A professional Rock, Paper, Scissors player can win a single game (trial) with p

Question

A professional Rock, Paper, Scissors player can win a single game (trial) with probability 0.77. The player is up against an opponent.

A)What is the probability that the player will win 2 games before the opponent wins 2 games?

Hint: You want to add several values of a negative binomial distribution u(m,k,p) with k being the number of games needed to win and p being the probability of winning. What values of m (the number of games played) do you want? Can the player get 2 wins in 1 game? Can the player get 2 wins over 2 games without the opponent getting 2 wins? How about 3 games? 4?

B)What is the probability that the player will win 1 game before the opponent wins 2 games?

Explanation / Answer

Rock paper scissor game:

It is a zero sum game, where signs of rock paper and scissor are made by the players using their fists.

It is generally played between two players

If one makes a rock and other makes a scissor, the one with rock wins

If one makes paper and the other makes a rock, the one with paper wins

And if one makes a scissor and the other makes a paper, the one with scissor wins

And in case if both the players make same sign, it is a tie

In the question, we are already provided with the probability (experimental probability) of wining the game by one player.

P(w) = 0.77, Hence P(lose) = 1 - 0.77 = 0.23

Part a ) probability that the player wins 2 games, before opponent wins 2 games

Using negative binomial distribution, we get:

k = number of games needed to win = 2 wins [2 wins of the opponent]

p= probability of win of the opponent = 0.23 [opponent wins, when the professional player loses]

m= total number of games played = 4 [ 2 wins of professional player and 2 wins of opponent]

Hence u(m,k,p) = m-1 Ck-1 * (p)^k * (1-p)^m-k   [Formula for Negative Binomial Distribution]

on plugging in the values of m, k and p in the equation above we get:

u(4,2,.23) = 3C1 * (.23)^2 * (.77)^2

u(4,2,.23) = 0.09401 [ANSWER]

Can the player get 2 wins in 1 game: in one game, as there is one round only there can be only one win. it is possible that a tie happens, but then immediately a second round is played to decide the winner.

Can the player get 2 wins over 2 games without the opponent getting 2 wins: Yes it is possible for a player to get two cosnecutive wins over two games. suppose if first he shows a paper and the opponent shows a stone, the player wins. In the second game, the player shows a scissor and the opponent shows a paper. Again the player wins. Even if a tie comes in between the two players, the opponent does not get to win.

So in three games, a player can get 2 wins without the opponent getting 2 wins , if tie happens in one of the game.

And so is the case with four games.

Part b) probability that the player will win 1 game before the opponent wins 2 games:

u(3,2,.23) = 2C1 * (.23)^2 * (.77)^1

u(3,2,.23) = 0.0815 [ANSWER]

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