AA and KK

[mcozzy1]

If you hold KK, what are the odds that someone else at a 10 person table has AA?

[DesertCat]

roughly 1/(1-(219/220)^9) to 1

[Kaz The Original]

What is that in normal person speak?

[Stallion714]

Slightly better than 1 in 220.

[BruceZ]

[ 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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=417 383&Forum=probability&Words=inclusion-exclusion&Match=Entire% 20Phrase&Searchpage=0&Limit=25&Old=1year&Main=4169 81&Search=true#Post417383). 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".

[jtr]

Nice post Bruce. I worked this out with much less precision than you some time ago and have always kept "1 time in 20" in my head as a rough approximation. Nice to see I was close.

[DyessMan89]

[ 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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=417 383&Forum=probability&Words=inclusion-exclusion&Match=Entire% 20Phrase&Searchpage=0&Limit=25&Old=1year&Main=4169 81&Search=true#Post417383). 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. /images/graemlins/wink.gif

[K C]

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 /images/graemlins/smile.gif - although I do appreciate your math /images/graemlins/smile.gif

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 /images/graemlins/smile.gif

KC
http://kingcobrapoker.com

[BruceZ]

[ 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 /images/graemlins/smile.gif - although I do appreciate your math /images/graemlins/smile.gif

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 /images/graemlins/smile.gif

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.

[badbrownie]

[ 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)

[BruceZ]

[ QUOTE ]
[ 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.

[/ QUOTE ]

That is why it is an approximation. It would be exact if it could only happen to one player, that is, if the players hands were "mutually exclusive". If it can happen to more than one player, then this approximation will have the effect of over counting the deals where it happens to more than one player. Method number 5 in my above post is the way to adjust for this over counting to get an exact answer by the inclusion-exlusion principle.


[ QUOTE ]
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)

[/ QUOTE ]

I guess we have to flog the horse some more because what you are describing in general is also an approximation. It is only exact in the case where the events are independent, as in your dice example. If the events are not indpendent, then it is not exact to simply multiply these probabilities x times. In the pocket pair problem, this method would actually yield a less accurate result than multiplying the single player probabilty by the number of players. The probability that someone holds AA is not
1 - (1 - 1/221)^10, but it is close. In general, the inclusion-exclusion principle can always yield an exact result, but it is often not necessary to be this precise.