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#3
12-31-2004, 04:30 AM
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Re: AA and KK
What is that in normal person speak?
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#4
12-31-2004, 05:39 AM
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Re: AA and KK
Slightly better than 1 in 220.
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#5
12-31-2004, 07:05 AM
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Re: AA and KK
[ QUOTE ]
roughly 1/(1-(219/220)^9) to 1 [/ QUOTE ] Errors with this approximation: 1. The odds of a single player having AA is 220-to-1 = 1/221 with 52 unseen cards. Ignoring for a moment that we actually have 50 unseen cards (see 2), the probability of a single player NOT having AA with 52 unseen cards would be 1 - 1/221 = 220/221, not 219/220. 2. But we have 50 unseen cards, not 52. The probability of a single player having AA with 50 unseen cards is 6/(50*49/2) = 6/1225. The probability of not having AA is 1219/1225. 3. With the above corrections, this approximation would become 1/(1-(1219/1225)^9) = 1 in 23.1 = 22.1-to-1. Not 24.8-to-1 as you stated. Note the difference of 1 between odds notation ("-to-1") and probability ("1 in "). 4. A better approximation since only 2 players can have AA is: 9*6/1225 = 1 in 22.69 = 21.69-to-1. 5. The EXACT answer is: 9*6/1225 - C(9,2)/C(50,4) =~ 1 in 22.77 or 21.77-to-1. This is obtained from the inclusion-exclusion principle. The first term counts all deals with at least a single AA, and the second term subtracts off all deals with two AA counted twice by first term. C(9,2) is the number of ways to choose 2 players out of 9, which evaluates to 9*8/2 = 36, and C(50,4) is the number of ways to choose 4 cards out of 50, which evaluates to 50*49*48*47/(4*3*2*1). There is only 1 combination of 4 cards which is 4 aces. Many examples of problems like this can be found in the probablity forum, especially in the older archives. Search for my username, or "inclusion-exclusion".
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#7
01-01-2005, 12:40 AM
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Re: AA and KK
[ QUOTE ]
[ QUOTE ] roughly 1/(1-(219/220)^9) to 1 [/ QUOTE ] Errors with this approximation: 1. The odds of a single player having AA is 220-to-1 = 1/221 with 52 unseen cards. Ignoring for a moment that we actually have 50 unseen cards (see 2), the probability of a single player NOT having AA with 52 unseen cards would be 1 - 1/221 = 220/221, not 219/220. 2. But we have 50 unseen cards, not 52. The probability of a single player having AA with 50 unseen cards is 6/(50*49/2) = 6/1225. The probability of not having AA is 1219/1225. 3. With the above corrections, this approximation would become 1/(1-(1219/1225)^9) = 1 in 23.1 = 22.1-to-1. Not 24.8-to-1 as you stated. Note the difference of 1 between odds notation ("-to-1") and probability ("1 in "). 4. A better approximation since only 2 players can have AA is: 9*6/1225 = 1 in 22.69 = 21.69-to-1. 5. The EXACT answer is: 9*6/1225 - C(9,2)/C(50,4) =~ 1 in 22.77 or 21.77-to-1. This is obtained from the inclusion-exclusion principle. The first term counts all deals with at least a single AA, and the second term subtracts off all deals with two AA counted twice by first term. C(9,2) is the number of ways to choose 2 players out of 9, which evaluates to 9*8/2 = 36, and C(50,4) is the number of ways to choose 4 cards out of 50, which evaluates to 50*49*48*47/(4*3*2*1). There is only 1 combination of 4 cards which is 4 aces. Many examples of problems like this can be found in the probablity forum, especially in the older archives. Search for my username, or "inclusion-exclusion". [/ QUOTE ] I think someone needs a hobby. [img]/images/graemlins/wink.gif[/img]
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#8
01-01-2005, 07:49 AM
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Re: AA and KK
The real answer though is: on certain sites, it's DAMN likely - lol.
Here's a simpler way to do this stuff for those who don't have a math degree [img]/images/graemlins/smile.gif[/img] - although I do appreciate your math [img]/images/graemlins/smile.gif[/img] We know that the odds of any one player having a particular pocket pair is roughly 1 in 220. You're going to want to know what the chances are at a table of 10 though of course. Multiply that by 10 players and you get 10 in 220, which works out to 21:1 against it happening. Bottom line here is this doesn't happen enough to worry about too much [img]/images/graemlins/smile.gif[/img] KC http://kingcobrapoker.com
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#9
01-01-2005, 08:52 AM
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Re: AA and KK
[ QUOTE ]
The real answer though is: on certain sites, it's DAMN likely - lol. Here's a simpler way to do this stuff for those who don't have a math degree [img]/images/graemlins/smile.gif[/img] - although I do appreciate your math [img]/images/graemlins/smile.gif[/img] We know that the odds of any one player having a particular pocket pair is roughly 1 in 220. You're going to want to know what the chances are at a table of 10 though of course. Multiply that by 10 players and you get 10 in 220, which works out to 21:1 against it happening. Bottom line here is this doesn't happen enough to worry about too much [img]/images/graemlins/smile.gif[/img] KC http://kingcobrapoker.com [/ QUOTE ] In fact that's all I'm doing in my approximation (see point 4) except that since you hold KK you should multiply by 9 instead of 10, and the probability of any one player having AA is 6/1225 rather than 1/221. The 1/221 comes from 6/1326 for a full deck. So we get 9*6/1225 = 21.7:1. You can do this for any number of opponents. You have solved a different problem, namely, the probability that AA was dealt to any of 10 players.
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#10
01-01-2005, 07:56 PM
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Re: AA and KK
[ QUOTE ]
We know that the odds of any one player having a particular pocket pair is roughly 1 in 220. You're going to want to know what the chances are at a table of 10 though of course. Multiply that by 10 players and you get 10 in 220, which works out to 21:1 against it happening. [/ QUOTE ] Can this be right? It seems to me that while your answer is close to Bruce's your methodology is flawed. If the chance of something happening is 1 in 220 for 1 player it does not follow that it is 10 in 220 for 10 players. To demonstrate this with a simple example, the chance of throwing a coin and getting heads is 1 in 2. But the chance of throwing 2 coins and one of them being heads is not 2 in 2. However, perhaps your methodology is perfectly valid for roughly calculating the probability of one of x players hitting a y in z opportunitiy where z is much greater than x and y. So the chance of one of 10 players hitting a 1 in 220 can be calculated your way (roughly) because 220 is so much larger than 1 and 10. In general the way to calculate the chance of something happening at least once in x times is to calculate the chance of it NOT happening in one try, then raise that number to the xth power and subtract the result from 1. So...(to flog a dying horse) the chance of not rolling a 6 in 10 tries is... 1 - ((5/6)^10)
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