There's a game where you are given two fair six-sided dice and asked to roll. If the sum of the values on the dice equals seven, then you win $21. However, you must pay $5 to play each time you roll both dice. Do you play this game? And in follow-up: What is the probability of making money from this game?

In order to win $21, you need to come up with (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) on two dice. There are 6*6=36 possible outcomes, so the chance of wining in 6/36=1/6, in which case you actually earn 21-5=$16. The chance of loosing is of course 5/6 and you earn -$5. So, your expected earning is: 1/6*16 + 5/6*(-5) = -1.5 So, you are expected to loose money in average!

In terms of the follow-up question, it's unclear what interval we're considering. If we're interested in a single trial, then the probability of making money is 1/6. Let's say that we're willing to play the game 4 times. If we win at least once within 4 trials we'll win money (losing 3 times results in a loss of $15 and winning 1 time results in a gain of $16). So, let's reframe this as the probability of getting at least 1 success in 4 trials. We can treat this as binomial problem. If N is the number of trials, X is the number of successes, and P is the probability of success, then (N choose X) * P ^ X * (1 - P) ^ (N - X) will give us the probability of getting exactly 1 success in 4 trials. However, we want the probability of at least 1 success. The probability of at least 1 success in 4 trials = 1 - the probability of no successes in 4 trials. So, we can use 1 - ( (4 choose 0) * (1/6)^0 * (5/6)^4), which equals ~ .52.

This can be treated as a binomial problem. Each time throwing 2 dices is one trial, and the chance of sum=7 is 1/6. X ~ B(n,p), where p=P(sum=7)=1/6. Suppose there were n trials and x wins, to make money 21*x>5n, i.e. x>5n/21. Let m be the smallest integer greater than 5n/21. P(making money) = sum of ( (n choose i) * p**i * (1-p)**(n-i)) over i=m to n

For the follow up questions, as mentioned above, in order to make money from the game you would need to win in with 4 or less rolls. So we need to calculate the probability of winning in 4 or less rolls. Or we can find probability of losing first and we can subtract it from 1. Since the sum of their probabilities must equal 1. Probability of losing in 1 rolls is 5/6, Probability of losing in 4 rolls is (5/6)^4 = 0.48 Probability of winning = 1-0.48=0.52

For the second question with a series of n try say you win m times (lose n-m times). To be an overall winner you need m * 16$ - (n-m) * 5$ >0 -> 21m$> 5n$ -> m> 5/21* n for a series of 21 double dice throws you need at least 5 {6,6} double to win binomial says the probability of k wins in n tries (n,k)p^k*(1-p)^(n-k) in our case, p =1/6 and we should sum over all k's where k> 5/21n sum((n,k)(1/6)^k*(5/6)^(n-k), k , ceil(5n/21),n) for say n = 21; k starts from 5 and we will get ~0.26495. 21% chance to win for n goes inf this goes 0

probability of winning = 1/6 probability of losing = 5/6 min sample games = 60 as np>=10 and n(1-p)>=10 is satisfied 60 trials => cost => $5 * 60 = $300 to play Expected number of wins = 10 wins => earn => $21 * 10 = $210 Expected losses = 50 losses => earn => $0 Losses = $300-$210 = $90 for playing 60 games