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#1
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Re: *** Official joining every $4/180man until I win one ***
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I think 50% to succeed sounds about correct for one session of all of them. [/ QUOTE ] Considering that you've won exactly 1 out of the 208 4/180s you've played, a reasonable guess for the number you'd have to play to have a 50% chance of winning 1 is about 144. Quite a session. |
#2
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Re: *** Official joining every $4/180man until I win one ***
[ QUOTE ]
[ QUOTE ] I think 50% to succeed sounds about correct for one session of all of them. [/ QUOTE ] Considering that you've won exactly 1 out of the 208 4/180s you've played, a reasonable guess for the number you'd have to play to have a 50% chance of winning 1 is about 144. Quite a session. [/ QUOTE ] That can't be right at all IMO. If OP has a clue what he is doing, he ought to be 50% to win 1 in 90 or less 180s. Look, if the MTT was a lottery he would win 1 in 180 on average. So if he was playing based on pure luck alone he ought to be 50% to win 1 in 90. Because OP posts on 2+2 alone makes him a much bigger favorite than that to win one. I'd say he is at least 50% to win 1 in 60 and close to 100% (on average) to win 1 in 120. Sherman |
#3
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Re: *** Official joining every $4/180man until I win one ***
[ QUOTE ]
[ QUOTE ] Considering that you've won exactly 1 out of the 208 4/180s you've played, a reasonable guess for the number you'd have to play to have a 50% chance of winning 1 is about 144. Quite a session. [/ QUOTE ] That can't be right at all IMO. If OP has a clue what he is doing, he ought to be 50% to win 1 in 90 or less 180s. Look, if the MTT was a lottery he would win 1 in 180 on average. So if he was playing based on pure luck alone he ought to be 50% to win 1 in 90. Because OP posts on 2+2 alone makes him a much bigger favorite than that to win one. I'd say he is at least 50% to win 1 in 60 and close to 100% (on average) to win 1 in 120. Sherman [/ QUOTE ] I was basing my calculation on his actual record in 4/180s (1 win in 208 attempts). I agree that it's entirely possible -- even likely -- that his actual chance of winning is better than 1/208. If his chance to win was 1/180, he would be 50% to win at least 1 if he played 125. *NOT* 90, as you suggested. |
#4
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Re: *** Official joining every $4/180man until I win one ***
Forgive me, but this makes no sense to me.
The probabilities ought to be additive. That is, if his chance of winning a single 180-person MTT is 1/180, he chances of winning 1 of 2 180-person MTTs ought to be 2/180. And so forth. So his probability of winning 1 180 person tournament given 90 tournaments ought to be exactly 90/180 or 50%. Please explain how I am wrong. Sherman PS - FWIW, I know what you were doing by using his current data, I am just pointing out that unless he is a losing player (which he probably isn't) or is trying to lose, his probability of winning a single 180 tournament is surely better than 1 out of 200+. |
#5
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Re: *** Official joining every $4/180man until I win one ***
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The probabilities ought to be additive. [/ QUOTE ] Not correct. Each tournament is an independent event. Just because you didn't win 179 in a row doesn't mean you're guaranteed to win #180. As with any other grouping of independent events you have to calculate the risk of not winning one specific tournament: NW(1) = (180 - 1)/180 = 179/180 = 0.9944444 .... From there you can calculate the risk of not winning any of X tournaments: NW(X) = NW(1) raised to the power X = NW(1) * ... * NW(1) (that reads: NW(1) multiplied with itself X times). This gives us the chance of winning (at least) one out of X tournaments: CW = 1 - NW(X) We're looking for the number of (random outcome) 180/$4.40 to play to have exactly 50% chance of winning one: X = log(0.5) / log(NW(1)) = 124.4 So to have 50% chance of winning (at least) one tournament in a completely random universe (which doesn't seem too far off when it comes to $4.40 buy-ins) you have to play a little more than 124 of them. |
#6
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Re: *** Official joining every $4/180man until I win one ***
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[ QUOTE ] The probabilities ought to be additive. [/ QUOTE ] Not correct. Each tournament is an independent event. [/ QUOTE ] LOL. I'm an idiot. U win. |
#7
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Re: *** Official joining every $4/180man until I win one ***
if my math teacher would have told me that math could help me in poker, I would have actually paid attention.
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#8
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Re: *** Official joining every $4/180man until I win one ***
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So to have 50% chance of winning (at least) one tournament in a completely random universe (which doesn't seem too far off when it comes to $4.40 buy-ins) you have to play a little more than 124 of them. [/ QUOTE ] This is suggesting that I'm approximately 50% to win one if I play 124 in a row? That sounds about right. |
#9
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Re: *** Official joining every $4/180man until I win one ***
[ QUOTE ]
[ QUOTE ] So to have 50% chance of winning (at least) one tournament in a completely random universe (which doesn't seem too far off when it comes to $4.40 buy-ins) you have to play a little more than 124 of them. [/ QUOTE ] This is suggesting that I'm approximately 50% to win one if I play 124 in a row? That sounds about right. [/ QUOTE ] It's right if you really suck...somehow I doubt you do. IMO, you ought to be able to win 1 in 90-150 on average. |
#10
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Re: *** Official joining every $4/180man until I win one ***
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I'd say he is at least 50% to win 1 in 60 [/ QUOTE ] To be 50% to win 1 (or more) out of 60, you have to win 1.15% of the time. That's a little bit over 2 out of 180. |
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