Math Probability question Please help?

Question

please provide correct answer and explanation

Two players take turns throwing a die. The first one to roll a one wins. How big is the first players advantage? What are the probabilities of winning for each of the two players?

Answer 1

The first player has a 1/6 chance of winning on their first go.

The second player has a 5/6 * 1/6 chance of winning on their first go. 5/6 is the probability of the first player not winning and 1/6 is the probability of rolling a 1.

The first player has a (5/6)² * 1/6 chance of winning on their second go.

The second player has a (5/6)³ * 1/6 chance of winning on their second go.

Etc.

Adding up the probabilities for the first player gives: 1/6 + (5/6)² * 1/6 + (5/6)^4 * 1/6 + ...
This is a geometric series which has sum of (1/6) / (1-(5/6)²) = (1/6) / (1 - 25/36) = (1/6) / (11/36) = 6 / 11.

Adding up the probabilities for the second player gives: 5/6*1/6 + (5/6)³ * 1/6 + (5/6)^5 * 1/6 + ...
This is a geometric series which has sum of (5/36) / (1-(5/6)²) = (5/36) / (1 - 25/36) = (5/36) / (11/36) = 5 / 11.

So the first player has a 6/11 or 54.54% chance of winning.
And the second player has a 5/11 or 45.45% chance of winning.

Answer 2

The probability p of the first player winning must satisfy
p=P(first player wins)=P(first player wins first throw)
+P(first player does not win first throw)*P(second player does not win second throw)*P(first player wins)
=1/N+(1-1/N)*(1-1/N)*p
=1/N+(1-1/N)^2*p
where N is the number of sides on the die.
Solving
p=1/N+(1-1/N)^2*p,
we get
p=N/(2*N-1)
so the win probabilities for first and second player are N/(2*N-1) and (N-1)/(2*N-1) respectively.
This equals 6/11 versus 5/11 for the interesting case N=6. No matter what positive integer value N has, the first player does indeed have the advantage.

Answer 3

Player 1 --> 1/6 probability of winning on 1st roll.
Player 1 --> 5/6 x 5/6 x 1/6 probability of winning on his 2nd roll.
Player 1 --> (5/6)^4 x 1/6 probability of winning on his 3rd roll.
etc.

Now compare that to player 2:
Player 2 --> 5/6 x 1/6 probability of winning on his 1st roll.
Player 2 --> 5/6 x 5/6 x 5/6 x 1/6 probability of winnning on his second roll.
Player 2 --> 5/6 x (5/6^4) x 1/6 probability of winnning on his third roll.
etc.

Notice how Player 2's probability for a specific roll is always 5/6 of the Player 1's probability on that same roll.

If you were to sum up Player 1's probabilities they would be P.
If you were to sum up Player 2's probabilities they would be 5/6 P.

P + 5/6P = 100%
11/6 P = 100%
P = 6/11 * 100%

Player 1:
P = 6/11 = 54.5454%

Player 2:
1 - P = 5/11 = 45.4545%

Answer 4

The probability for any one to win on their turn is 1/6, because they have 6 possible rolls and only 1 can make them win.

I'm not sure what the first player's advantage is... hmm... Sorry!

It would probably be a 16.6...% advantage because that is the probability of rolling a 1 on any roll.

Answer 5

Each player has the same proberbility of winning if they each roll the die the same amount of times. They each have a one in six probability of winning

Answer 6

Each time a die is rolled, there is a 16.66% likelihood of hitting the number 1.

This would consequently be the advantage of the first to roll.

Answer 7

1/6 probablity of winning since there are 6 sides to the dice and only one side has the number 1.