Question

There's a game where you are asked to roll two fair six-sided dice. 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 the follow-up: If he plays 6 times what is the probability of making money from this game?

Answer 1

The game has a negative expected value of -$1.50, meaning you are expected to lose money in the long term and should not play. Calculating the probability of making money after playing 6 times involves using the binomial probability distribution.

Step-by-step explanation:

The game described involves rolling two fair six-sided dice with the intention of obtaining a sum of seven to win $21. However, you have to pay $5 each time to play. To decide whether to play, we must consider the expected value of the game.

There are 6x6=36 possible outcomes when rolling two six-sided dice, and 6 of these outcomes equal a sum of 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1). The probability of winning is therefore 6/36 or 1/6. If you win, your net gain is $21 - $5 = $16, because you pay $5 to play. If you lose (which happens with a probability of 5/6), your net loss is $5.

The expected value (EV) of one game is calculated as follows:

EV = (Probability of Winning x Net Gain) + (Probability of Losing x Net Loss)

EV = (1/6 x $16) + (5/6 x -$5) = $2.67 - $4.17 = -$1.50

Because the expected value is negative, you are expected to lose $1.50 on average for each game played, which means it is not a favorable game to play in the long term.

Regarding the follow-up question, the calculation of the probability of making money after playing 6 times requires binomial probability distribution since each roll is independent. With the probability of winning as 1/6 and losing as 5/6, we can calculate the probability of making money (winning at least once) over 6 plays.