Note: Not at the old Poker1 site. This 39part series of quizzes, originally published (20042006) in Poker Player, is based on the Mike Caro University of Poker library of research and advice. In each entry, Mike Caro presents 10 questions covering a category of poker, targeted for beginner, intermediate, or advanced players. Answers with explanations appear below each quiz, with the questions repeated for easy reference.
The MCU Targeted Poker Quiz series
(See the index to this series)
Odds (level: beginner)

What are the odds against beginning with three of a kind in sevencard stud
(a) 424to1;
(b) 1400to1;
(c) 220to1;
(d) exactly 190.7to1.

What are the odds against beginning with suited cards in hold ’em?
(a) 3.25to1;
(b) 3to1
(c) 4to1;
(d) 12to1.

What are the odds against beginning with a pair in hold ’em?
(a) 16to1;
(b) 25to1;
(c) 47to1;
(d) 3to1.

What are the odds against beginning with three suited cards in sevencard stud?
(a) 90to1;
(b) 42to1;
(c) 18.3to1;
(d) 5.7to1.

Your odds against starting with a pair in any poker game depend on the number of players who are dealt in:
(a) true;
(b) false.

What are the odds against starting with a pair of aces in hold ’em?
(a) 19.8to1;
(b) 220to1;
(c) 403to1;
(d) 1,279to1.

What are the odds against starting with a pair of deuces in hold ’em?
(a) 19.8to1;
(b) 220to1;
(c) 403to1;
(d) 1,279to1.

If the board is 872A in hold ’em and you hold 109, what are the odds against you making a straight on the last card (assuming you know nothing about the cards opponents may be holding)?
(a) 5.75to1;
(b) 11.25to1;
(c) 3.5to1;
(d) 4.75to1.

If you hold a pair to start with in hold ’em, what are the odds against flopping four of a kind?
(a) 1,003to1;
(b) 407to1;
(c) 80to1;
(d) 13to1.

If you start with a pair in hold ’em, what are the odds against flopping a straight or a flush?
(a) 2,540to1;
(b) 19to1;
(c) 402to1;
(d) none of the above.
Answers and explanations (with questions repeated for convenience)
Odds (level: beginner)

What are the odds against beginning with three of a kind in sevencard stud
(a) 424to1;
(b) 1400to1;
(c) 220to1;
(d) exactly 190.7to1.
Answer: (a). It’s 424to1 against beginning with three of a kind in sevencard stud. This type of problem is particularly easy. Think about it this way… The first card counts no matter what, because it must be some rank. After that, there are 51 cards left of the original 52, and since there are four cards of each rank to begin with, there are now three matching your first rank. Three out of 51 is the same as 1 out of 17 (divide 51 by 3, if you doubt this). So, there’s one chance in 17 that the first two cards will constitute a pair. If that happens (and only if that happens) do we care what the third card is – at least, for the purposes of this calculation. If you now have a pair, then there are 50 cards remaining that are unknown to you and only two that complete your three of a kind. That’s the same as one in 25. Now, we just multiply the chances – 17 times 25 = 425. That’s a universe of 425 chances and only one that qualifies as three of a kind. So, it’s 424to1 against.
I won’t go into this depth with the other answers, but it’s interesting to note that some calculations – like this one – can be relatively simple to reason out. When I did my collection of 50 poker tables for Doyle Brunson’s Super/System in 1977, some were extraordinarily complex and others were just as easy as this. But most of the poker literature of the day was oddly wrong, even when citing some of the simplest statistics. 
What are the odds against beginning with suited cards in hold ’em?
(a) 3.25to1;
(b) 3to1
(c) 4to1;
(d) 12to1.
Answer: (a). The odds against beginning suited in hold ’em are 3.25 to 1.

What are the odds against beginning with a pair in hold ’em?
(a) 16to1;
(b) 25to1;
(c) 47to1;
(d) 3to1.
Answer: (a). The odds are 16to1 against beginning with a pair in hold ’em.

What are the odds against beginning with three suited cards in sevencard stud?
(a) 90to1;
(b) 42to1;
(c) 18.3to1;
(d) 5.7to1.
Answer: (c). The odds are 18.3to1 against beginning with three cards of the same suit in sevencard stud.

Your odds against starting with a pair in any poker game depend on the number of players who are dealt in:
(a) true;
(b) false.
Answer: (b). It’s false. The odds against starting with a pair in any poker game are completely independent of the amount of cards dealt or the number of opponents. It makes no difference whether unknown cards remain in the deck or are dealt facedown to others – they are still unknown and must be accounted for as such.

What are the odds against starting with a pair of aces in hold ’em?
(a) 19.8to1;
(b) 220to1;
(c) 403to1;
(d) 1,279to1.
Answer: (b). It’s 220to1 against starting with a pair of aces in hold ’em.

What are the odds against starting with a pair of deuces in hold ’em?
(a) 19.8to1;
(b) 220to1;
(c) 403to1;
(d) 1,279to1.
Answer: (b). It’s also 220to1 against starting with a pair of deuces (or any other pair) in hold ’em. Aces are no harder to get than any other rank. We just care more about them, because we’ve arbitrarily decided they’re worth more.

If the board is 872A in hold ’em and you hold 109, what are the odds against you making a straight on the last card (assuming you assume nothing about the cards opponents may be holding)?
(a) 5.75to1;
(b) 11.25to1;
(c) 3.5to1;
(d) 4.75to1.
Answer: (d). It’s 4.75to1 against hitting an openend straight on the river in hold ’em (if you make no assumption about opposing cards).

If you hold a pair to start with in hold ’em, what are the odds against flopping four of a kind?
(a) 1,003to1;
(b) 407to1;
(c) 80to1;
(d) 13to1.
Answer: (b). It’s 407to1 against flopping four of a kind in hold ’em if you begin with a pair.

If you start with a pair in hold ’em, what are the odds against flopping a straight or a flush?
(a) 2,540to1;
(b) 19to1;
(c) 402to1;
(d) none of the above.
Answer: (d). It’s impossible to flop a straight or flush in hold ’em if you begin with a pair.
Hi Mike!
I’m having trouble with the math for Question 9. Could you spell it out? (I’m not sure how to express the probability that we have three chances to hit the two cards that we need.)
Thanks!
Hi, Brian —
Thanks for posting your first comment to Poker1.
There are many ways to calculate this, but the easiest is to think about how many combinations of three cards there could be to represent the flop.
That would be 50 unknown cards times 49 unknown cards after the first times 48 unknown cards after the second. Since these can arrive in six different orders and they’d all be the same flop combination, we divide by six.
The reason for the six ways is that the first card can be any of the three and then the second can be any of the remaining two, while the third is a given. So, 3 x 2 = 6 ways.
The formula for number of possible flop combinations, then, is (50 x 49 x 48) / 6 = 19,600.
How many of these include a matching pair with another card? Well, there’s only one matching pair left in the deck and 48 other cards it can be coupled with. So, there are 48 combinations that will make four of a kind.
Now just divide 19,600 by 48. That’s 408.33 and we’ll just say it’s 408, for simplicity. That means there’s one chance in 408 of flopping four of a kind when you hold a pair and 407 chances of not doing so.
Therefore, it’s 407to1 against flopping quads — answer (b).
Hope this helps.
Straight Flushes,
Mike Caro
Thanks for your reply! I understand how you did it, but I think I thought of another way to work it out that might help some people see it.
Suppose we have pocket AA (lucky us!). There are three ways we could flop quads: AAx, AxA, and xxA (where x is a random nonA). Each of these is equally likely (the cards don’t care what order they come out in), so we just calculate the probability of any one of these and multiply it by three.
The probability that the fist card will be an A is 2/50 and the probability that the second will be an A is 1/49, so the joint probability is the product, which is 2/2450 or 1/1225.
So we take this amount and multiply it by 3 to get 3/1225 or approximately 1/408 = 407:1 against.
Hi, Bryan —
That’s another way to do it — and a very good one.
That’s why I said, “there are many ways to calculate this.” You’ll eventually have a collection of statistics that are often reused — such as 19,600 flops for a specific hand.
Once you’ve done that, you’ll just say, hey, there’s one pair that can be paired with 48 cards to give me four of a kind. Divide 19,600 by 48 and… almost instant answer!
Straight Flushes,
Mike Caro
It’s most convenient for me to consider that a deck consists of 13 ranks and 4 suits. Therefore, if you have:
(p,p) and you want: p,p,x on the flop:
Then: you have one choice for the rank of that pair: that it match the rank of the pocket pair.
For the odd card, you have 12 choices for its rank.
Then: for the flopped pair, you have one choice for suits (i.e. if you hold a pair of red cards, the pair on the flop must be black cards). For the odd card, you have four choices for its suit.
Therefore: 1 X 12 X 1 X 4= 48
Pr= 48/19600= 2.449E3
Odds= 1/Pr – 1= 407.333 : 1
it seems like 4 runners of the same suit could produce a flush–Question 10
Hi, Harry —
Welcome to Poker1.
Four running cards of the same suit certainly could produce a flush, but you might be misunderstanding the terminology we’re using here.
We’re only considering the flop to be the traditional three cards, so the closest you could come to a flush after that would be four parts.
Straight Flushes,
Mike Caro