Choosing the dealer in poker – is dealing to the first ace a fair system?
In Texas Holdem, sitting in the dealer position is a strategic advantage. The dealer position generally acts last in betting and is not forced to post blinds.
For a game in progress, the dealer position rotates around the table after each hand. But at the start of the game, the dealer position is simply assigned to one player.
So who gets to be dealer initially?
In poker tournaments, everyone gets a fair shot at holding the dealer position as seats are assigned randomly.
In home games, an attempt is also made to assign the dealer spot randomly. There are many methods to choosing the dealer. One of the common methods is dealing to the first ace. It works like this: the host deals a card to each player, face up, and continues to deal until someone receives an ace. This player gets to start the game as dealer.
The question is: does dealing to the first ace give everyone an equal chance to be dealer? Is this a fair system?
Let’s dig into the math and find out.
A hunch
Dealing to the first ace seems like a fair system. The first ace is randomly in the deck and everyone has a chance to get it.
But while playing home poker games, I noticed something unusual. It seemed like people who were first dealt cards were more likely to be dealer. It struck me this system was not entirely fair.
So why should there be any bias?
Although dealing the cards is random, I realized the unfair part of the system. The process of dealing simply stops when the first aces is dealt. If the very first card is an ace, then that person is dealer and no one even gets a chance. That means players who are earlier dealt cards have a “first-mover” type of advantage.
(This aspect me of the famous dice puzzle which I solved before)
In conclusion, I suspected dealing to the first ace involved an edge. What are the exact odds?
The distribution of the first ace appearing on the kth card
To solve the problem, it is helpful to solve a related question: what is the probability that the first time an ace is dealt from the deck is the 1st, or 2nd, or 3rd, or the kth card?
Once this distribution is known, it will then be possible to calculate the odds a person will get the dealer by summing up the possible “winning” positions. But more on that later. For now, let’s calculate the probability distribution of the first ace being dealt in position k.
To begin, note that a standard deck has 52 total cards of which 4 are aces.
What are the odds an ace will be the 1st; card dealt? The probability is readily calculated as the number of aces divided by the total cards which is 4/52.
So far easy enough. Continuing, what are the odds an ace will first be dealt as the 2nd card from the deck? This happens only if the following two events occur:
(i) the first card dealt was not an ace (48/52) AND
(ii) the second card dealt is an ace (4/51)
I have written the probabilities at the end of each condition. The probability for (i) is the number of non-ace cards divided by the number of total cards, or 48/52. The probability for (ii) is similarly calculated but just slightly more complicated. The numerator is the number of aces which is obviously 4. The denominator is the number of cards still left in the deck. As one card was dealt for event (i), there are 51 cards remaining. And hence the probability for (ii) becomes 4/51.
Therefore, the probability for the first ace being dealt as the 2nd card from the deck is the product of these two events, which is (48 x 4) / (52 x 51).
We can continue the exercise to calculate the first ace appearing on the 3rd card. This only happens when three events occur:
(i) the first card dealt was not an ace (48/52) AND
(ii) the second card dealt was not an ace (47/51) AND
(iii) the third card dealt is an ace (4/50)
The probabilities for each event are calculated in the same fashion as above: the only tricky part is remembering to decrement the numerators and denominators to account for the cards already dealt out.
Putting these together, the probability for the 3rd card being the first ace is (48 x 47 x 4) / (52 x 51 x 50).
By now it is evident the probability calculation has a pattern. We can thus generalize the logic to calculate the first ace appearing on the kth card.
The specifics for this to happen are the following events:
(i) the first card dealt was not an ace (48/52) AND
(ii) the second card dealt was not an ace (47/51) AND
…
(k) the kth card dealt is an ace [4/(52-k+1)]
This calculation is straight-forward and again the only tricky part is the diminishing numerators and denominators.
Multiplying these event probabilities together yields the chance as [48 x 47 x ... (48 - k + 2) x 4] / [52 x 51 x 50 x (52 - k + 1)], for 1 > k > 49
There is a restriction on k because the process can theoretically continue until there are just 4 cards left in the deck, all of which are all aces. And then the next card must be an ace.
I went ahead and calculated the probability for each k and I thought a graph would be instructive. Here is what the distribution looks like:
The distribution is very gently falling because it is less and less likely it will take so many turns for the first ace to appear.
For completeness, here is a table with each probability explicitly written.
| Card dealt |
Probability 1st Ace |
| 1 | 7.69% |
| 2 | 7.24% |
| 3 | 6.81% |
| 4 | 6.39% |
| 5 | 5.99% |
| 6 | 5.61% |
| 7 | 5.24% |
| 8 | 4.89% |
| 9 | 4.56% |
| 10 | 4.24% |
| 11 | 3.94% |
| 12 | 3.65% |
| 13 | 3.38% |
| 14 | 3.12% |
| 15 | 2.87% |
| 16 | 2.64% |
| 17 | 2.42% |
| 18 | 2.21% |
| 19 | 2.02% |
| 20 | 1.83% |
| 21 | 1.66% |
| 22 | 1.50% |
| 23 | 1.35% |
| 24 | 1.21% |
| 25 | 1.08% |
| 26 | 0.96% |
| 27 | 0.85% |
| 28 | 0.75% |
| 29 | 0.65% |
| 30 | 0.57% |
| 31 | 0.49% |
| 32 | 0.42% |
| 33 | 0.36% |
| 34 | 0.30% |
| 35 | 0.25% |
| 36 | 0.21% |
| 37 | 0.17% |
| 38 | 0.13% |
| 39 | 0.11% |
| 40 | 0.08% |
| 41 | 0.06% |
| 42 | 0.04% |
| 43 | 0.03% |
| 44 | 0.02% |
| 45 | 0.01% |
| 46 | 0.01% |
| 47 | 0.004% |
| 48 | 0.001% |
Solving the original question
Now that we have the complete distribution, we can solve for the probability a particular player is assigned the dealer.
To see how this works, consider a poker game with just two players. Let’s say the first person dealt a card face up is “player 1″ and the other person is “player 2.”
When will player 1 be dealer? Player 1 is dealer if the first ace is dealt to him and not player 2. Which cards are potentially dealt to player 1? Player 1 gets the first card, then one card goes to player 2, but then he gets the third card, and so on. In other words, player 1 is the dealer precisely if the first ace appears in the odd-numbers positions 1, 3, 5, …, 47. And correspondingly, player 2 is the dealer if the first ace appears in any of the even-numbered positions 2, 4, 6, …, 48.
Using a spreadsheet it is easy enough to sum up those entries to find the probabilities. It turns out that player 1 gets to be dealer almost 52 percent of the time versus 48 percent for player 2. This might seem like a small edge, but realize this is worse of a bias than most casino games! Player 1 has a great advantage in this system.
The entire distribution
Similar calculations can be performed if the game starts with a different number of players. For illustration, I extended the calculations for games of 3 players up to 9 players (a full ring game).
The probability is again calculated based on the distribution of the first ace. In a 3-handed game, for example, the first person dealt is the dealer if the first ace appears on the turns 1, 4, 7, etc.; the second on turns 2, 5, 8, etc.; and the third on turns 3, 6, 9, etc. (This calculation was automated as I used a handy spreadsheet array formula to sum up the probabilities based on the turn modulo).
Here are the results.
The first table is about the probability the first player receiving a card gets the ace. Notice there is a definite edge over the fair odds of anywhere from 2 to 4 percent.
The second set of results shows player by player probabilities for receiving the first ace, split up by game size. The advantage becomes exceeding large in games with 7, 8, and 9 players where the first person receiving a card has almost double the chance of getting to be dealer initially compared to the last player.
| Players | Fair odds | Actual odds 1st player dealer | Edge |
| 2 | 50% | 52% | 2% |
| 3 | 33% | 36% | 3% |
| 4 | 25% | 28% | 3% |
| 5 | 20% | 23% | 3% |
| 6 | 17% | 20% | 3% |
| 7 | 14% | 18% | 4% |
| 8 | 13% | 16% | 4% |
| 9 | 11% | 15% | 4% |
| Three | Prob | Seven | Prob |
| 1 | 36% | 1 | 18% |
| 2 | 33% | 2 | 17% |
| 3 | 31% | 3 | 15% |
| 4 | 14% | ||
| Four | Prob | 5 | 13% |
| 1 | 28% | 6 | 12% |
| 2 | 26% | 7 | 11% |
| 3 | 24% | ||
| 4 | 22% | Eight | Prob |
| 1 | 16% | ||
| Five | Prob | 2 | 15% |
| 1 | 23% | 3 | 14% |
| 2 | 22% | 4 | 13% |
| 3 | 20% | 5 | 12% |
| 4 | 18% | 6 | 11% |
| 5 | 17% | 7 | 10% |
| 8 | 9% | ||
| Six | Prob | Nine | Prob |
| 1 | 20% | 1 | 15% |
| 2 | 19% | 2 | 14% |
| 3 | 17% | 3 | 13% |
| 4 | 16% | 4 | 12% |
| 5 | 15% | 5 | 11% |
| 6 | 13% | 6 | 10% |
| 7 | 9% | ||
| 8 | 9% | ||
| 9 | 8% |
In conclusion, dealing to the first ace is not a fair system.
(Off the top of my head, some fair methods are: using a random number generator, dealing everyone a card and allowing the high card to be dealer (with a runoff for ties), or picking numbers from a hat.)
Appendix: dealing WITH replacement
The math on this one got complicated because the deck size kept decreasing.
What if we instead replaced each time a card is dealt–that is, always sampled from a full deck?
The two-person case essentially becomes a version of the dice brain teaser where the odds of winning are 4/52 = 1/13. Using the same method as the dice puzzle, we can find the odds of the first person winning are 13/25 = 52 percent. This is very close to the case of replacement which has 51.98!
The lesson is it’s instructive to consider easier versions to get a ballpark estimate.
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