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............................... is because the winning expectation grows with each failure.....................................

Your math and logic are misplaced on this false assumption. Since you are always drawing from the same 42 # pool, previous failures have no bearing on any possible winning expectation. The odds of winning are still 1/42. Also 1/42 is not typed to represent a fraction but is to be shorthand for 1 in 42.
 
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AlexB said:
............................... is because the winning expectation grows with each failure.....................................

Your math and logic are misplaced on this false assumption. Since you are always drawing from the same 42 # pool, previous failures have no bearing on any possible winning expectation. The odds of winning are still 1/42. Also 1/42 is not typed to represent a fraction but is to be shorthand for 1 in 42.
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do u know the difference between odds and probability? odds are not represented in the form 1 in 42. they are expressed in the form 2 to 1 or in this case 1 to 41 as in 1 win to 41 losses.

my math and logic are not misplaced. ur inability to do proper math is. if everything remains constant, the expectation of of winning increases with every failure. thats why if u talk to professional gamblers who play carribean stud, they do not bet on the progressive jackpot unless it's over 200000 dollars. because the expectation of winning that jackpot increases as the jackpot increases after every failure. the probability means how probable it is that you will win. so it is more probable that u will win after every failure if everything remains constant. like if u played rock paper scissors and used rock every time. u should win once every 3 games and over a long period of time that should be expected. here, you have a 1 in 42 chance to win. but since u get 3 chances to win, u have a 3 in 42 chance to win since ur number remains constant. honestly man if u cant do this then refer the problem to a math teacher. you cant even tell the difference between odds and probability. u use those two interchangably when they are different. previous failures have no bearing on the future, that's why i said ur probability of winning is 1/14. i didnt say that u will definitely win if they ran this 14 times. i said u are expected to win once for every 14 times this set of 3 drawings occurs - if u keep the same number every time. just take this problem to a math teacher. he'll tell u the same thing i am saying.
 
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AlexB and Danktrees: You both need some help with the math.

I think the probability you are trying to determine is the probability of winning at least one time in three drawings from 42 numbers. The easiest way to calculate this is one minus the probability of winning none of the drawings, or 1 - [(41/42) x (41/42) x (41/42)] = 1 - (68,921/74,088) = 5,167/74,088, or approximately .06974.

This same probability (of at least one win) could also be determined by summing three other probabilities -- the probability of exactly one win, the probability of exactly two wins, and the probability of exactly three wins. These three probabilities are, respectively, 5,043/74,088, 123/74,088, and 1/74,088. The sum of the three is 5,167/74,088, the same answer we got the shorter way above.

This problem is equivalent to rolling a 42-sided die three times and asking what is the probability of rolling a "31" (or pick any other number from 1 to 42) at least once.

Or, even more simply, suppose there are only two numbers instead of 42. That's equivalent to flipping a coin. So what is the probability of at least one head in three flips of a coin? AlexB -- the logic you used for the raffle probability would say the answer is still 1/2. Danktrees, the logic you used would say it is 3 x (1/2) = 1.5!!! The actual answer is 7/8, which can be calculated as one minus the probability of no heads, or 1 - (1/2) x (1/2) x (1/2) = 7/8. Alternatively, you can enumerate the seven ways to obtain at least one head: HTT, THT, TTH, HHT, HTH, THH, HHH. Each of these ways has a probability of 1/8 and their sum is 7/8.

I hope this helps.
 
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AtLarge said:
AlexB and Danktrees: You both need some help with the math.

I think the probability you are trying to determine is the probability of winning at least one time in three drawings from 42 numbers. The easiest way to calculate this is one minus the probability of winning none of the drawings, or 1 - [(41/42) x (41/42) x (41/42)] = 1 - (68,921/74,088) = 5,167/74,088, or approximately .06974.

This same probability (of at least one win) could also be determined by summing three other probabilities -- the probability of exactly one win, the probability of exactly two wins, and the probability of exactly three wins. These three probabilities are, respectively, 5,043/74,088, 123/74,088, and 1/74,088. The sum of the three is 5,167/74,088, the same answer we got the shorter way above.

This problem is equivalent to rolling a 42-sided die three times and asking what is the probability of rolling a "31" (or pick any other number from 1 to 42) at least once.

Or, even more simply, suppose there are only two numbers instead of 42. That's equivalent to flipping a coin. So what is the probability of at least one head in three flips of a coin? AlexB -- the logic you used for the raffle probability would say the answer is still 1/2. Danktrees, the logic you used would say it is 3 x (1/2) = 1.5!!! The actual answer is 7/8, which can be calculated as one minus the probability of no heads, or 1 - (1/2) x (1/2) x (1/2) = 7/8. Alternatively, you can enumerate the seven ways to obtain at least one head: HTT, THT, TTH, HHT, HTH, THH, HHH. Each of these ways has a probability of 1/8 and their sum is 7/8.

I hope this helps.
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That's weird......I was gonna say the exact same thing.

Go figure......

:)
 
Atlarge

Thanks for the explanation and the work involved. I will think about it before being totally convinced.
 
AtLarge said:
AlexB and Danktrees: You both need some help with the math.

I think the probability you are trying to determine is the probability of winning at least one time in three drawings from 42 numbers. The easiest way to calculate this is one minus the probability of winning none of the drawings, or 1 - [(41/42) x (41/42) x (41/42)] = 1 - (68,921/74,088) = 5,167/74,088, or approximately .06974.

This same probability (of at least one win) could also be determined by summing three other probabilities -- the probability of exactly one win, the probability of exactly two wins, and the probability of exactly three wins. These three probabilities are, respectively, 5,043/74,088, 123/74,088, and 1/74,088. The sum of the three is 5,167/74,088, the same answer we got the shorter way above.

This problem is equivalent to rolling a 42-sided die three times and asking what is the probability of rolling a "31" (or pick any other number from 1 to 42) at least once.

Or, even more simply, suppose there are only two numbers instead of 42. That's equivalent to flipping a coin. So what is the probability of at least one head in three flips of a coin? AlexB -- the logic you used for the raffle probability would say the answer is still 1/2. Danktrees, the logic you used would say it is 3 x (1/2) = 1.5!!! The actual answer is 7/8, which can be calculated as one minus the probability of no heads, or 1 - (1/2) x (1/2) x (1/2) = 7/8. Alternatively, you can enumerate the seven ways to obtain at least one head: HTT, THT, TTH, HHT, HTH, THH, HHH. Each of these ways has a probability of 1/8 and their sum is 7/8.

I hope this helps.
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What?! AlexB isn't correct? Imagine that:rolleyes:

BTW can you break that down and just tell us what the chances/probabilities are for this raffle?:D
 
AtLarge said:
AlexB and Danktrees: You both need some help with the math.

I think the probability you are trying to determine is the probability of winning at least one time in three drawings from 42 numbers. The easiest way to calculate this is one minus the probability of winning none of the drawings, or 1 - [(41/42) x (41/42) x (41/42)] = 1 - (68,921/74,088) = 5,167/74,088, or approximately .06974.

This same probability (of at least one win) could also be determined by summing three other probabilities -- the probability of exactly one win, the probability of exactly two wins, and the probability of exactly three wins. These three probabilities are, respectively, 5,043/74,088, 123/74,088, and 1/74,088. The sum of the three is 5,167/74,088, the same answer we got the shorter way above.

This problem is equivalent to rolling a 42-sided die three times and asking what is the probability of rolling a "31" (or pick any other number from 1 to 42) at least once.

Or, even more simply, suppose there are only two numbers instead of 42. That's equivalent to flipping a coin. So what is the probability of at least one head in three flips of a coin? AlexB -- the logic you used for the raffle probability would say the answer is still 1/2. Danktrees, the logic you used would say it is 3 x (1/2) = 1.5!!! The actual answer is 7/8, which can be calculated as one minus the probability of no heads, or 1 - (1/2) x (1/2) x (1/2) = 7/8. Alternatively, you can enumerate the seven ways to obtain at least one head: HTT, THT, TTH, HHT, HTH, THH, HHH. Each of these ways has a probability of 1/8 and their sum is 7/8.

I hope this helps.
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I have to rep this post.

Nice cues in the raffle too, but I think you would have a better shot at filling 3 individual raffles instead. I love a nice wrapless cue. Just my honest opinion.
 
Pii said:
What?! AlexB isn't correct? Imagine that:rolleyes:

BTW can you break that down and just tell us what the chances/probabilities are for this raffle?:D
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You lied. You said previously that I was on "ignore." I knew you couldn't resist me.
 
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