Fiddler problems

Fiddler Problem: Don’t flip out
We calculate the probability using a Markov process method. We define the states \((a_1,a_2,a_3)\times (b_1,b_2,b_3)\) and define the transition probabilities from a state to other \(p_{ij}((a_1,a_2,a_3)\times (b_1,b_2.b_3)\rightarrow (c_1,c_2,c_3)\times (d_1,d_2,d_3))\), where each \(a,b,c,d\) are 0 or 1.
The tuples can be any of \(((0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1))\)
For our problem, we have 14 absorbing states, 7 when the game ends with either first player wins (corresponding to tuple (1,1,1)) and 7 when second player wins(corresponding to tuple (1,1,0)). We have that each transition probability is either 1/4 or 0.
For example: \(P((0,0,1)\times (0,1,1)\rightarrow (0,1,0)\times (1,1,0))=1/4\) \(P((0,0,1)\times (0,1,1)\rightarrow (0,1,1)\times (1,1,0))=1/4\)
\(P((0,0,1)\times (0,1,1)\rightarrow (0,1,0)\times (1,1,1))=1/4\)
\(P((0,0,1)\times (0,1,1)\rightarrow (0,1,1)\times (1,1,1))=1/4\)
but \(P((0,0,1)\times (0,1,1)\rightarrow (0,0,0)\times (1,1,0))=0\)
The transition probabilities form a transition probability matrix (t.p.m.) M that has dimension 64 x 64 and has the following form: \(\begin{bmatrix} I & 0 \\ R & Q \\ \end{bmatrix}\)

where \(R\) is the 50x14 t.p.m. from the absorbing states to the non-absorbing, and the \(Q\) is the 50x50 t.p.m. of the transient (non-absorbing) states

Then the solution is provided by the following theorem:
If \(a_{ij}\) is the probability that starting from transient state \(i\) we reach the absorbing state \(j\), then the matrix \(A=(a_{ij})\) is given by:
\(A=(I-Q)^{-1}R\). The matrix \(A\) has dimension 50x14. It can be split into two parts \(A_1\) with columns from 1 to 7 corresponding to reaching (1,1,1) and \(A_2\) corresponding to reaching (1,1,0). We calculate the total \(t_1\) of all elements in matrix \(A_1\) and respectively the total \(t_2\) for \(A_2\).
Then using the probabilities for the initial step we obtain that the probability that first player wins is \(1/8^2+7/8^2\cdot t_1\) and the probability that second player wins is \(1/8^2+7/8^2\cdot t_2\).
We obtain the first probability as 25439/71106~0.3578 and second probability as 45667/71106~0.6422.

Fiddler Problem: Can You Roll the Dungeon Master’s Dice?
Suppose we choose \(x\) and \(y\) from set \(A=[4,6,8,10,12,20]\). We find minimum of them, \(\min(x,y)\). Then the probability to pick two equal numbers is:
\(\begin{equation} \sum_{(x,y)\in A\times A}\frac{\min(x,y)}{36\cdot x \cdot y}=\frac{3}{32}\sim 0.09375 \end{equation},\) since we have 36 pairs.
Extra Credit Suppose we pick a tuple \((x,y,z)\in A\times A\times A\). Then we sort \((x,y,z)\) to get \((a,b,c)\), where \(a\leq b\leq c\). Now we cast the dice \(x,y,z\) to obtain three values. The probability that all three values are equal is: \(\begin{equation} p_1=\sum_{(x,y,z)\in A\times A\times A}\frac{a}{216xyz}=\frac{32753}{3110400}\sim 0.01053 \end{equation}\)
The probability that all three values are different is:
\(\begin{equation} p_3=\sum_{(x,y,z)\in A\times A\times A}\frac{a(b-1)(c-2)}{216xyz}=\frac{1150553}{1555200}\sim 0.73981 \end{equation} \notag\)

The probability that only two values are different out of three is :\(p_2=1-p_1-p_3\).
Therefore the expectation of the number of different values is:
\(E=p_1+2\cdot p_2+3\cdot p_3=\frac{8489153}{3110400}\sim 2.72928\).

 ```python   
import numpy as np;import random  
from decimal import Decimal, getcontext  
from fractions import Fraction
from myscript import List,dec
A = [4, 6, 8, 10, 12, 20]
p = [(x, y,min(x,y),Fraction(1,36)*Fraction(min(x,y),(x*y))) for x in A for y in A]
p=np.array(p)
w=sum(p[:,3]);print(w)
print(dec(w)
q=[(x,y,z,pr,Fraction(min(x,y,z),216*x*y*z),Fraction(pr)) for x in A for y in A for z in A
   if ((o:=List(sorted([x,y,z]))) and (pr:=Fraction(o[1]*(o[2]-1)*(o[3]-2),(x*y*z)*216)))]
q=np.array(q)
q1=sum(q[:,4]);print(q1);print(dec(q1))
q3=sum(q[:,5]);print(q3);print(dec(q3))
q2=1-q1-q3;print(q2);print(q1+2*q2+3*q3);print(dec(q2))
```

Fiddler On the Roof Problem:Could You Have Won the Super Bowl?
Let’s call your own team A and opponent team B. The probability team A can win in the first two times is:
\(P(A=7)*P(B=0 \mbox{ or } B=3)+P(A=3)*P(B=0)=1/3*2/3+1/3*1/3=1/3\) However A can win in the subsequent times if they draw up to this point. The probability they draw in the first two times is \(1/3*1/3+1/3*1/3+1/3*1/3=1/3\). And the probability A wins later on is \(2/3+1/9*2/3+1/9^2*2/3+\ldots =2/3*9/8=3/4\)
Hence, overall, A wins with probability: \(1/3+1/3*3/4=1/3+1/4=7/12\sim 0.583(3)\)
Extra Credit
The teams have to choose between scenarious (7 3 0) and (7 0). B knows the score after first part of game, so it can adjust its strategy accordingly. We have the followinng cases:
1a) if A chooses (7 0) and scores a 7, B clearly chooses (7 0). Then \(P(A \mbox{ wins}|A=7)=1/2+1/2*3/4=1/2*7/4=7/8\)
1b) if A chooses (7 0) and instead A scores a 0, if B chooses (7 0), then B wins with probability \(P(B=7)+P(B=0)*1/4=1/2+1/2*1/4=1/2*5/4=5/8\). If instead, B chooses (7 3 0), B wins with probability of \(2/3+1/3*1/4=2/3+1/12=9/12=3/4>5/8\). Therefore, B opts for the better alternative (7 3 0). In this case, \(P(A \mbox{ wins}|A=0)=1-3/4=1/4\)
Hence, overall, if A chooses (7,0), its probability of winning is \(P(A=7)*P(A \mbox{ wins}|A=7)+P(A=0)*P(A \mbox{ wins}|A=0)\) \(=1/2*7/8+1/2*1/4=7/16+2/16=9/16= 0.5625\)
2a) if A chooses (7 3 0) and scores a 7, B chooses (7 0) and A wins with probability 7/8, just as in case 1a)
2b) if A chooses (7 3 0) and scores a 3, if B chooses (7 0), B wins with probability \(1/2\). If B chooses (7 3 0), it wins with probability \(1/3+1/3*1/4=1/3*5/4=5/12<1/2\). Hence B opts for the better (7 0). Then A wins with probability \(1/2\)
2c) if A chooses (7 3 0) and scores a 0, B chooses (7 3 0). In this case, A wins with probability 1/4, just in case 1b).
So overall, if A chooses (7 3 0), its probability of winning is \(P(A=7)*P(A \mbox{ wins}|A=7)+P(A=3)*P(A \mbox{ wins}|A=3)\) \(+P(A=0)*P(A \mbox{ wins}|A=0)=1/3*7/8+1/3*1/2+1/3*1/4=13/24= 0.541(6)\).
In conclusion, since A wants to optimize its way of winning, it will therefore want the stategy (7 0), in which case its chance of winning is 0.5625.