Gambler vs Host

Probability Level pending

There is a game where a gambler bets on a number between $1$ to $6$ inclusive then the host rolls $6$ $six$ sided fare dices with numbers $1$ through $6$ on the faces of every dice. If the betted number does not show up or show up exactly on $2$ of the dices, the host wins otherwise (show up on $1, 3, 4, 5$ or $6$ of the dices) the gambler wins. If the game is played multiple times who has more probability of winning in the long run?

host impossible to determine gambler equal chance

1 solution

Saad Khondoker
Feb 20, 2021

the probability of the host winning is:

$^{6}C_{0}$ $*$ ( $\frac{1}{6}$ ) $^{0}$ $*$ ( $\frac{5}{6}$ ) $^{6}$ $+$ $^{6}C_{2}$ $*$ ( $\frac{1}{6}$ ) $^{2}$ $*$ ( $\frac{5}{6}$ ) $^{4}$ $=$ $0.536$ $=$ $53.6\%$

So the host has more probability of winning in the long run

Gambler vs Host

1 solution

0 pending reports