Probability Review: Aces, Who Has Aces?
One of the ‘good’ things about poker math is that almost always there are several ways to calculate a probability. However, there is a negative to that, namely, if one is used to method A and someone in a discussion uses method B, then that may be confusing.
In this article we present various methods for determining poker probabilities. We will answer the following question to show the methods:
What is the probability that you will be dealt at least one ace in a hold ’em hand?
Some methods will be fairly obvious and perhaps some may be puzzling. At the end, we will pose some problems and suggest you try to work out the answers several ways to help in understanding each method. Answers are provided at the end of the article.
As a reminder, for card deals, we define the probability of an event as the number of ways or outcomes the event can occur divided by the total number of outcomes. For a joint event, say the simultaneous occurrence of 2 events, A and B, we have Pr(A & B) = Pr(A) *Pr(B|A), where Pr(B|A) is the conditional probability of B given that A has occurred. If A and B are independent—the occurrence of one has no effect on the occurrence probability of the other-- then Pr(B|A) = Pr(B). As a result, for independent events, Pr(A & B) = Pr(A)*Pr(B).
For either of two events, A or B or both, Pr( A or B) = Pr(A) + Pr(B) – Pr(A & B ). If the events are mutually exclusive (cannot occur simultaneously), then Pr(A & B) = 0, so Pr(A or B) = Pr(A) + Pr(B).
Notation for dealing the hold ’em hand and looking for aces:
A1 = Ace on first card; A2=Ace on second card
X1 = any non-ace on first card; X2 = any non-ace on second card
Joint Probability Method: Ace on only the first card or Ace only on second card or Ace on both cards
Pr(Dealt an ace) = Pr(A1 & X2) + Pr(X1 & A2) + Pr(A1 & A2). (Note that the three joint events are mutually exclusive)
The deck has 4 aces and 48 non-aces. Therefore,
Pr(A1 & X2) = 4/52 * 48/51 = 0.0724 (48/51 is a conditional probability)
Pr(X1 & A2) = 48/52 * 4/51 = 0.0724
Pr(A1 & A2) = 4/52 * 3/51 = 0.0045
Pr(Dealt an ace)= 2*0.0724 + 0.0045 = 0.1493.
Complementary Method: One minus the probability no ace is dealt
Pr(Dealt an ace) = 1 - Pr(X1 & X2)
Pr(X1 & X2) = 48/52 * 47/51 = 0.8507
Pr(Dealt an ace) = 1- 0.8507 = 0.1493
Conditional Probability Method: Ace on first card or no ace on first card and ace on second card
Pr(Dealt an ace) = Pr(A1) + Pr(X1)*Pr(A2|X1)
= 4/52 + (48/52)*(4/51) = 0.1493
Inclusion-Exclusion Method: P(A1 or A2) = Pr(A1) + Pr(A2) - Pr(A1 & A2)
Pr(Dealt an ace) = 1/13 + 1/13 –(1/13)*(3/51) = 0.1493
Combinatorial Method: Pr = Number of combos with at least one ace/Total number of combos
C(n,r) = n!/[r!(n-r)!], where m! = m * (m-1) * (m-2) * . . . * 3 *2 * 1
Number of Single Ace Combos = C(4,1)*C(48,1) = 4*48 = 192 (1 of 4 aces and 1 of 48 non-aces)
Number of AA Combos = C(4,2) = 6 (2 of 4 aces)
Number of Hand Combos = C(52,2) = 1326 (2 of 52 cards in the deck)
Pr(Dealt an ace) = (192 + 6)/1326 = 0.1493
Permutation Method: Number of perms (ordered events) with at least an ace / Total number of permutations
P(n,r) = n!/(n-r)!
Number of Single Ace (AX) Perms =2* P(4,1)*P(48,1) = 2*4*48 = 384 (the 2 is for two positions the ace can occupy)
Number of AA Perms = P(4,2) = 12
Number of Hand Perms = P(52,2) = 52*51 = 2652
Pr(Dealt an ace) = (384 + 12)/2652 = 0.1493
Summarizing -- we used 6 different methods to calculate the probability you will be dealt an ace. It is advisable to use at least two methods in doing your own calculations for probability is a tricky animal and therefore a check can often reveal an error.
Problems
1. What is the probability you will be dealt a pair of aces?
2. What is the probability you will be dealt exactly one ace?
3. What is the probability villain has a pair of aces
a. given you don’t have an ace?
b. given you have one ace?
4. In a heads-up game, what is the probability neither player has an ace?
5. What is the probability the flop has exactly two aces?
6. What is the probability the flop has two cards the same suit?
7. Given you have 88, what is the probability you will flop a set or better (ignore straights and flushes)?
8. Given you have 88 and no overcard on the flop, what is the probability there will be an overcard by the river?
9. Given you hold 2 spades and the flop has 2 spades, what is the probability you hit a flush by the river.
10. What is the probability villain has a pair
a. given no other specification
b. given you do not have a pair?
c. given you have a pair?
Bonus Question. 8 is the median rank. In a heads-up match, you are dealt 8 X, X< 8. What is the probability your 8 outranks your opponent?
Answers (using one of the above methods; can be done other ways):
1. 4/52 * 3/51 =0.45%
2. 4/52 * 48/51 + 48/52 * 4/51 = 14.48%
3. a. 4/50 * 3/49 = 0.49%
b. 3/50 * 2/49 = 0.24%
4. C(48,4)/C(52,4) = 71.9%
5. C(4,2) * C(48,1)/C(52,3) = 1.30%
6. 4 * C(13,2) * C(39,1)/C(52,3)= 55.1%
7. 1 - C(48,3)/C(50,3) = 11.8%
8. 24/47 + 23/47 * 24/46 =76.6%
9. (9*38 + 38*9 + 9*8)/(47*46) = 0.35
10. a. 13 * C(4,2)/C(52,2) =0.0588
b. [11 * C(4,2) + 2 * C(3,2)]/C(50,2) = 0.0588
c. [12 * C(4,2) + 1 * C(2,2)]/C(50,2) = 0.0596
Bonus: Prob = Chance of 2 cards less than 8 = C(23,2)/C(50,2) = 11*23/1225 = 253/1225 =0.207
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