View Full Version : 10 10 UTG

08-16-2002, 06:20 PM
I would like to know what is the probability and how to calculate similar schenarios where the probability that of 9 random hands that if you have 10 10 UTG that one of these 9 hands is JJ QQ KK or AA.

Can someone suggest a computer program or book that explains or has tables to figure out situations like these.

Has anyone used Hold'Em Analyzer or another such program.

08-16-2002, 07:48 PM
It's really not that tricky. There are 52!/(2!*(52-2)!) = 1326 number of different holdem hands. If you hold AA you can have it 6 different ways. The same goes for any pair. So, AA, KK, QQ, JJ, that sums up to 24 ways.

24 out of 1326 for a probability of 1.81%

Now, instead of calculating the probability that one or more players have such a pair we will calculate the probability that none of them has one. The probability that one player does not hold a pair is 98.19% so the probability that none of the remaining 9 players have one is:

0.9819^9 = 0.8484

So, there is about an 85% chance that the players do not have such a pair. Consequently there is about a 15% chance that one of the players has such a pair.

Hope this didn't cause as many questions as it answered. /images/wink.gif

Greets, Andreas

08-16-2002, 08:14 PM
Actually it is 16.3%.

Remember, two cards are already gone.

But yes, very easy to compute.

Pretty interesting you are that far behind 1/6 of the time.

08-16-2002, 11:25 PM
It's very easy to compute the wrong answer, and a relative pain in the butt to compute the correct one:











= 22%.

The denominator will come up in other problems with 9 opponents, so it is worth remembering its value(2.2575x10^26).

Note that the odds of a given player having one of these hands is 24/1326=1.8%, but if the first 9 players do not have one of these hands, the chance that the final player has it is 4.3% or more than twice as great.

08-16-2002, 11:35 PM

08-17-2002, 12:19 AM
Please explain which factorials you were using for each of the 9 hands and why you multiplied them instead of adding them.

Also have you considered the cards UTG + 1 has could be anything including AJ and 33 one of them depletes two of the cards that could make the possible 4 hands the other does not. Please explain your numbers.

08-17-2002, 12:47 AM
I am multiplying the probability that the first player doesn't have one of the 24 hands, times the probability that the second player doesn't have it, etc. The denominator in each case is the total number of hands each player can make out of the remaining cards. The numerator is this number minus the 24 hands.

You are correct that depending on which cards each player receives, some of the 24 hands may no longer be possible. I am not taking this into account. That would make the problem even more difficult, and the result would be lower than I have reported.

08-17-2002, 02:00 AM
Let's try a different approach which should account for everything:

The total number of ways to deal 50 cards to 9 players (18 cards total) is:


We divide by 2 to avoid counting every player's possible hand twice.

The number of ways for exactly 1 player to get JJ-AA is:


Since there are 24 hands and 9 players that can get them. So the probability of exactly 1 player having JJ-AA is (24*9)/(50*49) = 8.8%.

The number of ways exactly 2 players can have JJ-AA is:


Since there are C(24,2) = 24*23/2 hands and

C(9,2) = 9*8/2 players that can get them.

So the probability of exactly 2 players having JJ-AA is C(24,2)*C(9,2)/(50*49*48*47) = 1.8%

We could go on like this computing probailities for exactly 3,4,5...players, but the rest of the terms become too small to worry about. So the total probability is 8.8% + 1.8% = 10.6%.

So in summary, the original posters ignored the effect that each players cards had on the others. My original post accounted for the fact that each successive player has fewer cards to choose from which made it more likely to draw one of the combinations, but it failed to take into account an even bigger effect which is that the exact cards drawn tend to block certain hands. The net result was that I moved the percentage about the right amount in the wrong direction. It should actually go down to 10.6% in light of the above.

08-17-2002, 02:52 AM
I'm not sure what you mean by "number of ways to deal 50 cards to 9 players". If we don't care about the order of the cards, but do care about the order of the players, then the number of ways to do this is P(50,18)/(2^9) = 50*49*...*33/(2^9). This is because each of the 9 hands can be ordered in 2 ways.

08-17-2002, 03:00 AM
Another problem here is how to come up with the number of ways of assigning 2 hands, given you've picked the 2 players. It's not C(24,2), because once you assign one of the players a hand, Qh Qd, say, then you've eliminated hands like Qh Qs, or Qd Qc. So, C(24,2) counts Qh Qs + Qh Qd, whereas this possibility is impossible in the problem.

08-17-2002, 03:07 AM
When I counted the number of ways that exactly 1 player could get JJ-AA, I am also counting ways that more than one player could get these hands, so 8.8% is too big. It is also not the number of ways that 1 or more can get these hands since then some of these would be double counted. So the correct answer must be less than 8.8% not 10.6%. I think 8.8% is a pretty close approximation since the probability of getting more than one is small. I will see if I can improve this further.

Also, when I divide by 2 to account for the order that each player receives his 2 cards, this should be 2^9, but that doesn't affect the answer since it cancels out.

08-17-2002, 03:19 AM
C(24,2) would become 24*19/2 since only 1 pair of the assigned denomination would remain.

In any case, my current thinking is to ignore the second term, and use the first term as an upper bound to the final answer.

08-17-2002, 03:43 AM
I'm making a hell of a mess here. The 2^9 doesn't cancel out because in the numerator it is 2^8 since the 24 is already order independent. That makes the first term 17.6%.

(9*24*48*47*...*33/2^8)/(50*49*48...*33/2^9) = 17.6%

The correct answer should be slightly less than this. So we have come full circle and are converging on the initial approximation.

08-17-2002, 05:11 AM
The way to find an exact solution to this problem is to use the inclusion-exclusion principle. Here's how:

The sample space consists of all the ways of assigning hands to the 9 players, where the orders of the cards in the hands don't matter, but the order of the players does. In other words, JhJd = JdJh, but this hand is not the same if seat 3 holds it vs. if seat 7 holds it.

As I mentioned in a previous post, the total number of elements in this space is P(50,18)/(2^9). This is because there are P(50,18) ways to distribute 18 cards to the players in order. Then, for each player, the cards have 2 orders, so we divide by 2^9. Another way to do this is C(50,2)*C(48,2)*...C(34,2), where the first factor is the number of ways of choosing 2 from 50, then another 2 from the remaining 48, and so on. This is also sometimes called a multinomial coefficient. For reasons that will make sense later, I'm calling this number

N(0) = P(50,18)/(2^9). This is about 2.3*(10^26).

Now, we have to count the total number of elements in which at least one hand has JJ, QQ, KK, or AA. Let spades = 1, hearts = 2, diamonds = 3, and clubs = 4. We consider the 24 events: J_12, J_13, J_14, J_23, J_24, J_34, Q_12, Q_13, Q_14, Q_23, Q_24, Q_34, K_12, K_13, K_14, K_23, K_24, K_34, A_12, A_13, A_14, A_23, A_24, A_34. Here, the event J_12 consists of all the ways of assigning the hands so that JsJh is dealt. The number we're trying to calculate is the number of elements in the union of all these events (i.e. number of ways that at least one JJ, QQ, KK, or AA hand is dealt). Inclusion-exclusion says, add up the number of elements in all of them together, not worrying about overlap and such. Then, take all possible combinations of 2 events, and for each of these pairs of events, find the number of ways of BOTH events happening (intersection). Subtract this from the first number. Then ADD back all the number of things in all intersections of combinations of 3 events. Then subtract for 4, add for 5, and so on. Fortunately, for us, if we take 9 or more of these events, the number of things in their intersection is zero (all the cards are taken up), so we only have to go to 8, not all the way down to 24. Thank goodness.

Some notation, just so we don't lose so much track of everything and get lost immediately. Let N(1) = number of ways of a given fixed, (but arbitrary) pair JJ, QQ, KK, or AA can occur, counting repetitions. It's important to count repetitions. Let me explain. The number of ways that JsJh can occur is 9*P(48,16)/(2^8). The reason is because there are 9 possible players that can receive the hand, and after giving the hand out, there are P(48,16)/(2^8) ways of assigning out the other hands, for exactly the same reason I gave above for the total size of the sample space. Now, the number of ways that JsJd can occur is identical, and also for QsQd, and so on. Note, it doesn't matter that we're counting certain elements more than once. That's the whole point of the inclusion-exclusion principle. It does your dirty work for you.

Getting back, since there are 24 possible events that we're considering, this means that

N(1) = 24*P(9,1)*P(48,16)/(2^8). This is about 4.0*(10^25).

Now, let N(2) = number of ways of a give pair of pairs JJ, QQ, KK, or AA can occur, again counting repetitions. Here, we have to make sure to get the number of combinations right. Some of the intersections (like JsJh + JsJd) simply can't occur, and have no elements. How many non-empty intersections are there? Here, we can think of all 16 of the face cards JJJJQQQQKKKKAAAA together as a single set. There are 24 ways to choose the first pair. Then we have something like JJJJQQQQKKAAAA left over. Now, there are only _19_ ways to choose the second pair, not 23. Altogether, there are 24*19/2 possible combinations of 2 of the original 24 events, (that don't have empty intersection). Now, we have to choose which players will get these, there are P(9,2) = 72 ways of doing this. The reason this is a permutation, not a combination, is because the ordered position of the players matters. Finally, we have to assign all the other hands, as above, there are P(46,14)/(2^7) ways to do this. So, altogether,

N(2) = (24*19/2)*P(9,2)*P(46,14)/(2^7). This is about 2.7*(10^24).

The pattern for the case of 2 generalizes to the rest of the cases. To find N(3), for example, there are 24*(18*14+1*18) = 6480. I guess this is a bit different than just the case of 2. Here the 24 is the first pair. The 18 is the number of ways of picking your second pair different rank from the first, in this case, there are 14 possible pairs left for the third pair. If you DO pick the same rank, there are 18 pairs left. The rest is exactly the same. Thus,

N(3) = (24*(18*14+1*18)/6)*P(9,3)*P(44,12)/(2^6). This is about 2.9*(10^22).


N(4) = (24*(18*(12*9+2*13)+1*18*13)/24)*P(9,4)*P(42,10)/(2^5). This is about 1.3*(10^21).

At this point, the numbers are becoming so small that they're negligible. (It's even better...since the terms form a decreasing alternating series, we know for certain that the final term is an error bound.)

So, our estimate is N(1)-N(2)+N(3)-N(4), which is about 3.715*(10^25). So, the estimated probability of one of the pairs being out is 3.715*(10^25)/(2.257*(10^26)), which is about 16.46%. So it agrees with your estimate from earlier. Moreover, the error bound is .0005%, so even this is definitely overkill.

I know this probably seems like a ridiculous amount of work to get roughly the same estimate. The first defense I'd give is that I'm just a geek and I enjoy trying to find exact solutions, whether it's practical or not. Also, the reason I carry some problems through like this, is that I'm very paranoid or skeptical about my own ability to "guestimate" how far my estimate is over, under, etc. This one turned out okay, but there are other problems, where finding a "rough estimate" by making various assumptions can really land you in a lot of trouble if you're not careful, and I don't seem to trust myself to know when I'm pushing things. In any case, this is a pain in the butt to do by hand or calculator, but the algorithm can easily be programmed on a computer that grinds through the number-crunching for you. I have a feeling this is how a lot of the numbers in odds tables are arrived at, by sophisticated algorithms with lots of computation, that we can understand conceptually, but computationally are only within reach of a chip.

08-17-2002, 06:50 AM
Looks good. I computed N(1) and left it at that. That is my first term which includes overlaps. Subtracting the 1% for N(2) gets us the rest of the way.

I'm amazed that the inital estimate of

1-(24/1225)^9 was so close. This would be correct if each hand were independent of each other, as if each player put back his cards before the next player selected his. I see no reason why it should give the correct answer here other than by sheer coincidence.

08-18-2002, 01:57 AM

08-19-2002, 06:18 PM
math geek,

Your reasoning looks good to me. You have one calculation error: N(3) is about 8.592 * 10^22. With that correction the answer becomes:

16.48 %

08-19-2002, 06:30 PM
sdplayer estimated the 10 10 UTG question by assuming independence. His result, 16.31% is very close to the actual answer 16.48%. Is there any way that we could have estimated the error caused by assuming independence without going through math geek's lengthy explanation of the exact answer?