Person $A$ and person $B$ have $8$ and $10$ fair coins. They both flipped their coins. What is the probability of person $B$ having more heads than person $A$? (I know how to do the question if number of coins are $8,9$ by using symmetry, but not in the case of $8,10$).
Asked
Active
Viewed 483 times
0
-
Check this: https://math.stackexchange.com/questions/146802/probability-of-5-fair-coin-flips-having-strictly-more-heads-than-4-fair-coin?rq=1 – Ty Jensen Oct 27 '20 at 17:48
-
@TyJensen the linked you provided is the person having 1 extra coin than the other and I know how to solve it by using probability symmetry. But having 2 more coins than your opponent is somewhat difficult and different... – tommy Oct 27 '20 at 20:03
-
did you read the remark in the link ? – Vasil Yordanov Oct 27 '20 at 22:09