Probability of winning a card game where you can re-draw

Question

Say you and your opponent both pick a card from a normal deck of cards (labeled 1 through 13). You can see your opponent's card. Whoever has the higher card wins the game. You have an option to redraw your card. What's a close estimate of your probability of winning?

Isn't this question exactly like asking what's the probability that I have 2 numbers (my card, and my redraw), someone else has 1 number. What's the probability that amongst my 2 cards I have a larger number? So it would just be 2/3?

Answer 1

Not quite, consider the case where you both drew a card from the 13 card deck, and you lost. Now when you get a chance to redraw, a card that would cause you to lose, the card you have, is no longer avaliable to draw and your chances to win are slightly higher. It would be 2/3 if you had to shuffle your card back into the deck before drawing again.

Answer 2

Note: I am assuming if the "you" player "opts to redraw a card," they do not reshuffle their first card back into the deck before drawing.

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If there are only cards 1 to 13 (a single card for each rank) so that ties are impossible then yes the answer is exactly $2/3$ and your reasoning is correct.

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For a normal deck there are $\binom{52}{3} = 22100$ possible [unordered] triplets of cards. For each triplet, there are $6$ ways to assign them to "opponent's card, your first card, your second card."

• $\binom{13}{3} \cdot 4 \cdot 4 \cdot 4 = 18304$ of these triplets consist of 3 distinct ranks. For each of these triplets you win in 4 of the six configurations, and lose in the other two.
• $13 \cdot 12 \cdot \binom{4}{2} \cdot 4 / 2 = 1872$ of these triplets have two distinct ranks where the higher rank appears twice. For each of these triplets, you win in 4 of the six configurations, and tie in the other two.
• $13 \cdot 12 \cdot \binom{4}{2} \cdot 4 / 2 = 1872$ of these triplets have two distinct ranks where the lower rank appears twice. For each of these triplets, you win in 4 of the six configurations, and lose in the other two.
• $13 \binom{4}{3} = 52$ of these triplets have a single rank appearing thrice. You tie in all of the six configurations.

Thus, the probability of winning, tying, and losing in a single iteration of the game are respectively \begin{align} \frac{18304 \cdot 4 + 1872 \cdot 4 + 1872 \cdot 4}{22100 \cdot 6} &\approx 0.6651 \\ \frac{1872 \cdot 2 + 52 \cdot 6}{22100 \cdot 6} &\approx 0.03059 \\ \frac{18304 \cdot 2 + 1872 \cdot 2}{22100 \cdot 6} &\approx 0.3043 \end{align}

Vague intuition: since tying is possible, it makes sense that the probability of tying "steals" some probability from the original $2/3$ and $1/3$ probabilities of winning and losing. The player with two cards still has a huge advantage however, so most of the "stolen" probability comes from the the $1/3$ rather than the $2/3$.

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If the setup of the game is such that in the event of the tie, the game is restarted (so all cards are reshuffled into the deck and a new game is started), then the probabilities of winning and losing are \begin{align} \frac{18304 \cdot 4 + 1872 \cdot 4 + 1872 \cdot 4}{22100 \cdot 6 - (1872 \cdot 2 + 52 \cdot 6)} &\approx 0.6861 \\ \frac{18304 \cdot 2 + 1872 \cdot 2}{22100 \cdot 6 - (1872 \cdot 2 + 52 \cdot 6)} &\approx 0.3139 \end{align}

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Given that the question asks "What is a close estimate for the probability" rather than "What is the probability", I think you were intended to do the argument you provided in your post. It is not rigorous, but as my above breakdown shows, the introduction of multiple ranks does not drastically change the probabilities.