In this simple playing card game, is the expected probability constant?

Question

I came up with a simple game using a standard 52-card deck. This is a 1-on-1 game involving a player and a dealer.

The rules are as follows.

1. Player makes a bet.
2. One card is dealt to the player.
3. One card is dealt to the dealer.
4. If the player's card is higher than the dealer's, player is paid 1:1. If not player loses their bet.

I did the numbers to determine if the house edge is acceptable. There are $52\times 51=2652$ total possible deals. The number of cases the player wins can be found by subtracting all ties(in which the player loses) then dividing it by half. Pick one of the $13$ ranks, and assign one suit to player and another to dealer. Thus the number of ties is $13\times 4\times 3=156$. The player wins $\frac{2652-156}{2}=1248$ cases. So the probability of the player winning is $\frac{1248}{2652}\times 100\approx 47.0588\%$. So it is a feasible game.

I sat down with my friends and played this game. Because we are lazy we decided not to shuffle the deck after every single deal. This meant that the deck gets shorter as cards are used. I wondered whether this benefited the player or the house. So I calculated the probability of the player winning after one deal. Now that $50$ cards are left in the deck there are $50\times 49=2450$ possible deals. There are two cases: whether the first deal was a tie or not.

1. First deal was a tie

   • There are 12 ranks that still have 4 suits and one rank 2 suits. $12\times 4\times 3 + 1\times 2\times 1 = 146$. $\frac{2450-146}{2}=1152$ wins. $\frac{1152}{2450}\times 100\approx 47.0204\%$.

2. First deal was not a tie

   • 11 ranks have 4 suits left while 2 ranks 3 suits. $11\times 4\times 3 + 2\times 3\times 2 = 144$. $\frac{2450-144}{2}=1153$ wins. $\frac{1153}{2450}\times 100\approx 47.0612\%$.

Notice how the probability tilts against the player very slightly if the first deal was a tie and in favor of if not a tie. Now let's find the expected win rate.

1. First deal was a tie: $\frac{156}{2652}\times\frac{1152}{2450}$
2. First deal was not a tie: $\frac{2496}{2652}\times\frac{1153}{2450}$

When I added the two together I got $47.0588\%$. Thinking the difference between the first deal and the second was within my calculator's precision, I multiplied both by $2652\times 2450$.

1. First deal win: $1248\times 2450 = 3057600$
2. Second deal win: $156\times 1152 + 2496\times 1153 = 3057600$

No, it was not a precision problem. Surprisingly the expected value was exactly the same. I intuitively thought after a deal it would be closer to 50%. Would the expected house edge stay the same until the final, 26th deal? Or will it change after enough deals? Am I missing something that could easily show that the expected value never changes?

Answer 1

Yes, probability of winning $k$-th round for player is the same as probability of player winning the first round. If deck is well-shuffled, it doesn't matter if you take the first and second card from it, or 7th and 13th - distribution of pairs is the same. The same principle applies in many scenarios: if you have $n$ balls and $m \leq n$ people, each drawing a ball, your probability to get any fixed ball is $\frac{1}{n}$ regardless if you are the first to draw, the last, or anywhere in the middle.

But knowledge who won the first round indeed affects probabilities of who wins the second. To make it to the extreme: assume there are just $4$ cards, pair of twos and pair of threes. Then if there is a tie in the first round, there for sure will also be a tie in the second, and if there were no tie, there also will be no tie in the second.