A/K vs QQ

[NutJob]

The rough math he is doing really isnt that bad, and certainly close enough for most pot odds calculations. For example, you say dont just take the 24% and divide by two...getting 12%, when 13% is the right answer. Is the difference between 7.333/1 pot odds and 6.692/1 pot odds really going to change his outcome?

I agree that its a stretch to extrapolate that to the flop/river and turn though. If my math is any good, the probability of hitting AT LEAST one A or K (I assume he wouldnt complain if he hit a boat instead of just a pair) is 1-C(44,5)/C(50,5), or 49%, rather than the 60% he estimated.

When it comes to more complex situations like how often does AK turn into a pair, straight or flush, invest in a simulator...you're better off spending you time at the tables and earning the cost of the simulator back in an hour.

[doormat]

A simple way to calculate this is:
success = 1 - (chance of failure)
So if you don't know what your opponent holds, there are 47 unseen cards after the flop and 6 that help you. Therefore you fail when on the first card after the flop you miss 41 of 47 times, and on the second (river) you miss 40 of 46 times. So your chance of success is 1 - (41/47) * (40/46) = .2414, or 24.14%.

If you know your opponent has 2 queens, your chance of hitting is 1 - (39/45) * (38/44) = .2515 or 25.15%.
doormat

[Legend27]

Yeah I was just saying since 1 card on the river gives you a 12% matching A/K then a total 5 cards should give you a 60% chance. Which I know isn't right, I just wanted to know why.

[BruceZ]

Yeah I was just saying since 1 card on the river gives you a 12% matching A/K then a total 5 cards should give you a 60% chance. Which I know isn't right, I just wanted to know why.

It's because if you just multiply 12% by 5 to get the odds of catching A/K in 5 cards, you are effectively double counting the times you get 2 A/Ks, triple counting the times you get 3, quadruple counting the times you get 4, and quintuple counting the times you get 5. You can only do this when it is not possible to hit A/K on more than one card (mutually exclusive events). I sometimes use your method to do a quick and dirty approximation in cases where I just want to get a rough feel for the odds, or as a sanity check before doing a more exact computation so I know what kind of numbers to expect. This can be surprisingly accurate in many cases, especially if you subtract 3% for each card, so in this case I would estimate 60% - 5*3% = 45%.

The normal way to compute this exactly is to take 1 minus the odds of not catching in 5 cards:

1 - C(44,5)/C(50,5) = 48.7%

[Cyrus]

Thanks for your input.

"The rough math he is doing really isn't that bad, and certainly close enough for most pot odds calculations. For example, you say dont just take the 24% and divide by two...getting 12%, when 13% is the right answer."

I'm thinking that perhaps someone who takes the probability p of hitting with 1 card to come and, to get the probability of hitting with k cards to come, multiplies it by k should not be led down this path. When straight arithmetic is employed for not-simple situations, it can get very misleading.

"The probability of hitting AT LEAST one A or K (I assume he wouldnt complain if he hit a boat instead of just a pair) is 1-C(44,5)/C(50,5), or 49%, rather than the 60% he estimated."

I agree with that figure, which should be of more practical help IMHO than the probabilities he requested (ie of hitting only a straight, etc). But the 49% is "close" to that 60% figure by accident. 60% is 12% multiplied by 5, when there are 5 cards to come. Legend27 should assume that there are 9 cards to come : 9 * 12% = ?

"When it comes to more complex situations like how often does AK turn into a pair, straight or flush, invest in a simulator."

Long before I had www.twodimes.net on my Favorites list, I had purchased a couple of those, Mike Caro's and PokerWiz, and I have gotten my money's worth.

[BruceZ]

Here's a simpler example to illustrate what is going wrong, and also to show that you can use your method as a starting point to get the exact answer.

Suppose you want to know the probability that an A,K, or Q will appear on the next flop, before any hole cards are dealt. There are 12 of these cards in all, so the chance of any 1 card being A,K,Q is 12/52. You can't just say the chance of getting one of these cards on the flop is 3*12/52 = 9/13 because again you would be double counting all the times that exactly 2 cards were A-Q, and you would be triple counting the times that all 3 were A-Q. If you knew the probability of exactly 2 flopping, you could subtract this off since you double counted it. Suppose you tried to use the same method of computing the probability of exactly 2 of these cards flopping. You would say that the chance of getting A-Q on 2 specific cards is 12*11/(52*51), and this is exactly correct. Since there are 3 ways to choose the 2 cards out of the 3 cards on the flop, your method would multiply this by 3 to get 3*12*11/(52*51). Now the problem is that you triple counted all the times that all 3 flopped A-Q, just as you did when you computed 9/13 in the beginning. You don't want to subtract all of these off since then you wouldn't be counting them at all. So now we have to add back the probability of all 3 flopping A-Q. This is easy to compute, it's just 12*11*10/(52*51*50). So now we can finally use these results to get the probability we are after, the probability of flopping A,K, or Q. It is:

9/13 - 3*12*11/(52*51) + 12*11*10/(52*51*50) = 55.29%

In summary, we started by easily computing 9/13, and we said oops, that double counts all the times we flop 2, and triple counts all the times we flop 3. Then we subtracted the second term which subtracted the times we flopped 2 that we double counted, and we said oops, now we subtracted too many for the times we flopped 3. Then we computed that and added it back on to get the final answer. This is called the inclusion-exclusion principle. For this particular problem, it is a complicated way to compute the answer. For some other more difficult problems, it is the only practical way to compute the answer. In many problems we do not use this method to compute the exact answer, but we only carry out the calculation to as many terms as needed for an acceptable approximation. Note that the terms become smaller, so if we decide not to compute all the terms, we know how close our result is to the final answer.

Here is the way you would normally compute this probability exactly. We take 1 minus the probability of not flopping A,K, or Q. This is:

1 - C(40,3)/C(52,3) = 55.29% exactly as before.

Similarly, you could use your method of taking 5*12% = 60% as a starting point to compute the probability of hitting A/K in 5 cards when you hold AK. Here is how to do that:

60% -
C(5,2)6*5/(50*49) +
C(5,3)*6*5*4/(50*49*48) -
C(5,4)*6*5*4*3/(50*49*48*47) +
6*5*4*3*2/(50*49*48*47*46)
= 48.7% same as we got in the last post

If you don't understand the details of this method right away, that's OK. The important point was to show why the 60% overestates the probability that we are after, and that we would have to modify this to get the correct answer. While there is a much easier method to get the answer in these particular problems, the method presented here is extremely important and essential for advanced problems.

By understanding this principle, you can very quickly answer the following question which otherwise would not be obvious. If you hold AA in a 10 handed game, what is the probability that someone else holds AA with you? The probability of any particular player holding AA when you hold AA is 1/(50*49/2) = 1/1225. The chance that one of the other 9 players holds AA is just 9/1225. Here we DID just multiply by 9, but in this case this gives the exactly correct answer. Do you see why? It is because it is impossible for more than one other player to hold AA.

[Cyrus]

Similarly, you could use your method of taking 5*12% = 60% as a starting point to compute the probability of hitting A/K in 5 cards when you hold AK. Here is how to do that:

60% - C(5,2)6*5/(50*49) + C(5,3)*6*5*4/(50*49*48) - C(5,4)*6*5*4*3/(50*49*48*47) + 6*5*4*3*2/(50*49*48*47*46)

= 48

Is that the short cut?

[img]/forums/images/icons/cool.gif[/img]

[Legend27]

Alright that makes alot more sense now, thanks for the help.

[Tekari]

when you are saying C(47,2)=1081 what math is involved to get the 1081?

thanks

[BruceZ]

C(47,2) = 47*46/2! = 1081.

That is, there are 47 ways to pick the first one times 46 ways to pick the second one, and then we divide by 2! = 2 ways to get the same cards in a different order. In general,

C(n,k) = n! / [ (n-k)!*k! ] = n*(n-1)*(n-2)*...*(n-k+1) / [k*(k-1)*(k-2)*...*1]