(Biginner Forum) Poker Math Quiz

Would somebody please come up with an occasional Poker Math Quiz for us beginners(and then maybe intermidiate questions), the quizes on the General Theory Forum are just WAY WAY WAY too hard.


Thanks

You have As Js:


The flop comes Qs 8h 2s


Q1. What percentage of the time will you complete your flush draw by the river?


Q2. What ratio is this? IE 4:1


Is this too easy, just right, or too hard?


Derrick

Q1. What percentage of the time will you complete your flush draw by the river? 35%


Q2. What ratio is this? 1.85:1


I memorized a cheat sheet, so I don't know how they arrived at these answers. I understand how to figure out the percentage for making a four flush and outside straight on the turn, but I don't know how to figure it out for both the turn & river.


For the turn only is: 9/47 = 1.9 or 19%


the odds are: 38:9 = 4.2:1


Anyway, your questions are at least approachable for me personally. Thanks

I like the idea of beginner math quizzes. Thank you Derrick for trying your hand at one. The only problem with it was that it was a number I think every beginner just memorizes, but you're on the right track and I hope you post more of them.


Another idea is to have an occasional basic strategy quiz or, even better, a "Read this Hand" quiz, where the play on each street is described and we are asked to read a player's hand. Reading hands is something every beginner struggles with, and it is a common subject on this forum.


Thanks to all who take the time to post a quiz.


Brett

Q1: You have a 4 card flush on the flop what are your chances of completing the flush?


S1a: Let X be a non flush card, and let F represent a flush card. The following combinations will give you a flush: <table> <tr><td>X-F</td></tr> <tr><td>F-X</td></tr> <tr><td>F-F</td></tr> </table>


There are 52 cards in the deck, 2 in your hand, and 3 on the flop. There are 47 unseen cards of which 9 are flush cards and 47-9=38 are non-flush cards.


The chance that the turn is a blank (X) AND the river is a flush card is worked out below:


38/47 * 9/46 = 0.158 or 15.8%


Similarly the chance that the turn is a flush card AND the river is a non flush card is: 9/47 * 38/46 = 0.158 or 15.8%


Finally the chance that the cards come F-F are 9/47 * 8/46 = 0.033 or 3.3%


Summing all the possible solutions together you have .3499 or 35%


S1b: Another way to solve this problem is to determine the chance of not completing the flush which means the turn and river would be blanks. If you find the odds that this will occur you can subtract that from 100% and get the odds that a flush will come.


Chance of X-X: 38/47 * 37/46 = 0.650 or 65%. Therefore the chance of completing the flush is 100% - 65% = 35%.


S2: Ratios are basically the number of cards that help vs. the number of cards that don't. For instance the ratio of completing the flush on the turn can be found as follows:


There are 9 cards out of 47 that help you and 38 cards out of 47 that don't help you. Your chance of completing the flush on the turn is 38:9 against or 4.22:1.


Now, since you have calculted a percentage of 35% to complete you can say there are 35 times out of 100 that you will complete the flush, and there are 65 times out of 100 that you will fail to complete the flush. Therefore your ratio is 65:35 against or 1.86:1.


A real world lesson to the above: We have seen that your chance of completing a 4 card flush is about 35%. This problem is good if you have the A high flush, but if you have 8-7 suited you only want a 3 card flush on board, so you must only consider the first 2 scenarios in solution 1a as viable outs. Your chance of completing a 4 card flush and not having a 4 flush on board is 15.8% + 15.8% = 31.6% or 68.4:31.6 or 2.16:1


There is also a 3rd way to determine your odds of completing a hand called combinations. I am not as good at this, so maybe someone else will show you.


I hope this helps,


Derrick Ashworth

"Reading hands is something every beginner struggles with,.."


I know I do.

Wow, thanks!


I had no idea there was that much to it, plus you're really good at simplifying this stuff, thanks again.

I'm confused, which is nothing new.


I thought you did it this way:


there are 47 cards left after the flop. The odds of getting that last flush card on any given chance is 47/9, which is 5.22:1.


In percentages, this is 1/6.22 = 16.08%.


The chance of hitting the flush with two chances is .1608 X .1608 = 25.86%.


Or am I an idiot?


Er, thanks, Greg

Greg,


This is a very common misconception. Consider that the turn is 1 event, and the river is another. The outcome of the river event depends on the outcome of the turn event.


Here is an easier way to explain mutual exclusion.


Say you have a full deck of cards and you want to draw an A. Your odds of drawing an A are 4/52... relatively obvious. Now say you draw a card and it is the 2c... you throw it away. Now what are your chances of drawing an A... 4/51. If you had returned the card to the deck the chance of drawing an A would again have been 4/52, but the one event had changed the chance of the subsequent event.


An Example: You have a full deck of cards and you want to draw an A on your second draw from the deck. The card you draw first is not returned to the deck, what are your chances of drawing an A on your second draw?


Solution:


There are 2 ways that this could happen, you draw an A on card 1 and card 2 OR you don't draw an A on card one and you draw an A on card 2.


4/52 * 3/51 + 48/52 * 4/51 = 204 / (52*51) = (4 * 51) / (52*51) = 4/52


Pretty obvious answer but not so obvious a solution.


Another thing I noticed about your post is that you have 9/47 chance of catching a flush on the turn... you are correct on this, but that does not equate to 5.22:1 chance of hitting your flush card. Think of a ratio as follows:


(The number of cards that help you (9)):(The number of cards that hurt you 47-9=38) = the ratio of 9/47 9:38 or 4.22:1.


I hope this shows you where you have been misinformed.


Derrick

Ok, this line is what helps me the most:


(The number of cards that help you (9)):(The number of cards that hurt you 47-9=38) = the ratio of 9/47 9:38 or 4.22:1.


I was looking at it backwards (and hence made up charts to memorize incorrectly!) I really thank you for pointing this out to me. Now it's back to the calculator!


Another question, though: I have a cheat sheet that lists the number of outs in one column, the odds against getting the card you need, e.g. 4.22:1, and what percentage this represents. Provided this is done correctly (hehe) is this a good thing to do? I use it so that say, given I have a flush draw, and I'm 4.22:1 to make it, I'll need this much money in the pot for my call or whatever to be correct.


Meaning, is the rote memorization of this stuff helpful?


Thanks, Greg

It is very important to be able to approximate your outs when you are deciding what to do with a hand.


So, IMO, this is a good chart to continue with.


If you keep track of the number of bets in the pot you will get more use out of the ratio chart for instance if you know there are 13 SB in the pot and you have to call 2 SB then you need 13:2 (6.5:1) to call.


Derrick

For some reason, I have always had an easier time figuring this stuff out using combinations. It is easier for me to visualize. Just an example of looking at first part of this question using combinations.


Making a flush by the river..... At this point, we have seen 5 cards (two in hole plus three on the board), leaving 47 cards. We will be seeing two more cards, so there are a total of 47*46/2*1 = 1081 possible combinations. If we take out all of the suited cards we need (9 remaining flush cards) there are 38*37/2*1 = 703 combinations that result in a missed flush. Thus 703/1081 = 65% of the time we miss, or 1 - .65 = 35% of the time we make our flush by the river.


BTW, I think there are actually some scenarios that can only be solved using combinations, but I can't remember where the other method breaks down of the top of my head.

I have a table of Hold 'em odds for any number of outs, with one and two cards to come, but I got it from Ken Warren's book, which I know has been widely panned, and I've noticed differences between some of his figures and those S&M have printed. I have tried calculating it independently, but trusting my math is dangerous.


Does anyone have a reliable outs table they could share?


Brett

I could email you one, if you want.


Derrick

Derrick,


I would love a copy of your odds table. My email is kubbfan@earthlink.net.


Thanks! Greg

I received your email with the odds table. I appreciate it.


Greg

Please send one, if you don't mind.


BTW, I worked on the quiz questions above, and just saw your answer to Question #1. This is doing wonders to fine tune my understanding of these calculations. Thanks for writing these quizzes.


Brett

Is it possible to send one table copy to me? Please Thanks

Thanks,


Derrick

It's an ongoing effort that you'll continue to try and improve on for the rest of your poker playing career.

You make 2 mistakes


first mistake is this 16% times 16 % is 2,58% not 25%


second mistake is that you have find out all the posibilities: they are: You can make the flush on the turn 9/47 You don't make the flush on the turn 38/47 If you don't make the flush on the turn you have 9/46 ( you've seen one more card so 47-1) and 37/46


Now you can calculate all the posibilties individual and you wil com to the answer, but there is aan more easy way.


or you make the flush or you don't if you know the odds that you don't make it you als know the change that you do make it because you know that together it should be 100% The change that you don't make the flush are 38/47 * 37/46=0,65 or 65% so the change that you do make it is 100%-65%=35%

Well there's a 3th way but it's more or less the same as the other one: *how many cards are left 47 *in how many different ways can you take 2 cards out of 47 *how many of these combinations contains at least one flushcard or how many of these combinations don't contain a flushcard


you can take 2 cards out of 47 ,in 47 above 2 ways this is a math term what means this.


47!/(2!*45!)=1081 ways (e.g. 5!=5*4*3*2*1=120) Notice that 47!/45! is 47*46*45*....*1/45*44*43...1 so that's the same as 47*46 Because you can make this hand in two ways AB and BA which are in this example the same (we're not interested in the order) we have to divid this by 2! because you can take 2 cards from 2 cards in to different ways AB and BA


Now you have to calculate in how many way you can take 2 cards from 47 which don't contain you flushcard this is 38 above 2=38!/(2!*36!)=703 So the change that you don't make the flush are 703/1081=0,65 of 65% so the change you do make it is 100%-65%=35%