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Legend27
06-04-2003, 01:08 PM
What is the % chance of A/K matching either the A or K on the turn and river? I don't wanna know if it's 6:1 odds or whatever, I just wanna if it's like a 20% chance or whatever it is.

Jimbo
06-04-2003, 05:52 PM
About a 27% chance. Actually I always thought that was why you used your nickname of Legend27, guess I was wrong. /forums/images/icons/smile.gif

Legend27
06-05-2003, 01:59 AM
27 is just my lucky #. Thanks for replying.

Cyrus
06-05-2003, 02:27 AM
"I don't wanna know if it's 6:1 odds or whatever, I just wanna if it's like a 20% chance or whatever it is."

IMHO, knowing how to convert odds to percentages and vice versa, rather quickly too, is more important than knowing my chances to hit. I mean, I should learn to do the first and then move on to know the other stuff. JMHO.

Odds of 6:1 against is equivalent to hitting (1/7)% of the time, ie 1/7=0.14 or 14%. The probability of hitting 20% of the time is equivalent to having 80%-20% odds against, which is being a 8:2 or 4:1 underdog.

"What is the % chance of A/K matching either the A or K on the turn and river?"

With 6 outs (3K+3A), the probability of hitting the turn and/or the river is around 24%.

This does NOT mean that your AK will win over QQ abt 24% of the time in that situation ! One of the two cards to come might be an Ace (or a King) but the other card a Queen. Ouch.

Here's the table with the probabilities of hitting your outs, when you have the turn and the river coming up:

<pre><font class="small">code:</font><hr>

TURN AND RIVER COMING UP

NUMBER OF PROBABILITY OF HITTING
OUTS ONE OR TWO OUTS

1 4.3 %
2 8.4 %
3 12.5 %
4 16.5 %
5 20.4 %
6 24.1 %
7 27.8 %
8 31.5 %
9 35.0 %
10 38.4 %
11 41.7 %
12 45.0 %
13 48.1 %
14 51.2 %
15 54.1 %
16 57.0 %
17 59.8 %
18 62.4 %
19 65.0 %
20 67.5 %
21 69.9 %

</pre><hr>

Legend27
06-05-2003, 05:28 AM
n/m

Cyrus
06-05-2003, 02:34 PM
Here's how the probabilities in the Table posted above were calculated. Take, for example, the probability of having 6 outs and hitting 1 or 2 outs on the Turn and the River.

C(6,2)=15 is the number of 2-card combinations that those 6 outs can produce. That means there are 15 combinations of Turn+River whereby you hit your outs in both cards.

For the number of hitting only 1 out (and get 1 blank), you have 6 outs and (52 - 2cards in your hand - 3cards on the flop -6 outs)= 41 blanks. Combining the outs with the blanks gives you 6*41=246 combinations whereby you hit only 1 out on Turn+River.

So, we have 15+246=261 combinations producing either 1 or 2 outs on Turn+River.

The total combinations of all the remaining cards that can be seen on Turn+River are
C(47,2)=1081.

Therefore, the probability of getting a combination of Turn+River that hits you with 1 or 2 outs is 261/1081=0.241443108233117483811285846438483 or 24.1%.

[It is irrelevant in this calculation whether you hit your 1 out on the Turn or on the River. The probability of hitting 1 out in either Turn or River is the same, before the Turn card is dealt.]

Jimbo
06-05-2003, 03:08 PM
Cyrus thanks for confirming that my about 27% was correct and your 24.1% did not take into account the fact that you knew the other player holds QQ.

Cyrus
06-06-2003, 02:11 AM
The original question seemed clear enough to me ("What is the % chance of A/K matching either the A or K on the turn and river?"). The poster however had "A/K vs QQ" in his post's title. So, I guess, this could be interpreted two ways : We want the chance of hitting our AK (1) not knowing our opponent holds QQ, and (2) knowing our opponent holds QQ.

The probability for the first case, where we don't know anything actually about our opponent's hand, was provided above, and is 24.1%. The whole Table in the "Addendum" post, in fact, is based on ignoring what our opponent holds. Which is the usual situation at the tables.

The probability for the second case is indeed higher than the first, because we know that our opponent holds QQ which means he holds neither Ace nor King! The probability, of our AK hitting the Turn and/or the River, is actually the same whether our opponent holds QQ or J5. No matter what he holds, as long as he doesn't have A or K.

Thanks to Jimbo for helping me clarifying this.

--Cyrus

PS : AK is of course a dog against QQ in the situation we examine, i.e. when neither has hit the flop. With a raggedy flop that helps neither hand, e.g. 9 6 3 rainbow, QQ is a 3.2:1 favorite against AKo, and 2.5:1 against AKs.

Legend27
06-06-2003, 06:33 AM
Let me start my reply off by saying i'm not good at math.

So if you have a 24% of matching the A or K on the turn and river then lets say we saw the turn and it was not an A or K then on the river A/K would have 6 outs and about a 12% of matching?

So if A/K has a 12% of matching on the river, wouldn't that mean A/K has a little less than a 60% chance of matching the A or K after 5 cards were dealt? (5 x 12%)
I know that that is not right, I just want to know why my thinking is flawed.

Here's a random question: You hold A/Ks. You are not playing against an opponet. What % of the time does A/Ks match the A or K, get the straight, or get the flush? What I mean is what is that total %?

Cyrus
06-06-2003, 12:28 PM
"I'm not good at math."

That makes two of us. Everybody else here is very good.

"If you have a 24% [chance] of matching the A or K on the turn and river then let's say we saw the turn and it was not an A or K, then on the river A/K would have 6 outs and about a 12% of matching?"

Don't divide that 24.1% in half! Here's the correct (and quite straightforward) calculation : You have 6 outs and 1 card to come. (52 - 2cards in your hand - 4cards on the board)=46 unseen cards, and 6 outs. 6/46=0.13 or 13%.

(This assumes that you don't know what your opponent holds or doesn't hold.)

"So if A/K has a 12% of matching on the river, wouldn't that mean A/K has a little less than a 60% chance of matching the A or K after 5 cards were dealt? (5 x 12%)"

I'm sorry but I don't know what you mean. Are you perhaps saying that, with "5 cards to be dealt", the chance of hitting your AK is 60%? It's not. You shouldn't multiply the number of cards to come with the probability of hitting with one card to come!

NutJob
06-06-2003, 12:53 PM
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
06-06-2003, 12:54 PM
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
06-06-2003, 09:28 PM
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
06-06-2003, 11:06 PM
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
06-06-2003, 11:57 PM
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 (http://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
06-07-2003, 12:48 AM
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
06-07-2003, 01:56 AM
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?

/forums/images/icons/cool.gif

Legend27
06-07-2003, 07:35 AM
Alright that makes alot more sense now, thanks for the help.

Tekari
06-18-2003, 04:46 PM
when you are saying C(47,2)=1081 what math is involved to get the 1081?

thanks

BruceZ
06-18-2003, 11:15 PM
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]

Cyrus
06-19-2003, 11:22 AM
Tekari,

The shortcut (more shortcuts! /forums/images/icons/smile.gif) to the formula provided by BruceZ in his response, and the way to calculate all combos, is the following:

You want to know how many combinations you can make by picking 5 different items every time from a total of 47 items. That's C(47,5).

What you do is you put in the denominator the number 5!, or 5factorial, in other words 5*4*3*2*1. The numbers in the multiplication are of course 5.

Start the factorial of 47 in the numerator, only cut off the sequence after 5 numbers again. In other words, the numerator will be 47*46*45*44*43.

Which means that C(47,5) = (47*46*45*44*43) / (5*4*3*2*1) = 1,533,939.

To calculate C(47,2) by hand the same thing would have been done --- but I do this on Excel. /forums/images/icons/smile.gif

Take care.

--Cyrus

PS : This simplification is extensively presented also in Sklansky's "Getting The Best Of It".

PPS : Note that when you are faced with a monster such as C(47,42), which asks you to get the total number of combinations with each combination having 42 items out of a total of 47 items, this is actually equal to C(47,5). That 5 is simply 47-42=5. It's much easier to calculate the latter than the former, and they give the same result!
In general, for every n&gt;k, C(n,k)=C(n,m), whereby m=n-k.

BruceZ
06-19-2003, 11:55 AM
Tekarki,

Like I already mentioned:

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

This is the "shortcut" of "cutting off the sequence" in the numerator, defined precisely. Nothing more to the "shortcut" or of course I would have presented it. Nothing more needed other than an understanding.

-Bruce

BruceZ
06-19-2003, 12:32 PM
Actually, if your calculator has a factorial key (!) but no key for C(n,k), then the above so-called shortcut is anything but. The real shortcut is n!/k!/(n-k)!. That's a lot shorter than multiplying a bunch of numbers.

Cyrus
06-20-2003, 03:00 AM
I referred to "a shortcut to the formula given by BruceZ", meaning the complete formula for combinations, C(n,k)= n! / [k!*(n-k)!] , which BruceZ posted (DeGroot p.28). BruceZ went on to simplify this in mathematical terms, C(n,k) = n*(n-1)*(n-2)*...*(n-k+1) / [k*(k-1)*(k-2)*...*1]. I gave a practical step-by-step, a summary actually of what Sklansky provides in "Getting The Best Of It".

So, yes, the shortcut is as described above by the notable contributor to this forum. (Also, in the formula posted by BruceZ /forums/images/icons/grin.gif )

--Cyrus

[I see that BZ has taken some kind of vow not to respond directly to my posts ( /forums/images/icons/frown.gif ) or even address me by name ("certain posters"). I'll follow suit this time and respond to my own post, too. Maybe there's some method to this.]