6
$\begingroup$

In a game players take turns saying up to 3 numbers (starting at 1 and working their way up) and who every says 21 is eliminated. So we may have a situation like the following for 3 players:

Player 1: 1,2

Player 2: 3,4,5

Player 3: 6

Player 1: 7,8,9

Player 2:10,11

Player 3:13,14,15

Player 1: 16,17,18

Player 2: 19,20

In which case Player $3$ would have to say $21$ and thus would be eliminated from the game. Player $1$ and $2$ would then face each other.

In the case of a two player game, the person who goes second can always win by ensuring the last number they say is a multiple of $4$.

Let us say we are in an $N$ player game, and that I am the $i^{\text{th}}$ player to take my turn. Is there any strategy that I can take to make sure I will stay in the game? For example in the simple case of a $2$ player game the strategy would be 'try to end on a multiple of $4$ and then stay on multiples of $4$'.

$\endgroup$
4
  • 6
    $\begingroup$ Doesn't seem so (unless there are 19+ players and you're less than 8th). Other players can cooperate against you; even two players can say 2-6 numbers, ensuring the last number is multiple of 5. $\endgroup$ Dec 18, 2015 at 8:33
  • $\begingroup$ I suggest looking at the game "Nim" $\endgroup$ Aug 5, 2016 at 18:14
  • $\begingroup$ What happens after a player is eliminated? The remaining players count from $P+1$ to infinity? $\endgroup$
    – bof
    Nov 5, 2016 at 11:41
  • 1
    $\begingroup$ @bof No they simply start back from 1 and play the exact same game just with one less player. $\endgroup$ Nov 5, 2016 at 11:47

2 Answers 2

7
$\begingroup$

Community wiki answer so the question can be marked as answered:

As Abstraction pointed out in a comment, there is no such strategy (except in extreme cases) because the other players can cooperate against you; they have a wider range of options, collectively, than you do, so you can't use the sort of strategy that you can use in the two-player version.

$\endgroup$
1
$\begingroup$

You can use a strategy but only by the means on where you stand in the order of people, there is complex maths involved in probabilities but if numbers chosen ($1$-$3$) by each player are randomised then certain positions have a much lower chance of being forced into saying “$20+1$”

This was figured out using a simulation with that all players do not want to be landed on $21$ and therefore I used the assumption of if a player finishes on $17$, the next player will always bust the succeeding player by saying $18$, $19$, $20$

Let $1^{st}$ person=$A$, $2^{nd}$=$B$, $3^{rd}$=$C$, $4^{th}$=... Let first letter be most likely to say $21$ and last be least likely (preferable position)

Number Of Total Players

2=B,A

3=B,A,C

4=C,D,B,A

5=A,B,E,C,D

6=E,F,D,A,C,B

7=D,E,C,F,B,G,A

8=C,D,B,E,A,F,G,H

9=B,C,A,D,I,E,F,H,G

10=A,B,J,C,I,D,E,H,F,G

The more people there are, the more significant the benefits are of choosing a good starting position I.e. with $10$ people if you stand in position G (7th place) you have a $0.01\%$ chance of being chosen in comparison to the worst position A (1st place) with a chance of $31.88\%$

I don’t believe there is much of a pattern other than positions G and H are consistently good places to stand.

[Edit]: After more research as a basic strategy let n be number of people, c be current number before you, and b be the number you go bust/lose (I.e. 21) Then if c+2(n-1)>b-3 then choose 3 numbers to say as this will likely cause a bust before it returns to you

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .