There is an infinite deck of regular playing cards (2-10, JQKA, four suits). The dealer deals face down one card to you and one card to me. The player with the higher card dealt wins.

We each put in $10 as an initial bet.

Now before you look at your cards, you can pay $x to have the option of replacing your card with a new (random card) if you wish, after you view it.

How much would you pay for this option? What is the new probability that you win?

Edit; in the event you tie, the dealer throws out both of your cards and deals both of you a new card

Edit2 : all suits are the same.

  • $\begingroup$ You haven't said what it means to win. How do you win? Are aces high? $\endgroup$ Sep 24, 2018 at 21:16
  • $\begingroup$ Not sure it matters, but you should clarify how ties are handled. Of course, depending on what you mean by "infinite deck" ties may or may not be possible. I take it to mean that each of the standard $52$ cards are equally likely to be drawn, regardless of whatever has been drawn before...but you might mean something else. $\endgroup$
    – lulu
    Sep 24, 2018 at 21:17
  • $\begingroup$ We don't need an infinite deck! Just two regular decks, so that my card is drawn from one deck, and my opponent's* is drawn from the other. This achieves the same independence as the "infinite deck". $\endgroup$ Sep 24, 2018 at 21:18
  • $\begingroup$ @ZubinMukerjee no, it doesn't. possibly three regular decks, so that your replacement card is drawn from the third deck. $\endgroup$
    – CR Drost
    Sep 24, 2018 at 21:31
  • $\begingroup$ @CRDrost Why can't you put your card back in the first deck, shuffle it, and draw the replacement card? :P $\endgroup$ Sep 24, 2018 at 21:32

3 Answers 3


Without the option, this is a fair game.

If the option is priced on the cheap side of fair, we buy the option and the option favors the player. If the option is rich, we don't buy the option.

Assuming the option is cheap, we plan to exercise if our first card is in $[1,7]$ and we won't exercise if the first card is in $[8,13]$ (number from 1-13 is simpler than translating face-cards to numerals)

Your chance of getting any particular number after exercising the option is:

$P(x) = \begin{cases} \frac{7}{169} &x\in [1-7]\\\frac{20}{169}&x\in [8-13] \end{cases}$

Your chance of winning is for any given $x$ is $\frac {x-1}{13}$ your chance of loosing is $\frac{13-x}{13}$ and you have a $\frac {1}{13}$ chance of a push.

If the option is free. $E[X] = 10 (\sum_\limits{i=1}^7 \frac {7(i-1)}{13^3} + \sum_\limits{i=8}^{13} \frac {20(i-1)}{13^3}) - 10(\sum_\limits{i=1}^7 \frac {7(13-i)}{13^3} + \sum_\limits{i=8}^{13} \frac {20(13-i)}{13^3}) = 10(\frac {7\cdot21+20\cdot57 - 7\cdot 63 - 20\cdot 15}{13^3}) = \frac{546}{13^3}$

Paying for the option, this is a fixed cost regardless of how the game comes out.

You should be willing to pay up to, but no more than $\frac{546}{13^3}\approx \$2.485$

Note that we have ignored the push. If the option is close to fair value the push can be safely ignored. But if the option were free, our expected gain would be greater by a factor of $\frac{13}{12}$

  • $\begingroup$ I think "expensive" makes more sense than "rich", otherwise good answer, +1 $\endgroup$ Sep 25, 2018 at 1:40
  • $\begingroup$ @ZubinMukerjee "rich" is the term that finance people use. $\endgroup$
    – Doug M
    Sep 25, 2018 at 1:52

If you are allowed one switch, your winning probability is $p(52) = \frac{5}{8}$, so you should be willing to pay up to 2 dollars for the right to switch.

$p(N) = \sum_{i=1}^{N} \frac{1}{N} \max(\frac{i-1}{N} + \frac{1}{2N}, \frac{1}{2}).$


$p(2n) = \frac{5}{8}$

$p(2n+1) = \frac{5}{8} - \frac{1}{(2n+1)^2}$

(I assumed 10 dollars payoff and 52 values, I see that it might be 20 and 13, but the principle and formula should be the same)

For N=13 and a total pot of 20 you should be willing to bet at most $10 ( 2 - \frac{1}{p(13)}) = \frac{840}{211} = 3.981...$.

  • $\begingroup$ You have assumed all the $52$ cards are ranked, but OP has clarified that suits do not matter. $\endgroup$ Sep 25, 2018 at 0:11

Problem Statement

So to be clear the problem I'm about to solve is... Both have \$10 in and can win the full \$20. Loser gets nothing. One player is offered the option to be shown a card and if he would like to, switch to that card. The player does not get to look at his own card before switching. How much should the player offered the option be willing to pay for the option?

The wording of the question is ambiguous. This is the way I read it. The solution does not contradict the others solutions, but solves a different and similar game to the ones they solved. That being said, this may actually be the game OP wanted.


Thinking now that the two games are symmetric. The final pricing should be the same no matter if you (1) view the new card and not your original one and (2) view the original card and not the new one.

The Base Case

In the base case (no pay, fair game), it is trivial to show probability of winning a round is $\frac{6}{13}$, probability of losing a round is $\frac{6}{13}$, and probability of a tie is $\frac{1}{13}$. Generally probability of winning if we replay when we tie is...

P(win after any number of replays)$=P_W + P_W \cdot P_T + \cdots = \frac{P_W}{1-P_T} $

in the base game this evaluates to... $\frac{6/13}{1-1/13}=\frac{1}{2}$.

The Fun Case

Optimal Strategy

Consider that you have already paid to switch and are shown a card. Should you switch to it?

If you are shown a 2, the probability that the 2 is higher than your card is 0. If you are shown a 3, the probability that the card is higher than your card is $\frac{1}{13}$.

If you are shown an $i$, the probability that the card is higher than your card is $\frac{i-2}{13}$ for $i \in \left\{2,3,\ldots,10,11=J,12=Q,13=K,14=A \right\}$.

That makes your probability of getting a higher card than your card greater than $\frac{1}{2}$ if the card shown to you is a 9, 10, Jack, Queen, King, or Ace.

The optimal strategy is to switch if the card shown to you is a 9, 10, Jack, Queen, King, or Ace.

Probability of Winning if you Pay and Play Correctly

You'll switch if you see a 9 or higher. I'll write games as (a,b,c) exmaple (9,2,10) if your opponent is dealt a 9, you are dealt a 2, and the card you are shown is a 10. In this case where you paid you'll switch to the 10 and win.

Probability of winning with optimal strategy...

The games that you will win given you play right and paid are...

$\left( 2,3,2 \right), \ldots ,\left(2,A,2 \right),$

$\left( 2,3,3 \right), \ldots ,\left(2,A,3 \right), \ldots$

$\left( 2,3,8 \right), \ldots ,\left(2,A,8 \right),$

Above are the games that you win when the opponent draws a 2 and you don't see a card greater than 8 (you don't switch). There are $12 \cdot 7$ possibilities above.

$\left(2,2,9 \right) ,\ldots ,\left(2,A,9 \right),$

$\left(2,2,10 \right) ,\ldots ,\left(2,A,10 \right),\ldots$

$\left(2,2,A \right), \ldots ,\left(2,A,A \right),\ldots$

The second block contains the games that you win when the opponent draws a 2 and you see a card greater than or equal to 9 and switch. There are $13 \cdot 6$ possibilities above.

There are clearly more cases but now we are on to the pattern.

Generally there are $\left(14-i\right) \cdot 7$ ways that you win (if the opponent draws card $i$ where $i$ is in $\left\{2,3,\ldots,10,11=J,12=Q,13=K,14=A \right\}$) and you don't see a card better than 8 so you don't switch.

If the opponent draws an $i\in \left\{2,3,\ldots,8 \right\}$ and you see a card better than 8 and switch there are $13\cdot 6$ ways you win for each $i$.

If the opponent draws an $i \in \left\{9,10,11=J,12=Q,13=K,14=A \right\}$ and you see a card better than 8 and switch there are $13 \cdot \left(14-i \right)$ ways that you win.

Overall there are...

$\sum_{i=2}^{14} \left[ \left(14-i \right) \cdot 7 \right] + \sum_{i=2}^{8} \left[ 13\cdot 6 \right] + \sum_{i=9}^{14} \left[ 13 \cdot \left(14-i \right) \right]$

$=546 + 546 + 195=1287$

ways to win the game if you pay and play right. In the above notation there are $13^3=2197$ games. Your probability of a win if you pay and play right on the first round is...

$P(Action \, win)=\frac{1287}{2197}=\frac{99}{169}$

Probability of a Tie if you Pay and Play Correctly

You tie these games...

$\left( 2,2,2 \right), \ldots ,\left(2,2,8\right),$

$\left(3,3,2 \right), \ldots ,\left(3,3,8 \right),\ldots$

$\left(8,8,2 \right), \ldots ,\left(8,8,8 \right),$

There are $7 \cdot 7$ no-switch games above.

$\left( 9,2,9 \right) , \ldots , \left(9,A,9 \right),$

$\left( 10,2,10 \right) , \ldots , \left(10,A,10 \right),\ldots$

$\left( A,2,A \right) , \ldots , \left(A,A,A \right)$

There are another $13 \cdot 6$ switch games above.

Overall the probability that you tie if you pay and play correctly is $\frac{127}{2197}$

Probability of Loss

By law of total probability the probability of a loss is $1-\frac{127}{2197} - \frac{1287}{2197}=\frac{783}{2197}$

Probability of Generally Winning with Advantage: Ad only Applied to First Round

$P=\frac{1287}{2197}+\frac{127}{2197} \cdot \frac{6}{13} + \frac{127}{2197}\cdot \frac{1}{13} \cdot \frac{6}{13} + \cdots = \frac{2701}{4394}$

Probability of Generally Winning: Pay once, Ad Applied Every Round

$P=\frac{1287}{2197}+\frac{127}{2197} \cdot \frac{1287}{2197} + \left( \frac{127}{2197} \right)^2 \frac{1287}{2197} + \cdots = \frac{143}{230}$

Pricing the Advantage

If the advantage only applies to the first round...

$E[W_{Ad}]-E[W_0]=\frac{2701}{4394} \cdot \$20 + \left( 1-\frac{2701}{4394} \right) \$0 - \frac{1}{2} \cdot \$20 - \frac{1}{2} \cdot \$0 = \frac{5040}{2197} \approx \$ 2.29$

If the advantage applied every round...

$E[W_{Ad}]-E[W_0]=\frac{143}{230} \cdot \$20 + \left( 1-\frac{143}{230} \right) \$0 - \frac{1}{2} \cdot \$20 - \frac{1}{2} \cdot \$0 = \frac{56}{23} \approx \$ 2.43$

Closing Comment

The \$2.29 number doesn't change round to round, so if the option is offered again the next round or any round after a tie it should still be taken at \$2.29. If the \$2.43 were to be paid, it would offer the advantage perpetually. These numbers could be reached if you had one and not the other with some geometric properties.


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