Dice

[Guest]

Hello, I have a problem with the following task:
- A gambling game is played in which a player bets \$ 10 the bank and rolls three dice. If a six falls, the bank returns it pledge. For two sixes, the bank pays the player \$ 50, and for three - \$ 250. What is the average profit of the player?
I've come this far:
After calculating what for zero 6s we have 5/6; 5/6; 5/6 and for one 6s we have 1/6; 5/6; 5/6 and for two 6s we have 1/6; 1/6; 5/6 and for three 6s we have 1/6; 1/6; 1/6 i used the formula and in the end thats what i got.
EX=-10*125/216+0*25/216+40*5/216+240*1/216=-3.75
Is it correct and if not can you please tell me how to solve it because the task is for tomorrow. Thank you in advance!

[Guest]

There are totally 6³= 216 outcomes in a roll. For every roll that has one dice show up 6, you will get a 1c1*5p2+1c1*5repeats= 25 for each dice. So total is 75.

For every roll that shows up two sixes, 3 dice has a 3c2 combos to turn up in 2 dice, and the other die can be 1 to 5. So 3c2*5 = 15.

For every roll that shows up three sixes there is only 1 case .

Hence you are looking at 216-75-15-1=125 for no six. Your expected value function should look like:

-10 * 125/216 + 0 * 75/216 + 50 * 15/216 + 200 * 1/216

[Guest]

There are totally 6³= 216 outcomes in a roll. For every roll that has one dice show up 6, you will get a 1c1*5p2+1c1*5repeats= 25 for each dice. So total is 75.

For every roll that shows up two sixes, 3 dice has a 3c2 combos to turn up in 2 dice, and the other die can be 1 to 5. So 3c2*5 = 15.

For every roll that shows up three sixes there is only 1 case .

Hence you are looking at 216-75-15-1=125 for no six. Your expected value function should look like:

-10 * 125/216 + 0 * 75/216 + 50 * 15/216 + 200 * 1/216

My mistake, -10 * 125/216 + 0 * 75/216 + 40 * 15/216 + 240 * 1/216

[Guest]

Hello, I have a problem with the following task:
- A gambling game is played in which a player bets \$ 10 the bank and rolls three dice. If a six falls, the bank returns it pledge. For two sixes, the bank pays the player \$ 50, and for three - \$ 250. What is the average profit of the player?
I've come this far:
After calculating what for zero 6s we have 5/6; 5/6; 5/6 and for one 6s we have 1/6; 5/6; 5/6 and for two 6s we have 1/6; 1/6; 5/6 and for three 6s we have 1/6; 1/6; 1/6 i used the formula and in the end thats what i got.
EX=-10*125/216+0*25/216+40*5/216+240*1/216=-3.75
Is it correct and if not can you please tell me how to solve it because the task is for tomorrow. Thank you in advance!

In your answer, 125/216 + 25/216 + 5/216 + 1/216 is not 1. You cannot have a gambling payout rule that doesn’t cover 100% of the probability.

[Guest]

A die has 6 faces, one of which is a six. The probability of a six on any single die is 1/6. The probability of anything other than a six is 5/6.

The probability of throwing three sixes is (1/6)^3.

The probability of two 6s and anything other than a 6, in that order, is (1/6)^2(5/6)= 5/6^3. But, writing "S" for a 6 and "N" for anything else, "SSN", "SNS", and "NSS" are two 6s and another number and have the same probability. The probability of "two sixes and a non-six", in any order, is 3(5/6^3)= 5/(2(6^2))= 5/72.

The probability of a 6 and then two non-6s is (1/6)(5/6)^2= 25/6^3. Again "SNN", "NSN", and "NNS" all have the same probability so the probability of a 6 and two non-sixes, in any order, is 3(25/6^3)= 25/(3(6^2))= 25/72.


So you start by paying \$10.
There is a probability of 1/6^3= 1/216 of getting three 6s and you win \$250. That is an expected value of $250/216= \$125/108= $1.16. There is a probability of 5/72 of getting two 6s and you win \$50. That is an expected value of \$50(5/72)= \$3.47. There is a probability of 25/72 of getting one 6 and you win \$10. That is an expected value of \$10(25/72)= \$3.47.

Your expected winnings will be 1.16+ 3.47+ 3.47- 10= 8.10- 10= -\$1.90.

You can expect to lose an average of $1.90 each time you play.

[ChanelLeuschke]

One of a die's six faces is a 6. A six appears 1/6 of the time on a single die. Five out of six chances are that it will be a number other than six. geometry dash

The odds of getting three sixes are (1/6)^3.

In that specific sequence, the odds of getting two 6s followed by any other number are 5/6^3. However, when "S" is written as a 6 and "N" as any other number, the three words "SSN", "SNS", and "NSS" are equivalent to two 6s and another number, thus their probabilities are identical. Whatever the sequence, the likelihood of "two sixes and a non-six" is 3(5/6^3)= 5/(2(6^2))= 5/72.

Having a 6 followed by two non-6s has a probability of 25/6^3 (1/6)(5/6)^2. The likelihood of a 6 and two non-sixes, in any sequence, is 3(25/6^3)= 25/(3(6^2))= 25/72, as the probabilities of "SNN", "NSN", and "NNS" are identical.

[Guest]

Your approach is correct, but there's a small mistake in your calculation. Let's correct it step by step.

First, let's denote:
• p0 = probability of getting zero sixes
• p1 = probability of getting one six
• p2 = probability of getting two sixes
• p3 = probability of getting three sixes

Now, we have:

p0=5/6x5/6x5/6 (no sixes on any of the three dice)

p1=3×(1/6×5/6×5/6) (one six and two non-sixes, but the six can occur in three different positions)

p2=3×(1/6×1/6×5/6) (two sixes and one non-six, but the non-six can occur in three different positions)

p3=1/6×1/6×1/6 (three sixes)

Now, let's calculate the expected profit (EP):

EP=−10×p0+0×p1+50×p2+250×p3

Substituting the values of p0, p1, p2, and p3:

EP=−10×(5/6)^3+0×3×(1/6)×(5/6)^2+50×3×(1/6)^2×(5/6)+250×(1/6)^3

Now, let's calculate each term:

EP=−10×125/216+0+50×75/216+250×1/216
EP=−1250/216+3750/216+250/216
EP=(3750−1250+250)/216
EP=2750/216
EP≈12.73

So, the average profit of the player is approximately $12.73.

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