT blackjack team. Good flick. And the answer is not what I would have deduced. Did you get it right? Quoted:
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I had to take a probability class to get my undergrad EE degree. The professor loved this little mental test: Imagine that you are on Let's Make a Deal and Monty Hall tells you that you can pick either door #1, #2, or #3 and that behind one of them is a new car and behind the other two are goats. You choose door #2 but before he opens it, he tells you that he is going to show you where one of the goats is and opens door #1 -- revealing a goat. Do you have a higher, lower, or the same probability of picking the door with the car behind it if you change your guess from door #2 to door #3? Show your work. You have been watching the movie 21 haven't you? Never heard of it. Like I said, I took a class. Did 21 come out before I took the probability class in 2001? Movie came out in 2008. In the movie Kevin Spacey asks the same question to his students in a class at MIT. The movie is about the MIT blackjack team. Good flick. And the answer is not what I would have deduced. Did you get it right? I took statistics around the same time and had the same question. |
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There is a real answer here --- provable with math and everything! Quoted:
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I had to take a probability class to get my undergrad EE degree. The professor loved this little mental test: Imagine that you are on Let's Make a Deal and Monty Hall tells you that you can pick either door #1, #2, or #3 and that behind one of them is a new car and behind the other two are goats. You choose door #2 but before he opens it, he tells you that he is going to show you where one of the goats is and opens door #1 -- revealing a goat. Do you have a higher, lower, or the same probability of picking the door with the car behind it if you change your guess from door #2 to door #3? Show your work. Oh lord. Plane on a treadmill time. There is a real answer here --- provable with math and everything! You always switch. Showing you a door doesn't change the original probability. Your first door has 1/3 of winning, the other two have 2/3 chance of winning ... even if you have seen a goat in one of the other two. The above is an insane mind bender. One well known mathematician even argued that switching didn't matter until he ran a simulation. |
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You always switch. Showing you a door doesn't change the original probability. Your first door has 1/3 of winning, the other two have 2/3 chance of winning ... even if you have seen a goat in one of the other two. The above is an insane mind bender. One well known mathematician even argued that switching didn't matter until he ran a simulation. Quoted:
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I had to take a probability class to get my undergrad EE degree. The professor loved this little mental test: Imagine that you are on Let's Make a Deal and Monty Hall tells you that you can pick either door #1, #2, or #3 and that behind one of them is a new car and behind the other two are goats. You choose door #2 but before he opens it, he tells you that he is going to show you where one of the goats is and opens door #1 -- revealing a goat. Do you have a higher, lower, or the same probability of picking the door with the car behind it if you change your guess from door #2 to door #3? Show your work. Oh lord. Plane on a treadmill time. There is a real answer here --- provable with math and everything! You always switch. Showing you a door doesn't change the original probability. Your first door has 1/3 of winning, the other two have 2/3 chance of winning ... even if you have seen a goat in one of the other two. The above is an insane mind bender. One well known mathematician even argued that switching didn't matter until he ran a simulation. I never quite understood this. Once you know what is behind one of the three doors, how is the probability of winning anything other than 1/2 for either of the unopened doors? |
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I never quite understood this. Once you know what is behind one of the three doors, how is the probability of winning anything other than 1/2 for either of the unopened doors? Quoted:
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I had to take a probability class to get my undergrad EE degree. The professor loved this little mental test: Imagine that you are on Let's Make a Deal and Monty Hall tells you that you can pick either door #1, #2, or #3 and that behind one of them is a new car and behind the other two are goats. You choose door #2 but before he opens it, he tells you that he is going to show you where one of the goats is and opens door #1 -- revealing a goat. Do you have a higher, lower, or the same probability of picking the door with the car behind it if you change your guess from door #2 to door #3? Show your work. Oh lord. Plane on a treadmill time. There is a real answer here --- provable with math and everything! You always switch. Showing you a door doesn't change the original probability. Your first door has 1/3 of winning, the other two have 2/3 chance of winning ... even if you have seen a goat in one of the other two. The above is an insane mind bender. One well known mathematician even argued that switching didn't matter until he ran a simulation. I never quite understood this. Once you know what is behind one of the three doors, how is the probability of winning anything other than 1/2 for either of the unopened doors? It makes more sense when you see a visual. It's because the host of the game is influenced by what door he must open, rather than being random. 
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