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Old 04-22-2009, 08:09 PM   #126
LargeLouster
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by UbinTook View Post
What is the correct method determining the odds relative to hands played?
You just divide the odds of the event happening by the number of chances you get at it. Odds of AA on consecutive hands are 1/221^2=48,880:1; so if you play 50 hands, you're getting 49 chances of getting aces twice in a row, so 48,880/49=~996:1. One way to understand it is to realize that you playing 50 hands is the same as if 49 guys played 2 consecutive hands one time each. I'm not sure how DocOfDan got his numbers, but I believe they are incorrect, since, for example, for 100 hands it would be 48,880/99=~492:1.
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Old 04-23-2009, 03:01 PM   #127
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Re: How is being delt aces twice in a row not over 40,000-to-1?

Thanks, here is where my thinking may be wrong...in 100 trials, there are 50 chances of getting AA TWICE correct?
so 100 trials would be 48841/50 or 916:1
hand 1+ hand 2 = chance 1
hand 2+ hand 3 = chance 2
etc
And if the number of hands played say (or is going to be played) is an odd number, then the number of chances is (n-1)/2 or 99-1/2=48 or 1017:1
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Old 04-23-2009, 04:03 PM   #128
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by LargeLouster View Post
You just divide the odds of the event happening by the number of chances you get at it. Odds of AA on consecutive hands are 1/221^2=48,880:1; so if you play 50 hands, you're getting 49 chances of getting aces twice in a row, so 48,880/49=~996:1. One way to understand it is to realize that you playing 50 hands is the same as if 49 guys played 2 consecutive hands one time each. I'm not sure how DocOfDan got his numbers, but I believe they are incorrect, since, for example, for 100 hands it would be 48,880/99=~492:1.
this doesnt seem correct. Say you were dealt more than 48880 hands. According to your method as you play more hands the chances of getting dealt aces becomes less as you surpass the 48880 mark.

I look at the problem as a binomial distribution which i think is really the correct way to do it. The binomial distribution tells you the chances of gettin n success in x trials given the probability of success is p. that seems like the simplest way to do this.
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Old 04-23-2009, 04:27 PM   #129
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by drzen View Post
Mason, I would like to offer you a bet. I will deal you two hands in a row. If both are aces, I will pay you 1000/1. Given such a hugely +EV bet, you will surely want to put some money on, amirite?
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If you meant that if the two hands were not AA, he had to pay you, he'd have to be crazy to accept.
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I'm gonna enroll in a typing class.
You might also want to see if they offer a course on reading between the lines. Mathematical corrections aside, most of the debate itt can be understood as a battle between those who can (and are willing to) do this and those who can't (or won't).

It's a point of irony that drzen (who is probably a great lover of paradox) chooses to convey his point, which is in support of the opinion of the latter group, in a way that is more likely to be understood by the former.
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Old 04-23-2009, 05:07 PM   #130
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by junwagh View Post
this doesnt seem correct. Say you were dealt more than 48880 hands. According to your method as you play more hands the chances of getting dealt aces becomes less as you surpass the 48880 mark.

I look at the problem as a binomial distribution which i think is really the correct way to do it. The binomial distribution tells you the chances of gettin n success in x trials given the probability of success is p. that seems like the simplest way to do this.
According to my method, the more hands you play, the more likely it is to happen. In fact, when you are dealt hand #48,882, the probability becomes 1.0, since you now have had 48,881 chances of something that has a probability of 1/48,881, and (1/48,841)/48,841=1. However, the probability of being dealt aces twice in a row says nothing about when it will occur, simply that the more hands you deal (The Law Of Large Numbers, as distinguished from The Law Of Large Lousters ), the closer and closer your actual results will approach the mathematical expression 1/48,881. You could just as easily get AA on the first two hands you are dealt as you could go 48,881 hands without even getting AA once.

Consider a simpler problem, flipping a coin. The probability of heads is 1/2, so mathematically, the probability of heads is 1.0 every two flips. But that is only a mathematical expression of what you can expect to happen over time and doesn't demand that the coin alternate between heads and tails. However, the more times you flip the coin, the more your results will get closer and closer to the probability of heads as 1/2. I see your point, but again, if your chances of getting AA twice in a row are 1/48,881, then, at a ten player table, aren't the chances that one of the players will get AA twice in a row ten times as great, or (1/48,881)/10?

Last edited by LargeLouster; 04-23-2009 at 05:25 PM.
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Old 04-23-2009, 05:37 PM   #131
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by UbinTook View Post
Thanks, here is where my thinking may be wrong...in 100 trials, there are 50 chances of getting AA TWICE correct?
so 100 trials would be 48841/50 or 916:1
hand 1+ hand 2 = chance 1
hand 2+ hand 3 = chance 2
etc
And if the number of hands played say (or is going to be played) is an odd number, then the number of chances is (n-1)/2 or 99-1/2=48 or 1017:1
I think that if you are dealt 100 hands, there are 99 chances of getting AA twice in a row. Hand 1 & 2, hand 2 & 3 and so on up to 100. I don't see why you would divide 99 or (99-1) by 2?
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Old 04-23-2009, 05:46 PM   #132
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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You might also want to see if they offer a course on reading between the lines. Mathematical corrections aside, most of the debate itt can be understood as a battle between those who can (and are willing to) do this and those who can't (or won't).

It's a point of irony that drzen (who is probably a great lover of paradox) chooses to convey his point, which is in support of the opinion of the latter group, in a way that is more likely to be understood by the former.
What debate? I simply gave a mathematical explanation of why it would or would not make sense to accept the bet. As I said, mathematically analyzing the proposed bet was no different than comparing pot odds to the odds of making a hand. I'm not interested in any "debate" between drzen and Mason, ironic or otherwise. As far as your second paragraph, I have no idea what you are talking about, either on or between the lines.
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Old 04-23-2009, 06:16 PM   #133
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by LargeLouster View Post
According to my method, the more hands you play, the more likely it is to happen. In fact, when you are dealt hand #48,882, the probability becomes 1.0, since you now have had 48,881 chances of something that has a probability of 1/48,881, and (1/48,841)/48,841=1 However, the probability of being dealt aces twice in a row says nothing about when it will occur, simply that the more hands you deal (The Law Of Large Numbers, as distinguished from The Law Of Large Lousters ), the closer and closer your actual results will approach the mathematical expression 1/48,881. You could just as easily get AA on the first two hands you are dealt as you could go 48,881 hands without even getting AA once.

Consider a simpler problem, flipping a coin. The probability of heads is 1/2, so mathematically, the probability of heads is 1.0 every two flips. But that is only a mathematical expression of what you can expect to happen over time and doesn't demand that the coin alternate between heads and tails. However, the more times you flip the coin, the more your results will get closer and closer to the probability of heads as 1/2. I see your point, but again, if your chances of getting AA twice in a row are 1/48,881, then, at a ten player table, aren't the chances that one of the players will get AA twice in a row ten times as great, or (1/48,881)/10?
typo correction: On line 5, paragraph 1, (1/48,841)/48,841=1 should just be 48,841/48,841=1
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Old 04-23-2009, 06:35 PM   #134
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by LargeLouster View Post
I think that if you are dealt 100 hands, there are 99 chances of getting AA twice in a row. Hand 1 & 2, hand 2 & 3 and so on up to 100. I don't see why you would divide 99 or (99-1) by 2?
I see it now.
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Old 04-23-2009, 06:41 PM   #135
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Re: How is being delt aces twice in a row not over 40,000-to-1?

So if you play 100 hands the probability is 916:1
Meaning, one time out of 916 separate 100 hand sessions, you will get AA dealt to you twice in a row.
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Old 04-23-2009, 07:49 PM   #136
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by LargeLouster View Post
According to my method, the more hands you play, the more likely it is to happen. In fact, when you are dealt hand #48,882, the probability becomes 1.0, since you now have had 48,881 chances of something that has a probability of 1/48,881, and (1/48,841)/48,841=1. However, the probability of being dealt aces twice in a row says nothing about when it will occur, simply that the more hands you deal (The Law Of Large Numbers, as distinguished from The Law Of Large Lousters ), the closer and closer your actual results will approach the mathematical expression 1/48,881. You could just as easily get AA on the first two hands you are dealt as you could go 48,881 hands without even getting AA once.

Consider a simpler problem, flipping a coin. The probability of heads is 1/2, so mathematically, the probability of heads is 1.0 every two flips. But that is only a mathematical expression of what you can expect to happen over time and doesn't demand that the coin alternate between heads and tails. However, the more times you flip the coin, the more your results will get closer and closer to the probability of heads as 1/2. I see your point, but again, if your chances of getting AA twice in a row are 1/48,881, then, at a ten player table, aren't the chances that one of the players will get AA twice in a row ten times as great, or (1/48,881)/10?
This is my last typo correction. I'm just a lousy typist and that's that. On the final line, (1/48,881)/10 should just be 48,841/48,841.
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Old 04-23-2009, 08:07 PM   #137
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by UbinTook View Post
So if you play 100 hands the probability is 916:1
Meaning, one time out of 916 separate 100 hand sessions, you will get AA dealt to you twice in a row.
I don't think so. Again, the chances of getting AA twice in a row is 1/48,841. If you play 100 hands you have 99 chances of getting AA twice in a row, which is 44,841/99= ~1/493 or expressed as odds, 492:1 against. If you play 916 separate 100 hand sessions, mathematically you are just playing 91,600 hands. If you divide 91,600 by 48,841 you get ~1.875 so your odds of getting AA twice in a row are ~1.875:1 in your favor. Just keep in mind that probabilities and odds are mathematical expressions of what you can expect over time, not absolutes. You could get AA on hands 1 & 2, then again on hands 4 & 5 and then not get them again for a million hands. But, over time, your results will always approach what the probabilities and odds predict. If that weren't true, Las Vegas wouldn't exist.

Last edited by LargeLouster; 04-23-2009 at 08:21 PM.
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Old 04-23-2009, 08:55 PM   #138
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Re: How is being delt aces twice in a row not over 40,000-to-1?

il take 44,840 to 1 lol book me for 5$ the maybe i can play the main event this year lol
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Old 04-23-2009, 09:02 PM   #139
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by LargeLouster View Post
What debate?
I was using your comment to make a point about the thread, the majority of which is occupied by bickering over a problem that strictly speaking asks for one solution (the probability of being dealt aces twice in a row) while being understood contextually almost certainly is asking something else (the probability of being dealt aces again after just having been dealt two aces or, more elegantly, the chance of being dealt two aces twice in a row over the duration of a session).

Quote:
I simply gave a mathematical explanation of why it would or would not make sense to accept the bet.
Overlooking the fact that this was already obv to everyone?

Look at how you started your "explanation":

Quote:
Originally Posted by LargeLouster View Post
The way you say it, you'll pay 1,000 to 1 if the 2 hands are AA, you don't say he has to pay anything if they aren't, so of course he would accept.
The only thing worse than a nitpick is one that doesn't even make sense. The proposal (which wasn't serious in the first place!) was stated in a perfectly normal manner and would be understood even by people who aren't particularly familiar with poker or other kinds of gambling. If the payout is 1000:1 then the 1 must represent a non-zero amount, right?

Quote:
As I said, mathematically analyzing the proposed bet was no different than comparing pot odds to the odds of making a hand.
Let's look at it the way you wrote it before:

Quote:
Nothing personal, but you need to learn a little bit about probabilities, especially comparing pot odds (actual and implied) with the odds of making a hand.
I can only understand this comment in one of two ways: either as a troll, or as a show of willful ignorance with regard to the spirit in which drzen's proposal was offered.

Quote:
I'm not interested in any "debate" between drzen and Mason, ironic or otherwise.
Then why chime in? If you think the majority of 2p2ers are going to be impressed by your ability to perform simple math, you are wrong.

Quote:
As far as your second paragraph, I have no idea what you are talking about, either on or between the lines.
Okay, I fell for it once but next time someone else is going to have to feed you.
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Old 04-23-2009, 09:31 PM   #140
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by themuppets View Post
I was using your comment to make a point about the thread, the majority of which is occupied by bickering over a problem that strictly speaking asks for one solution (the probability of being dealt aces twice in a row) while being understood contextually almost certainly is asking something else (the probability of being dealt aces again after just having been dealt two aces or, more elegantly, the chance of being dealt two aces twice in a row over the duration of a session).



Overlooking the fact that this was already obv to everyone?

Look at how you started your "explanation":



The only thing worse than a nitpick is one that doesn't even make sense. The proposal (which wasn't serious in the first place!) was stated in a perfectly normal manner and would be understood even by people who aren't particularly familiar with poker or other kinds of gambling. If the payout is 1000:1 then the 1 must represent a non-zero amount, right?



Let's look at it the way you wrote it before:



I can only understand this comment in one of two ways: either as a troll, or as a show of willful ignorance with regard to the spirit in which drzen's proposal was offered.



Then why chime in? If you think the majority of 2p2ers are going to be impressed by your ability to perform simple math, you are wrong.



Okay, I fell for it once but next time someone else is going to have to feed you.
It's so nice that you know how people do or do not interpret things, that you know what is and is not obvious, that you know how something was or was not intended, that you don't like a nitpik while you go through my post almost line by line, and that finally, you state your opinions as though they are absolute truths or proven conclusions without ever either backing up your statements or even making any sense at all. What can I say except dude, you have major issues.

Last edited by LargeLouster; 04-23-2009 at 09:55 PM.
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Old 04-24-2009, 08:43 AM   #141
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Re: How is being delt aces twice in a row not over 40,000-to-1?

I was on the final table yesterday at Harrah's Atlantic City- 40 people in the tourney - they don't use new cards for the final table - I was dealt the Ace of Diamonds and the Ace of hearts pre flop- won the hand - the very next hand I was dealt the Ace of Spades and the Ace of clubs also prefolp. What are the odds of getting all four aces dealt in just two hands??

The very next hand I was dealt two kings lol

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Old 04-24-2009, 02:22 PM   #142
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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I was on the final table yesterday at Harrah's Atlantic City- 40 people in the tourney - they don't use new cards for the final table - I was dealt the Ace of Diamonds and the Ace of hearts pre flop- won the hand - the very next hand I was dealt the Ace of Spades and the Ace of clubs also prefolp. What are the odds of getting all four aces dealt in just two hands??

The very next hand I was dealt two kings lol

If you don't care in what order you get the aces, it's 1/(4/52*3/51*2/52*1/51)=1/293,046=293,046:1 against. If the aces have to come in exactly the way you got them, it's 1/(1/52*1/51*1/52*1/51)=1/7,033,104 or 7,033,104:1 against.
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Old 04-24-2009, 02:48 PM   #143
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Re: How is being delt aces twice in a row not over 40,000-to-1?

[QUOTE=LargeLouster;10199436]This is my last typo correction. I'm just a lousy typist and that's that. On the final line, (1/48,881)/10 should just be 48,841/48,841. [/QUOTE

Damn it! Should be 48,841/10
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Old 04-24-2009, 06:42 PM   #144
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by LargeLouster View Post
If you don't care in what order you get the aces, it's 1/(4/52*3/51*2/52*1/51)=1/293,046=293,046:1 against. If the aces have to come in exactly the way you got them, it's 1/(1/52*1/51*1/52*1/51)=1/7,033,104 or 7,033,104:1 against.
Yet again: 293,046:1 against should be 293,045:1 against

7,033,104:1 against should be 7,033,103:1 against

Don't know why I bother since I suspect no one even notices the typos (except maybe themuppets guy)
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Old 04-24-2009, 08:02 PM   #145
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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Originally Posted by LargeLouster View Post
According to my method, the more hands you play, the more likely it is to happen. In fact, when you are dealt hand #48,882, the probability becomes 1.0, since you now have had 48,881 chances of something that has a probability of 1/48,881, and (1/48,841)/48,841=1. However, the probability of being dealt aces twice in a row says nothing about when it will occur, simply that the more hands you deal (The Law Of Large Numbers, as distinguished from The Law Of Large Lousters ), the closer and closer your actual results will approach the mathematical expression 1/48,881. You could just as easily get AA on the first two hands you are dealt as you could go 48,881 hands without even getting AA once.

Consider a simpler problem, flipping a coin. The probability of heads is 1/2, so mathematically, the probability of heads is 1.0 every two flips. But that is only a mathematical expression of what you can expect to happen over time and doesn't demand that the coin alternate between heads and tails. However, the more times you flip the coin, the more your results will get closer and closer to the probability of heads as 1/2. I see your point, but again, if your chances of getting AA twice in a row are 1/48,881, then, at a ten player table, aren't the chances that one of the players will get AA twice in a row ten times as great, or (1/48,881)/10?
first off, the probability of getting aces twice in a row will never be one. It will approach one in the long run but never reach one. Second, my point still holds. Yes, if u divide 48800 by 1 you will get 1, but what if u divide 48800 by one million. THat number is smaller than 1. But wait, that doesnt make sense because according to large louster the probability should be approaching one as more trials occur.

Also I don't know if this was mentioned before but technically if u have n hands there can be n-1 trials in this experiment since in 2 hands there is onli 1 chance to get aces twice, in three hands thers two chances, etc...

Your thinking is so flawed. First off you do not specify the parameters of the problem. The proabiltiy of getting heads AT LEAST ONCE in two flips isnt even one. Thats absurd. It's 3/4. How can you have posted so many times in this thread and say such ludicrous things. Stop filling these people's heads with nonsense.

Last edited by junwagh; 04-24-2009 at 08:28 PM.
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Old 04-24-2009, 09:06 PM   #146
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Re: How is being delt aces twice in a row not over 40,000-to-1?

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first off, the probability of getting aces twice in a row will never be one. It will approach one in the long run but never reach one. Second, my point still holds. Yes, if u divide 48800 by 1 you will get 1, but what if u divide 48800 by one million. THat number is smaller than 1. But wait, that doesnt make sense because according to large louster the probability should be approaching one as more trials occur.

Also I don't know if this was mentioned before but technically if u have n hands there can be n-1 trials in this experiment since in 2 hands there is onli 1 chance to get aces twice, in three hands thers two chances, etc...

Your thinking is so flawed. First off you do not specify the parameters of the problem. The proabiltiy of getting heads AT LEAST ONCE in two flips isnt even one. Thats absurd. It's 3/4. How can you have posted so many times in this thread and say such ludicrous things. Stop filling these people's heads with nonsense.
These are the things you're saying:

1. first off, the probability of getting aces twice in a row will never be one. Nobody said it would; e1 has said it is 1/48,841.

1. what if u divide 48800 by one million. THat number is smaller than 1. But wait, that doesnt make sense because according to large louster the probability should be approaching one as more trials occur. if you deal one million hands, you divide one million by 48,841 (not 48800) not the other way around; also, I said that the more hands you deal, the closer and closer your results will come to 1/48,841, not to 1, as you mistakenly think, if you want to call what your saying thinking.

2.. Yes, if u divide 48800 by 1 you will get 1 Call me a maniac, but it seems to me that if you divide 48,800 by 1, you will get 48,800, not 1.

3. The proabiltiy of getting heads AT LEAST ONCE in two flips isnt even one. Thats absurd. It's 3/4 I never said, as you put it, AT LEAST ONCE. If you want to ask and answer your own questions, go ahead, but please leave me out of them.

4. You don't need me to fill your head with nonsense, it seems quite full as it is; but that's OK, I still like you because you're an even worse typist than me

Last edited by LargeLouster; 04-24-2009 at 09:22 PM.
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Old 04-24-2009, 09:41 PM   #147
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Re: How is being delt aces twice in a row not over 40,000-to-1?

1. you explicitly said
"In fact, when you are dealt hand #48,882, the probability becomes 1.0, since you now have had 48,881 chances of something that has a probability of 1/48,881, and (1/48,841)/48,841=1."

1.you explicitly said
"According to my method, the more hands you play, the more likely it is to happen. In fact, when you are dealt hand #48,882, the probability becomes 1.0, since you now have had 48,881 chances of something that has a probability of 1/48,881, and (1/48,841)/48,841=1.

2. Sorry that was a typo on my part. My point still holds. According to you if you play one million hands the probability of getting aces is 100000/48800 which is impossible since probability can never be greater than 1, which is the point I;m trying to make.

3.
I put in the at least because your problem is incomplete otherwise. The probability of getting heads once is actually 1/2. whereas the probability of getting heads at least once is 3/4.

4. You explicitly said

"According to my method, the more hands you play, the more likely it is to happen. In fact, when you are dealt hand #48,882, the probability becomes 1.0, since you now have had 48,881 chances of something that has a probability of 1/48,881, and (1/48,841)/48,841=1. However, the probability of being dealt aces twice in a row says nothing about when it will occur, simply that the more hands you deal (The Law Of Large Numbers, as distinguished from The Law Of Large Lousters ), the closer and closer your actual results will approach the mathematical expression 1/48,881. You could just as easily get AA on the first two hands you are dealt as you could go 48,881 hands without even getting AA once.

Consider a simpler problem, flipping a coin. The probability of heads is 1/2, so mathematically, the probability of heads is 1.0 every two flips. But that is only a mathematical expression of what you can expect to happen over time and doesn't demand that the coin alternate between heads and tails. However, the more times you flip the coin, the more your results will get closer and closer to the probability of heads as 1/2. I see your point, but again, if your chances of getting AA twice in a row are 1/48,881, then, at a ten player table, aren't the chances that one of the players will get AA twice in a row ten times as great, or (1/48,881)/10?"

proving it is ur head filled with nonsense.
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Old 04-25-2009, 10:33 AM   #148
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Re: How is being delt aces twice in a row not over 40,000-to-1?

What are the odds being dealt Kings/Aces 5 times in a row? About a year ago i got dealt Kings 5 times in a row playing a tourny on t6poker, unbelievable.
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Old 04-25-2009, 12:26 PM   #149
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Re: How is being delt aces twice in a row not over 40,000-to-1?

Quote:
Originally Posted by junwagh View Post
1. you explicitly said
"In fact, when you are dealt hand #48,882, the probability becomes 1.0, since you now have had 48,881 chances of something that has a probability of 1/48,881, and (1/48,841)/48,841=1."



1.you explicitly said
"According to my method, the more hands you play, the more likely it is to happen. In fact, when you are dealt hand #48,882, the probability becomes 1.0, since you now have had 48,881 chances of something that has a probability of 1/48,881, and (1/48,841)/48,841=1.

2. Sorry that was a typo on my part. My point still holds. According to you if you play one million hands the probability of getting aces is 100000/48800 which is impossible since probability can never be greater than 1, which is the point I;m trying to make.

3.
I put in the at least because your problem is incomplete otherwise. The probability of getting heads once is actually 1/2. whereas the probability of getting heads at least once is 3/4.

4. You explicitly said

"According to my method, the more hands you play, the more likely it is to happen. In fact, when you are dealt hand #48,882, the probability becomes 1.0, since you now have had 48,881 chances of something that has a probability of 1/48,881, and (1/48,841)/48,841=1. However, the probability of being dealt aces twice in a row says nothing about when it will occur, simply that the more hands you deal (The Law Of Large Numbers, as distinguished from The Law Of Large Lousters ), the closer and closer your actual results will approach the mathematical expression 1/48,881. You could just as easily get AA on the first two hands you are dealt as you could go 48,881 hands without even getting AA once.

Consider a simpler problem, flipping a coin. The probability of heads is 1/2, so mathematically, the probability of heads is 1.0 every two flips. But that is only a mathematical expression of what you can expect to happen over time and doesn't demand that the coin alternate between heads and tails. However, the more times you flip the coin, the more your results will get closer and closer to the probability of heads as 1/2. I see your point, but again, if your chances of getting AA twice in a row are 1/48,881, then, at a ten player table, aren't the chances that one of the players will get AA twice in a row ten times as great, or (1/48,881)/10?"

proving it is ur head filled with nonsense.
After rereading all your posts, I'm convinced that instead of worrying about the probability of being dealt AA twice in a row, you should consider joining an organization called AA. Also, here's an easy one. What is the probability that someone like you, who has an extensive vocabulary, would be incapable of combining those words into at least one coherent sentence? The answer is approximately, exactly 1/6.7 billion, because the world's population is estimated to be 6.7 billion and you are definitely one of a kind.
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Old 04-25-2009, 12:33 PM   #150
LargeLouster
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Re: How is being delt aces twice in a row not over 40,000-to-1?

Quote:
Originally Posted by Koit View Post
What are the odds being dealt Kings/Aces 5 times in a row? About a year ago i got dealt Kings 5 times in a row playing a tourny on t6poker, unbelievable.
The probability of being dealt Kings 5 times in a row is the same as the probability of being dealt any pocket pair 5 times in a row, which is 1/221^5=1/527,182,965,101, or odds of 527,182,965,100:1 against.

Last edited by LargeLouster; 04-25-2009 at 12:51 PM.
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