# Probability of winning a game by rolling a dice twice. [closed]

Ram plays a game of tossing a die with 2 faces twice. There is a number written on each face of the die (these numbers are not necessarily distinct). To win, Ram must get the number 1 on the first roll and the number 1 on the second roll of the die.

After rolling the die Ram gets 1 on the first roll and 2 on the second roll.

Find the probability that Ram will win the game. Assume that Ram gets each face of the die with same probability on each toss and that tosses are mutually independent.

• A die with two faces can also be called a coin. You can call 1=heads and 2=tails. The numbers may not necessarily be distinct means there is a chance that the coin you are flipping is a two-headed coin or a coin with two tails, but the problem then goes on to say that the coin has different sides, so each flip is a 50% chance of getting "heads". Thus, in general, there is a $\dfrac{1}{2}\cdot \dfrac{1}{2} = \dfrac{1}{4}$ chance of winning a single game, but as saulspatz mentioned, there is a 0% chance he will win the current game (as he has already lost). – InterstellarProbe Jun 6 '18 at 17:45
• I think he has lost give the information in the question. However, I believe what you were meant to ask is not a trick question, but rather he wins as soon as he rolls two consecutive $1$s. Is that right? I feel stupid now. – Tony Hellmuth Jun 7 '18 at 2:00