First Ace Problem

**Problem:**
One way to determine who deals first in a Texas Hold-Em poker tournament is for one player to start dealing cards to each of the others.
The first person to get an Ace turned up starts the deal.
What are the probabilities of getting an Ace, given your seating postion relative to the dealer?
How do the probabilities change for different numbers of players?
Who has the advantage and disadvantage?

**Solution:**
First, consider the odds of drawing an Ace on the first turn of a card.
To compute the odds for this, simply divide the number of Aces in the deck (4) by the number of cards in the deck (52).
This comes out to about 0.0769, or 7.69% (I'll use percents from here on out.)
So much for the easiest case.

What are the odds of drawing an Ace on the second card?
For this to happen, you must first *not* get an Ace on the first draw *and* you must get an Ace on the second draw.
The odds of not getting an Ace on the first draw is 48 / 52.
The odds of getting an Ace on the second draw are 4 / 51 (since one card is already gone, right?)
The odds of both of those happening is the product of the odds of each of the two events, (48 * 4)/(52 * 51), or 7.24%.

To figure out the odds of each of the remaining possibilities, continue to multiply the odds of each event. For the odds of getting the first Ace on the third draw? It's (48 / 52) * (47 / 51) * (4 / 50) = 6.81%.

Your odds of getting further into the deck are always decreasing.
If you continue the whole way to 49 cards turned over, the numbers gradually get to zero.
The odds of drawing the *first* Ace on the 50th, 51st or 52nd draw are zero.
I used Excel to create the following table showing the odds for all the possible draws.

Card Position in the Deck |
Odds of Drawing the First Ace |

1 | 7.692307692308% |

2 | 7.239819004525% |

3 | 6.805429864253% |

4 | 6.388770892973% |

5 | 5.989472712162% |

6 | 5.607165943300% |

7 | 5.241481207868% |

8 | 4.892049127343% |

9 | 4.558500323206% |

10 | 4.240465416936% |

11 | 3.937575030012% |

12 | 3.649459783914% |

13 | 3.375750300120% |

14 | 3.116077200111% |

15 | 2.870071105365% |

16 | 2.637362637363% |

17 | 2.417582417582% |

18 | 2.210361067504% |

19 | 2.015329208607% |

20 | 1.832117462370% |

21 | 1.660356450272% |

22 | 1.499676793794% |

23 | 1.349709114415% |

24 | 1.210084033613% |

25 | 1.080432172869% |

26 | 0.960384153661% |

27 | 0.849570597470% |

28 | 0.747622125773% |

29 | 0.654169360052% |

30 | 0.568842921784% |

31 | 0.491273432450% |

32 | 0.421091513528% |

33 | 0.357927786499% |

34 | 0.301412872841% |

35 | 0.251177394035% |

36 | 0.206851971558% |

37 | 0.168067226891% |

38 | 0.134453781513% |

39 | 0.105642256903% |

40 | 0.081263274541% |

41 | 0.060947455905% |

42 | 0.044325422477% |

43 | 0.031027795734% |

44 | 0.020685197156% |

45 | 0.012928248222% |

46 | 0.007387570413% |

47 | 0.003693785206% |

48 | 0.001477514083% |

49 | 0.001846892603% |

This chart shows how the probabilities drop as you get further into the deck. We've come along way, but the problem we just solved still isn't quite what we're after.

Since we are interested in the odds of a person sitting at the table to get dealt the first Ace, keep in mind that any single person's odds are really the sum of all the possible draws that person may have. So, on a table with 8 players, the person sitting in position 1 (just to the left of the dealer) may get the first Ace on the first card drawn from the deck, or the 9th card drawn from the deck, or the 17th, and so on.

So, to figure out the odds for any one person sitting at the table, it depends on 2 things:

- the number of player positions at the table and
- how many seats away from the dealer you are.

The table below shows the probabilities for each seating position, based on the number of players anywhere between 2 and 10.

Number of Players Seated |
|||||||||

Seating Position |
2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

P 1 | 51.982270% | 36.001478% | 28.030289% | 23.261612% | 20.097516% | 17.846154% | 16.169545% | 14.871179% | 13.842460% |

P 2 | 48.019208% | 33.281005% | 25.930372% | 21.534768% | 18.616677% | 16.542986% | 14.996768% | 13.801828% | 12.854373% |

P 3 | 30.718995% | 23.951981% | 19.905808% | 17.220427% | 15.312217% | 13.890110% | 12.791209% | 11.919845% | |

P 4 | 22.088836% | 18.372518% | 15.903962% | 14.151261% | 12.845507% | 11.838582% | 11.037030% | ||

P 5 | 16.926771% | 14.664327% | 13.057531% | 11.860744% | 10.937298% | 10.204082% | |||

P 6 | 13.498569% | 12.028442% | 10.933604% | 10.089574% | 9.419152% | ||||

P 7 | 11.062887% | 10.061871% | 9.291717% | 8.680395% | |||||

P 8 | 9.243328% | 8.541878% | 7.985964% | ||||||

P 9 | 7.838212% | 7.335488% | |||||||

P 10 | 6.722689% |

As the second table shows, it is *always* best to be the first to draw and worst to be the last (which is the dealer, in this case).
The difference in the odds are smaller when there are fewer players.
With 2 players, the difference is less than 2%.
With 10 players, the person to draw first is more than twice as likely to get the first Ace than the dealer is!

**Sanity Check:**
To make sure it's all consistent, I used Excel to add up the probabilities in the second column of the first table and found the answer to be 100.00147751%.
Thousandths of a percent is within the bounds of expected error limits (that's another proof).
I did the same for each of appropriate columns in table 2 and came up with the same results.