Thumb Guessing Games Theory

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The Thumb Guessing Game (Tebak Jempol)
The thumb guessing game, is a simple number guessing game using the number of players’
thumbs. This game can be played by at least two person, and up to as many as you like, though
the more the player is, the more complicated the strategies. The rule are simple: every player
have one turn each round, where a round is the number of turns until the first player plays again.
In each turn, each player must guess a number of thumbs will be raised by other players. But,
other players are not obligated to raise a thumb or two, they can choose to not raise, raise one, or
raise two. If a player cannot answer correctly, then he must wait for his next turn. If the players
can correctly answer the number of thumbs up, they can retract one of their hand, and only
playing with one thumb. A winner player get another turn. If a player have retracted both of
his/her hand, than it is the winner. The game ends when we get a loser, or the last man standing.
So this is like a sort of reverse battle royale, where the last man standing is the loser, not the
winner.
Players:  = 1, 2, … , 
Strategy sets:  =  +   =  + −1
 =1  for ,  ∈ ,  ≠ 
Payoffs: The best payoff for each player is that a total point of remaining available thumbs where a player
leave a game in the game, or simply:
−1
 =
 ;  ≠ 
 =1
Let’s assume a very simple form of the game, the one-on-one thumb guessing game. In this game, the
maximum number of thumbs is 4, so the strategy is to choose between 0 up to 4. A reasonal choice for
each player is choosing 0-2 if he raised 0 thumb, 1-3 if he raised 1 thumb, and 2-4 if he raised 2 thumbs,
or simply  +   = −1
 =1  ;  ≠ . The game in sequential form is like this.
Player 1
0
0
1
1
2
0
2
1
2
0
1
2
Player 1
0
0
Player 1 win
1
1
2
0
1
Player 1 win
2
Player-j’s responses
Player-i’s guess
 − 
 + 
 =

 + 
 ≠

−1
 =1 
 = 
−2
 =1 
, max  − 1+
0 , max( )+
 − 
 ≠ 
0 , max( )+
−2
 =1 
−1
 =1 
−2
 =1 
, max  − 1+
−2
 =1 
Player-j’s responses
Player-i’s guess
 − 
 =  − 1
 − 1 + 
 =

−1
 =1 
 − 1 + 
 ≠

0 , max( ) − 1+
,
 − 
−2
 =1 
 ≠  − 1
0 , max  − 1+
−2
 =1 
−1
 =1 
,
−2
 =1 
−2
 =1 
This game yields no dominant strategy since it is probabilistic games, so we will use mixed strategy
instead. Assuming that 0 = 0  − 1 + 
0  − 1 + 
 =
 =  and 0  − 1 + 
 as players increasing, and  0 =  0
 0
 − 
 =
 =  − 1 =  and  0
 − 
 , 0  − 1 + 
 ≠
 − 
, with
 ≠  − 1 , with
 ≠  − 1 = 1 −  with  < 1 − . We get:
−1
×

 = 1 − , and assuming that  < 1 −
 =  − 1 ,  0
 − 
 ≠
−1
 + 1 −  × 0 =  × 0 + 1 −  ×
 =1

 =1
=
1
2
−2
 × max  − 1 +
−2
 + 1 −  × max  +
 =1

 =1
−2
=  × max  +
−2
 + 1 −  × max  − 1 +
 =1

 =1
−2
− + max  +
−2
 = max  − 1 +
 =1
 =1
=
This will yield the Nash Equilibrium of
 + 
∗
0
, ∗0 =
1
2
1 1
,
2 2
,
1 1
,
2 2
Now this is where it is get tricky. After this, we will find the best-response function, but, since p and q are
more likely to be less than half the more the player presents, the players-i and players-j will play their
own dominant strategies, or we can say that they will play according to their hunch, since the probability
is more diversed the more the players joined the game. Looking from the simmetry of outcome from the
next turn of the winning player, it will yield the same Nash Equilibrium of:
∗
1
, ∗1 =
1 1
1 1
, , ,
2 2
2 2
So in each round of the game, the standard deviation will be
1 1
1 1
∗

, ∗ =
, , ,
2 2
2 2
Yet, the more rounds are played, the higher the probability of expected guess will correct, so the player
will try to play with their hunch and shifting to less dominant strategy, by expecting the opponent
movement.
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