I need a Hold'em math lesson

[SoBeDude]

Hello all,

I'd like a statistics lesson on how to calculate the odds of a card falling with more than one card to come, or the odds of a given card or two hitting the flop:

For example:
I'm holding JJ against raisers and several opponents. What are the odds I'm going to see an overcard on the flop?

Now there are 12 overcards in the deck, so with one card its trivial...its 12/50 or 24%. (I can only see my two)

But with 3 cards to come, itsn't it 12/50 + 12/49 + 12/48?

To trivialize the example for the sake of clarity let us assume I'm drawing one card out of a deck of ten cards, A-10.

with 1 chance to draw I'm 1 in 10. with 2 chances I'm just over 2 in 10, with 3 I'm better than 3 in 10.

This seems to indicate I add my chances together. 1/10 + 1/9 + 1/8. So if this is correct, in my jacks problem 33.6% of the time I'll see an overcard to my jacks on the flop.

Additional math problem:
I know if I need two cards runner-runner I need to multiply the odds for each card together. But how do I calculate the odds of two cards hitting the flop?

In the above example with JJ, what are the odds of me seeing any two overcards to jacks on the flop? So any combination of A, K, and Q in three cards.

In the Jacks example, against two or three players, having two overcards on the flop means I'm in big trouble, expecially if there was a raise pre-flop.

I think this would also be good to help identify the odds of seeing trouble cards after the flop as well. Lets say I'm in the BB with 8-10 off and the flop comes 2 9 T rainbow. Odds seem good I'm going to be looking at an overcard or two soon, so against multiple players, what odds do I need to play?

Thanks in advance for any and all comments and math lessons!

-Scott

[pudley4]

</font><blockquote><font class="small">In reply to:</font><hr />
But with 3 cards to come, itsn't it 12/50 + 12/49 + 12/48?

[/ QUOTE ]

No. Let's look at it this way:

Y=The card is an overcard.
N=The card is not an overcard.

With 3 cards on the flop, there are 4 possible outcomes (we don't care about the order of the cards):

1-YYY (all 3 are overcards)
2-YYN (2 overcards)
3-YNN (1 overcard)
4-NNN (no overcards)

To figure out the probability of at least 1 overcard coming, we could do 2 things:

1-Figure out the probability of scenarios 1, 2 and 3 and add them together

or

2-Figure out the probability of scenario 4 and subtract that number from 1 (remember that the sum of all scenarios must equal one)

You can see that it's much easier to calculate one probability than it is to calculate three separate ones. The answer then is:

1-[(38/50)(37/49)(36/48)]=1-.4304=.5696. So about 57% of the time at least 1 overcard will come.

To verify we can calculate scenarios 1-3.

Scenario 1:

(12/50)(11/49)(10/48)=.0112

Scenario 2:

(38/50)(12/49)(11/48)=.0427. We need to multiply this by 3 because there are 3 possible ways this flop can occur - YYN, YNY, NYY. So we get an answer of .1280

Scenario 3:

(38/50)(37/49)(12/48)=.1435. This also needs to be multiplied by 3, so we get .4304.

Add these together and we get .4304+.1280+.0112=.5696, which matches our answer above.

[pudley4]

</font><blockquote><font class="small">In reply to:</font><hr />
To trivialize the example for the sake of clarity let us assume I'm drawing one card out of a deck of ten cards, A-10.

with 1 chance to draw I'm 1 in 10. with 2 chances I'm just over 2 in 10, with 3 I'm better than 3 in 10.

This seems to indicate I add my chances together. 1/10 + 1/9 + 1/8. So if this is correct, in my jacks problem 33.6% of the time I'll see an overcard to my jacks on the flop.


[/ QUOTE ]

First question - are you replacing the card(s) after you draw or not? It affects the answer.

Let's assume you are not replacing. The easiest way to figure this out is very intuitive. Since we will keep drawing until we choose the A, and we have the same chance of drawing any card at any time, we should see the A come up on the first draw exactly as often as on the second draw, exactly as often on the third, etc. So the probability of picking the A on any one specific draw is 10%.

To figure this out, we can look at it this way:

The probability of picking it on exactly the first try is 1/10=10%.

The probability of picking it on exactly the second try is: The probability of NOT picking it on the first try multiplied by the probability of picking it on the next try.

Not picking it the first time is 9/10. Picking in the second time is 1/9 (because there are only 9 cards left). Multiply them together and we get 9/10 * 1/9 = 1/10.

Continuing on to the 3rd try we get the same results. The probability of NOT picking it on try 1 is 9/10. The probability of NOT picking it on try 2 is 8/9. The probability of picking it on try 3 is 1/8. Multiply these together and you again get 1/10.

[pudley4]

Now let's look at it with replacing.

Now the probability of picking the A is 1/10 each time you draw. We can look at it exactly the same way as I showed in the JJ example - either calculate the probability of each scenario (picking it on try 1, picking it on try 2, etc), or calculate the probability of NOT picking it and then subtract from 1.

Let's do the 2nd one first.

The probability of NOT picking it on or before the 3rd try is: 9/10 * 9/10 * 9/10 = .729. Subtract this from 1 and we get .271.

Now calculate it the first way.

The probability of getting it on the first try is 1/10 = .1. The probability of getting it on the second try is 9/10 * 1/10 = .09. The probability of getting it on the third try is 9/10 * 9/10 * 1/10 = .081. Add them together to get .1+.09+.081 = .271

[Homer]

1) Probability of at least one overcard on flop:

Do this as one minus the probability of zero overcards coming on the flop.

The probability of the first card not being an overcard is 38/50. Given that the first card is not an overcard, the probability of the second card not being an overcard is 37/49. Given that the second card is not an overcard, the probability of the third card not being an overcard is 36/48.

So, the probability that you will NOT see an overcard on the flop is:
(38/50)*(37/49)*(36/48) = 43.04%

The probability of seeing at least one overcard is one minus this result:
1 - .4304 = 56.96%

2) Probability of two or more overcards on flop:

P(2 overcards) = C(3,2) * [(12/50)*(11/49)*(38/48)]

C(3,2) is the number of combinations of 3 items taking two at a time. This is needed because there are a number of ways to have 2 overcards out of 3 cards drawn (O for overcard and X for non-overcard). You could have:

OOX
XOO
OXO

So you can see that there are three ways to have exactly two overcards out of three cards drawn.

P(2 overcards) = 3 * [(12/50)*(11/49)*(38/48)] = 12.80%

P(3 overcards) = C(3,3) [(12/50)*(11/49)*(10/48)] = 1.12% (Note than C(3,3) equals one since there is only one way to have 3 overcards out of 3 cards drawn).

P(&gt;=2 overcards) = .1280 + .112 = 13.92%

-- Homer

[SoBeDude]

Thanks for the responses. I appreciate the clarification!

Now to start calculating...

-Scott