# games

## ticket risk in Ticket to Ride

Wed, 2016-08-31 19:47

I've been playing a lot of Ticket to Ride recently. Jenni and I play with the 1910 cards. There are 69 cards total (so maybe this is the "Mega" game?). Each card has a point value. A player may choose to draw three cards: what is the expected minimum value of the cards drawn? This value can be interpreted as an upper bound on the "risk" associated with drawing tickets (cards): the worst case scenario when drawing is that you lose this minimum value worth of points.

The values on the cards are distributed like this:
A=[0,1,0,3,3,5,7,7,4,6,7,4,5,2,1,2,3,0,4,3,2]
That is, there is 1 card worth 2 points, 0 cards worth 3 points, 3 cards worth 4 points, etc., up to a maximum of 21 points (on two cards).
A little PARI/GP code:
B=vector(21);\
B[1]=0;for(i=2,21,B[i]=A[i]+B[i-1]);\
C=vector(21);\
for(i=2,21,if(A[i]>0,\
C[i]=A[i]*binomial(69-B[i],2)+binomial(A[i],2)*(69-B[i])+binomial(A[i],3)));\
print(C);\

Yields the array C:
[0, 2278, 0, 6436, 5860, 8560, 9660, 7259, 3202, 3683, 2856, 1060, 860, 225, 91, 144, 136, 0, 74, 10, 0]

Out of the "69 choose 3" possible hands, 2278 have a minimum card value of 2, 6436 have a minimum value of 4, etc.

From this we can calculate the expected value:
sum(i=1,21,i*C[i]*1.)/binomial(69,3) = 7.0304424170706...