A competent logician is playing cards with four of his friends, so that there are a total of 5 people sat around the table. Each player has the same 5 cards, showing A, 2, 3, 4, and 5 (Ace being low). Each player then has to choose a single card from their hand.

The objective of the game is to choose the lowest unique card that is displayed in the round. If two or more people show the same numbered card, they will not win. If no one chooses a card unique from anyone else, then the round is repeated until someone has a uniquely numbered card.

Each of them has 55 seconds to decide which card they want to draw, at which point they must each in turn show what they have picked. No one else knows which card anyone else has picked until they are displayed at the end of the time limit, but they are allowed to communicate with each other during that period.

The competent logician has to come up with a logical solution for which card to choose to maximize his chance of winning. What card should he choose? If the round needs to be repeated, should he choose the same card again?

Note: Whilst I do have a specific "correct" answer in mind, it is not necessarily the one that I will choose, as any card could reasonably be chosen and win, or lose. I'll accept the answer that is accompanied by the best logical argument for the choice of card.

Edit: To clarify, he has never played this game with any of his friends before, so he has no idea how they will initially think to play the game. He is unable to make any assumptions about their possible strategies, he is on the same playing field as they are, he simply needs his own strategy that he believes will give him an edge.

Edit 2: I should also clarify that this is only for the first game that they play. Only one game (with potentially multiple rounds) needs to be played until there is a winner. Once the game begins to be repeated the strategies will evolve based on what each person sees that the others are choosing. This question is for the first game only.

  • 1
    $\begingroup$ So everyone knows which card they have picked, correct? $\endgroup$
    – user14352
    Nov 2, 2016 at 14:25
  • 2
    $\begingroup$ I'm thinking about playing this game with 4 or my colleagues and try different strategies. I have only one problem. A deck of cards has only 4 Aces and I need 5....and I only have a deck of cards. $\endgroup$
    – Marius
    Nov 2, 2016 at 14:56
  • 1
    $\begingroup$ @Mike If the question is for the first game only and he has no idea what the others will pick, I am highly skeptical there is an answer as you assert. Unless you make an assumption (like I did) they all play alike. $\endgroup$
    – kaine
    Nov 2, 2016 at 15:47
  • 5
    $\begingroup$ There is no true game theory here. $\endgroup$
    – paparazzo
    Nov 2, 2016 at 19:00
  • 2
    $\begingroup$ All I can think is this would be an excellent drinking game. Lowest unique card person wins everyone else drinks, if no uniques and round repeats then drink number goes up 1 per round $\endgroup$
    – DasBeasto
    Nov 3, 2016 at 13:45

10 Answers 10


Assuming each player wants to maximize their personal win probability for the single game...

I immediately act, flipping over my 2,3,4, and 5; and putting the Ace face down in the center. "I've played an Ace."

Each player now knows that they cannot win by playing an Ace, individually, their only chance of winning is to play a non-ace, and hope someone else plays an Ace. It's becomes something that loosely resembles the Prisoner's Dilemma among my opponents.

The reason I chose this approach is

We have a 55 second window in which to discuss, which none of the other suggestions use. I believe the time window is critical, although it admittedly does give others opportunities to use external incentives to encourage diving on the Ace-Grenade.

  • $\begingroup$ I was going to answer something very similar to this; assuming nobody is willing to lose on purpose of course (real people can be so spiteful :P) While an Ace is the obvious no-lose strategy choosing something like 2 or 3 could also be useful if you want to play with the prisoners dilemma more by leaving a potential win that everyone can see (and thus tie on) lessening the chance that someone decides to tie you out of spite. $\endgroup$
    – Lunin
    Nov 3, 2016 at 6:29
  • $\begingroup$ Once you reveal your 2,3,4 and 5 and anounce you are playing an ace I reveal 4 and 5, shuffle the remainding ace, 2 and 3 and anounce that I will play one of ace,2 and 3 with equal probability. $\endgroup$
    – Taemyr
    Nov 3, 2016 at 11:37
  • 3
    $\begingroup$ The same strategy could be done with a 2. Everyone else might think that they have a chance of winning with an A and so all pile on to that. $\endgroup$
    – Dr Xorile
    Nov 3, 2016 at 16:32
  • $\begingroup$ For fun, I plugged this into my calculations. This is definitely the best strategy. In my method everyone does what is best for them so they only do the correct thing and go with an A 3.4% of the time... You win 87% of the time. $\endgroup$
    – kaine
    Nov 3, 2016 at 18:48
  • 3
    $\begingroup$ I don't know about your friends, but with my friends, this strategy would have a precisely zero chance of winning. $\endgroup$ Jun 22, 2018 at 17:55

Assuming he respects his friends, he assumes they play just as he will.

In order for him to have the highest chance of victory, he needs to be able to select each card with a random probability. The probability of victory after selecting each card, therefore, is dependent on the probability he will select the other cards.

The probability of victory if he selects an A (a 1) is:

$$P_1 = (1-P_1)^4$$

If he selects a deuce (a 2) it is:

$$P_2 = ((1-P_1-P_2)^4+P_1^4+6P_1^2(1-P_1-P_2)^2+4P_1^3(1-P_1-P_2))$$

If any probability is higher than the others, they will chose that card more frequently. At some point there is a set of $(P_1,P_2,P_3,P_4,P_5)$ where increasing your likelihood of choosing one does not help you.

I was too lazy to solve this so build a quick simulation and got the following probabilities:

$$(P_1, P_2, P_3, P_4, P_5) = (37.2\%, 33.7\%, 19.7\%, 8.3\%, 1.0\%)$$

This all assumes, however, that all players play the same way. A single idiot at the table, for instance, changes everything.

"Simulation" Details

I calculated the probability of victory for each card as I did for A and 2 above based on the probability each card would be chosen. I set a field with (1,1,1,1,1) such that players will chose each slot with a probability of the number in that slot over the sum. If choosing one of those numbers was more likely to result in victory compared to the others ($P_x>$ average$+c$), that number in the field increased. Note: draws were considered later. The probability of a draw times .2 is added to each win probability. For experiment, I also only tried letting one card increase probability a round and let lower cards win in a draw. That did not change final results at all. Simulations each have 10,000 rounds but it appears convergence is before 1000.

  • $\begingroup$ Note, I have found a mistake in caculations. I did not take into account 3 people having a single number....damn $\endgroup$
    – kaine
    Nov 2, 2016 at 16:26
  • $\begingroup$ Mistake corrected and got updated results. Did another type of simulation where actual games were run and results were within 0.1% in value. $\endgroup$
    – kaine
    Nov 2, 2016 at 18:36
  • $\begingroup$ I'm surprised we have P5>0, but the math is sound. Upon further thought, it makes sense, because lower numbers are more likely to tie, so it's easier to be a loner with a 5. $\endgroup$
    – ffao
    Nov 2, 2016 at 22:35
  • 1
    $\begingroup$ @Jik just for you (sarcasm) I removed the (P1+P2+P3..) from the equations, calculated the probability of a draw for each one which is $P_d_x = P_x^4+6P_x^2( \sum (P_o^2))+4P_x( \sum (P_o^3))$. multiplied that by .2 and added it to each win. My previous answer did bias it further towards the 4s and 5s. The numbers in the answer will be edited shortly. $\endgroup$
    – kaine
    Nov 3, 2016 at 17:59
  • 1
    $\begingroup$ The technical term for the solution you found is a Nash equilibrium. I calculated it using a different method and got the same result, so I'm fairly certain you have the right solution. $\endgroup$ Nov 4, 2016 at 11:04

Looking at this from a statistical standpoint we know that we should never play a card higher than a 3. In all cases an A, 2 or 3 could be the lowest winning card with 5 people playing the game. (As an aside, what kind of crazy deck of cards are we using that there are 5 Aces!!)

There are 243 permutations of these three cards (we really don't care about 4 and 5, let the peasants play with those). Of these 243 permutations

63 result in a draw 80 A wins 60 2 wins 40 3 wins

But this is only half the story, we need to check the probabilities given that we play a certain card!

If we play an A

Win: 16 Lose: 44 Draw: 21

If we play an 2

Win: 12 Lose: 48 Draw: 21

If we play an 3

Win: 8 Lose: 52 Draw: 21


If we consider winning and drawing good results(not losing) then playing an A results in 45.6% "good" 2 results in 40.7% "good" 3 results in 36.25% "good" This means should always play an A as it gives up the best chance at not losing (and the best chance at winning too!)

But that answer is no fun, so lets look at it a different way

We don't want to "Not Lose" we want to win! So lets look at the probabilities of winning as ratios to each other 16:12:8. This says we should play an A twice as much as a 3. Well, whats the easiest way to do this with keeping the ratios. Lets scale up the ratios to add up to 55.

We get

24.4:18.3:12.2 with a little bit of bad math to make life easy, we can use 25:18:12. Why did we scale up to 55? Because there is a timer in the room presumably counting down from 55! That is how much time we have. So we can decide by looking at the clock when its our turn to choose. If the clock is in the range 55-31 throw and A. Between 30-13 a 2 otherwise a 3. (This strategy would also work and might be better if we scale to 60 instead and remember to bring a watch with a seconds hand to this fight).

  • 1
    $\begingroup$ Since you are playing with random picks and probabilities, you might have to consider 4 and 5 to be a winnable options. It could even be argued that it's a viable strategy to throw a 4 or a 5 (under the assumption that a player suspects everyone else will tie and no one will pick a 5 because it sounds illogical, being the highest number). $\endgroup$ Nov 2, 2016 at 19:19
  • $\begingroup$ @Possion Fish - After I posted this I had the same thought. So I ran the numbers again and it comes out with 24:14:8:5:4. This makes sense because the assumption that I made was comparable to the assumption that 3,4,5 and are all the same. So in the end adding in 4 and 5 essentially just takes away from 3, but doesn't really effect A or 2. (This is assuming you are ok with some really bad rounding and a large margin of error because of this). $\endgroup$
    – Scott
    Nov 2, 2016 at 19:41
  • 1
    $\begingroup$ That is your assumption. See you first sentence. Looking at this from a statistical standpoint we know that we should never play a card higher than a 3. If I can get you to fall for that false statistics then can dominate you. $\endgroup$
    – paparazzo
    Nov 2, 2016 at 20:17
  • 1
    $\begingroup$ Please don't use codeblocks for anything which is not actual code. MathJax should be used to typeset equations and mathematical text. $\endgroup$
    – Nij
    Nov 2, 2016 at 21:18
  • 1
    $\begingroup$ @Nij, sorry I am still learning my way around formatting. Its feels very different than what I am used to over on StackOverflow. Sorry :) $\endgroup$
    – Scott
    Nov 2, 2016 at 21:26

I would say

there is no winning or best strategy


For all we know, the four others are also competent logicians. And because this game has the same open information for all players and they all reveal at the same time, any 'best strategy' would be applied by all of them, so everyone reveals te same card and you never get a lowest unique card.

  • 3
    $\begingroup$ Not if it was a random strategy. ("Pick A with $x\%$ probability, pick 2 with $y\%$ probability...") $\endgroup$
    – Deusovi
    Nov 2, 2016 at 14:34
  • 1
    $\begingroup$ @Deusovi I'm not entirely sure but if everyone would use that same random strategy there is no way that that strategy favors one over the other, so in conclusion everyone would have 20% to win. But yes, that would be an improvement on a non-random strategy because that has a 0% to win for everyone $\endgroup$
    – Ivo
    Nov 2, 2016 at 14:42
  • 2
    $\begingroup$ Basic game theory: yes, the winning strategy will be one of "Play A x%" of the time, 2 y%, 3 z%..." randomly. Unfortunately, using this strategy will result in each player winning 20% of the time in the long run. $\endgroup$ Nov 2, 2016 at 18:24
  • 2
    $\begingroup$ Well, the idea is to find what those probabilities are. If one of the players deviates from that strategy then they will do worse. $\endgroup$
    – Dr Xorile
    Nov 3, 2016 at 16:30

Since they are his friends

He knows
(a) what level of thinker each is; and
(b) what they think of him.

His strategy should be

He needs to figure out if they would all be "level one" thinkers who don't think any different of him - then they will all choose their ace, whereupon he should choose any other card, for example his deuce.

But if there is a "level 2" thinker who would do that too, he should choose a deuce to make certain another round is played. If there are two such opponents he should play his trey; but if there are three he needs to play his ace to ensure another round.

Similarly if there were a "level 3" thinker who thinks there are "level 2" thinkers then he needs to pick so as to again either win or ensure another round is played.

This reasoning can go on.

So the answer is

It depends.

For a good review of this kind of game see this lecture given by Ben Polack at Yale were the class are to pick a number in the range 1 to 100 that is as close as possible to two thirds of the average number chosen (four people actually chose numbers over 67).

  • 1
    $\begingroup$ (You mean two-thirds?) $\endgroup$
    – Deusovi
    Nov 2, 2016 at 14:43

It is clear that all players will play 1, 2 or 3 or be worse off than before. So to simplify the analysis, we consider only probabilities of playing 1, 2, or 3 as $p_1$, $p_2$, and $(1-p_1-p_2)$.

We then look for a Nash Equilibrium for the three moves. In other words, assuming all the opponents play with a strategy of $(p_1,p_2,1-p_1-p_2)$, the expected value of the moves should be identical.

This turns out to be the case for $(0.4,0.38,0.22)$, which is then the best strategy (ie choose A: 40%, choose 2: 38%, and choose 1: 22%).

Does it work? Well, if everyone plays the same strategy, you would expect to lose 0.8 of the time and to win 0.2 of the time, so your expected game value is -0.6 (I'm saying 1 point for a win and -1 for a lose).

Here's a simulation where player A plays $(0.4,0.38,0.22)$, and the other 4 players play a random strategy (specified in the table). I simulate 1000 games and tally the score. I then note whether player A did better than -0.6 or not:

0.31,0.12,0.57 -0.318 True (1000 games)
0.12,0.67,0.22 -0.11 True (1000 games)
0.42,0.19,0.40 -0.43 True (1000 games)
0.14,0.34,0.53 -0.234 True (1000 games)
0.39,0.21,0.40 -0.568 True (1000 games)
0.33,0.54,0.13 -0.514 True (1000 games)
0.21,0.56,0.23 -0.37 True (1000 games)
0.81,0.09,0.10 0.138 True (1000 games)
0.32,0.31,0.38 -0.576 True (1000 games)
0.82,0.10,0.08 0.124 True (1000 games)
0.26,0.45,0.29 -0.54 True (1000 games)
0.74,0.18,0.08 -0.178 True (1000 games)
0.49,0.45,0.06 -0.47 True (1000 games)
0.46,0.49,0.05 -0.464 True (1000 games)
0.77,0.15,0.09 -0.114 True (1000 games)
0.52,0.35,0.13 -0.518 True (1000 games)
0.25,0.04,0.71 -0.024 True (1000 games)
0.05,0.50,0.45 0.144 True (1000 games)
0.42,0.22,0.36 -0.558 True (1000 games)
0.09,0.27,0.64 0.034 True (1000 games)
0.07,0.25,0.68 0.152 True (1000 games)
0.09,0.36,0.55 -0.09 True (1000 games)
0.13,0.19,0.67 -0.076 True (1000 games)
0.35,0.45,0.20 -0.568 True (1000 games)
0.38,0.07,0.55 -0.208 True (1000 games)
0.72,0.10,0.17 -0.07 True (1000 games)
0.12,0.64,0.24 -0.154 True (1000 games)
0.03,0.71,0.26 0.22 True (1000 games)
0.09,0.75,0.15 -0.026 True (1000 games)
0.32,0.42,0.27 -0.584 True (1000 games)
0.35,0.37,0.29 -0.602 False (1000 games)
0.43,0.37,0.20 -0.612 False (1000 games)

I found the last two of these after a few runs through. You'll note that the score is close to -0.6 and the opponent's strategy was close to optimal.

Suppose I were obnoxious enough to compare my strategy with some of the other solutions here:

0.40,0.40,0.20 -0.59724 True (100,000 games)
0.40,0.40,0.20 -0.59624 True (100,000 games)
0.40,0.40,0.20 -0.5902 True (100,000 games)
0.33,0.33,0.33 -0.57194 True (100,000 games)
0.33,0.33,0.33 -0.58054 True (100,000 games)
0.33,0.33,0.33 -0.5716 True (100,000 games)
0.42,0.36,0.22 -0.60038 False (100,000 games - Kaine's solution scaled proportionately)
0.36,0.31,0.32 -0.58286 True (100,000 games - Kaine's solution alternative)
0.42,0.36,0.22 -0.5978 True (100,000 games - Kaine's solution scaled, but this time I won phew 2 out of 3?)
0.42,0.36,0.22 -0.59694 True (100000 games - Kaine's solution scaled. Close one!)

Anyone want anything else? Put it in the comments!

Okay, so I will leave this here as a warning against over-simplifying problems!!! I believe Kaine's answer is correct. Using his strategy yields the following:

0.53,0.01,0.26,0.00,0.19 -0.12 True (1000 games)
0.05,0.63,0.23,0.02,0.08 0.01 True (1000 games)
0.08,0.21,0.31,0.36,0.03 -0.128 True (1000 games)
0.01,0.02,0.08,0.09,0.80 0.722 True (1000 games)
0.04,0.08,0.19,0.05,0.65 0.388 True (1000 games)
0.03,0.07,0.04,0.83,0.04 0.562 True (1000 games)
0.13,0.61,0.04,0.13,0.09 -0.21 True (1000 games)
0.02,0.43,0.34,0.09,0.13 0.048 True (1000 games)
0.37,0.01,0.14,0.03,0.45 -0.156 True (1000 games)
0.28,0.18,0.29,0.16,0.10 -0.534 True (1000 games)
0.21,0.07,0.65,0.01,0.05 -0.132 True (1000 games)
0.06,0.56,0.07,0.23,0.08 -0.064 True (1000 games)
0.02,0.12,0.07,0.64,0.15 0.34 True (1000 games)
0.51,0.16,0.03,0.19,0.11 -0.374 True (1000 games)
0.50,0.03,0.31,0.07,0.09 -0.228 True (1000 games)
0.01,0.11,0.52,0.33,0.03 0.354 True (1000 games)
0.09,0.20,0.09,0.25,0.36 -0.184 True (1000 games)
0.03,0.57,0.08,0.22,0.09 0.098 True (1000 games)
0.11,0.38,0.04,0.37,0.09 -0.278 True (1000 games)
0.86,0.04,0.03,0.02,0.05 0.282 True (1000 games)
0.04,0.01,0.01,0.58,0.36 0.604 True (1000 games)
0.00,0.08,0.24,0.46,0.22 0.38 True (1000 games)
0.02,0.24,0.06,0.62,0.06 0.342 True (1000 games)
0.15,0.09,0.62,0.05,0.09 -0.16 True (1000 games)
0.11,0.66,0.02,0.11,0.10 -0.046 True (1000 games)
0.40,0.09,0.05,0.12,0.34 -0.274 True (1000 games)
0.31,0.08,0.25,0.33,0.03 -0.456 True (1000 games)
0.17,0.02,0.06,0.48,0.28 -0.01 True (1000 games)
0.24,0.08,0.05,0.44,0.19 -0.266 True (1000 games)
0.03,0.49,0.26,0.09,0.14 -0.06 True (1000 games)
0.15,0.22,0.57,0.04,0.02 -0.286 True (1000 games)
0.22,0.03,0.40,0.28,0.06 -0.288 True (1000 games)
0.18,0.24,0.22,0.05,0.30 -0.432 True (1000 games)
0.11,0.51,0.27,0.08,0.03 -0.268 True (1000 games)
0.27,0.00,0.32,0.33,0.08 -0.272 True (1000 games)
0.32,0.14,0.14,0.17,0.22 -0.424 True (1000 games)
0.40,0.19,0.04,0.09,0.28 -0.514 True (1000 games)
0.20,0.11,0.06,0.29,0.34 -0.324 True (1000 games)
0.13,0.36,0.27,0.12,0.12 -0.426 True (1000 games)
0.04,0.41,0.24,0.14,0.17 -0.102 True (1000 games)
0.42,0.22,0.03,0.09,0.24 -0.504 True (1000 games)
0.54,0.32,0.00,0.10,0.04 -0.484 True (1000 games)
0.30,0.31,0.08,0.14,0.17 -0.578 True (1000 games)
0.05,0.32,0.19,0.09,0.35 -0.13 True (1000 games)
0.20,0.23,0.25,0.06,0.26 -0.434 True (1000 games)
0.15,0.46,0.02,0.18,0.18 -0.274 True (1000 games)
0.28,0.03,0.35,0.21,0.13 -0.302 True (1000 games)
0.01,0.08,0.28,0.47,0.15 0.308 True (1000 games)
0.14,0.22,0.34,0.08,0.22 -0.356 True (1000 games)
0.12,0.26,0.20,0.24,0.18 -0.314 True (1000 games)
0.362,0.314,0.191,0.097,0.036 -0.5986 True (100000 games)
0.400,0.380,0.220,0.000,0.000 -0.57296 True (100000 games)

Note the strategy essentially draws with itself (second last), and wins against my original strategy.

  • $\begingroup$ See my answer. If all other players only play 1, 2, 3 then you dominate playing 4, 5, 6. You have to be the only player on the card to win so any time you would win with 1, 2, or three then you would win with 4. $\endgroup$
    – paparazzo
    Nov 2, 2016 at 20:14
  • $\begingroup$ Hmmm. Interesting point. That does seem to be true! I'll have to think about this... $\endgroup$
    – Dr Xorile
    Nov 2, 2016 at 21:06
  • $\begingroup$ What I haven't figured out yet, is why Kaine's solution is not a Nash Equilibrium, in the sense that the 5 expected values for the game are not equal, where they are much closer with a solution of (0.405,0.39, 0.205,0,0). $\endgroup$
    – Dr Xorile
    Nov 2, 2016 at 23:16
  • $\begingroup$ What do you mean by "the five expected values?" The expected values for each of the five players? $\endgroup$ Nov 4, 2016 at 11:14
  • $\begingroup$ I'm a bit rusty on it, but I'm working on a Nash Equilibrium where you figure out the expected value for each turn based on the probabilities of the other player's turns. At a Nash Equilibrium all these should be equal. $\endgroup$
    – Dr Xorile
    Nov 4, 2016 at 14:05

As a logician, I can remark that there are many configurations of the choices of the four other players where I can't win :

  • A,2,3,4
  • A,A,2,3

and so on. I can remark that a configuration is a win if and only if there is a value that was not chosen by other players that is lower than any single choice by them. For example

  • 2,2,3,4 (A is a winning choice for me)
  • A,A,A,3 (2 is a winning choice here)

I can also remark that there are no draw configuration : there are no configuration where there are no winners if I choose right because if all my opponent are sure to lose, I have a winning move.

So it's not useful to consider losing configurations where I can not be smart, so let's consider only winning configurations :

Will there be an A in the choices of other players ?

I would say probably, it would be very strange that nobody try to win by playing the most obvious move. So let's say there is one, and as I suppose it's possible for me to win, let's say there are at least 2. I don't think there will be four, so it may be a winning configuration as :

  1. A,A,2,2
  2. A,A,X,X (X>2)
  3. A,A,A,X (X>2)

For the first one, 3 is winning, for the other it's 2. I would then choose 2 or 3... 2 seems a little better, as there are more winning configuration with this move, but there are more chances that another smart player chooses 2, so maybe 3 is a good guess.


Was going to post a comment on another answer but I ended up running some tests and enjoying looking at the numbers way too much, so I'll include my own answer instead!

Starting with the fact that we only really care about winnable combinations I also ran this simulation taking into account all winning plays rather than just the best one to try to suss out the best of the best from a pure odds perspective.

Given this if your opponents play randomly you should play

Ace every time.

This is shown clearly when you look at probabilities factoring in only winnable games:

With random play, 52.64% of games are winnable. Of those:
- Ace wins 77.81% of the time (40.96% of all games)
- 2 wins 44.99% of the time (23.68% of all games, 12.96% of which lower numbers also win)
- 3 wins 26.75% of the time (14.08% of all games, 13.12% of which lower numbers also win)
- 4 wins 15.81% of the time (8.32% of all games, all 8.32% of which lower numbers also win)
- 5 wins 12.16% of the time (6.40% of all games, all 6.40% of which lower numbers also win)

The numbers add up to over 100% because of the overlaps where more than 1 number would win.

Now that of course is assuming random play, but what if the other players only consider A-3 like many of the above answers (either due to the reasons listed in those or because every win they get another lower number could also get)?

If this is the case, and assuming they still choose randomly, then you should play

4 or 5! (your choice, I'd go with 5 for style)

This should apply regardless of their strategies (as long as they're all similar) because

Those plays win any time none of the other players win due to their strategies picking either all the same number or two pairs.

Lets look at those probabilities!

With the others using an A-3 strategy only 40.74% of games are winnable. Of those:
- Ace wins 48.48% of the time (19.75% of all games)
- 2 wins 36.35% of the time (14.81% of all games, 1.23% of which lower numbers also win)
- 3 wins 24.25% of the time (9.88% of all games, 2.47% of which lower numbers also win)
- 4 wins 63.65% of the time (25.93% of all games, all 25.93% of which lower numbers also win)
- 5 wins 63.65% of the time (25.93% of all games, all 25.93% of which lower numbers also win)

The interesting thing to note here is that in all cases where you'd choose the highest chance card, there is another card which would also win... it's just that we don't know which one, so this pick gives better chances.

Overall, if the person going in didn't know how his friends were going to play and only one round was being played I would give this strategy:

Start by choosing Ace considering even in the second scenario it wins almost half of all winnable games. If there is a tie and others are playing all in the A-3 range (due to the no additional win states and/or reasons mentioned in other answers), then shift to picking 4 or 5.

Note that I am aware that none of this uses the speaking or the time to impose logical quandaries on the others involving the time and what is declared as I am not assuming that the logicians friends are also logicians. Chances are though that the answer that @Mike.C.Ford is thinking of uses logical friends, talking, and the oddly specific 55 second time limit, but this is what you get when I nerd snipe myself wanting to know more about the probability space while reading other answers.

That being said, my initial reaction to this question was very similar to Sconibulus's answer to declare what card you're going to play before you play it so anyone who makes you lose must also lose willingly themselves, which depending on the group of friends could work great, fail spectacularly, or end up with a few less friends...


I recommend a random strategy

40% Ace, 40% 2, 20% 3


With 5 people, either someone picking ace wins, or at least two people pick aces and someone who picks 2 can win, or two people pick aces and two people pick 2, so picking 3 wins (least likely).


Exploit that others will think 1,2,3 are the winners
You win be being the only card
Look at the common cards that will be played

1122 you would win with 3, 4, 5
2233 you would win with 1, 4, 5
1133 you would win with 2, 4, 5
six ways to make any or the above three AB, AC, AD, BC, BD, and CD
1222 you would tie with 1
1112 you would tie with 2
4 way to make the above two
1111 you would win with 2, 3, 4, 5
2222 you would win with 1, 3, 4, 5
1 way to make the above 2
- ironically 4 and 5 are the most common winners
if you are going for the tie then 2
I would go randomly on 4, 5
If the others naively play 1, 2, 3 you win 20 and lose 8 playing 4 or 5
I would play a 2 once to win a 1111 and tie a 1112
Force them to get more random
Then just play 45 and give myself a 2:1 and it would be over in 1-6 hands
No way they figure me out and adjust in time
I am 10x better than random

  • 1
    $\begingroup$ Out of curiosity, why not play 4 every time? Statistically, what does splitting it evenly between 4, 5, and 6 get for you? (As an aside, there is no 6 in the question, it is 5 cards but this does not effect your logic). $\endgroup$
    – Scott
    Nov 2, 2016 at 20:19
  • $\begingroup$ @Scott Because you can play 5 and it would help to confuse them $\endgroup$
    – paparazzo
    Nov 2, 2016 at 20:25
  • $\begingroup$ Your math is a little off. There are (4) 1112, (4) 1113, (4) 2223, (6) 1122, (6) 1133 and (6) 2233 along with 1 each of all 4 matching (1111, 2222, 3333) for a total of 33 ways to win when playing a 4. BUT there are 48 ways to lose (12) 1123, (12) 1223, (12) 1233 and (4) 1222, (4) 1333 and (4) 2333. This means (not saying I agree, but sticking with your logic) you win 33 and lose 48 playing a 4 or 5. $\endgroup$
    – Scott
    Nov 2, 2016 at 20:53
  • $\begingroup$ @Scott I think those are the common hands you would see. Even with your numbers 33 to 48 is a whole lot better than 1/5. I might drop 1133 and 2233 down for like 16:8 or 2:1. 2:1 is a full 10x better than random. $\endgroup$
    – paparazzo
    Nov 2, 2016 at 20:57

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