Roll 3 dice

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lenamitkova

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Hello, I am writing to you in connection with one of the sample assignments I was given for homework. I tried several times to calculate it, but to no avail. I have calculated all possible combinations when throwing the sums from 3 to 18 points and their percentages and probabilities, but I cannot reach a final solution to the problem. I will be infinitely grateful to you! - 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:
3 = 1 + 1 + 1
4 = 1 + 1 + 2
5 = 1 + 1 + 3 = 2 + 2 + 1
6 = 1 + 1 + 4 = 1 + 2 + 3 = 2 + 2 + 2
7 = 1 + 1 + 5 = 2 + 2 + 3 = 3 + 3 + 1 = 1 + 2 + 4 8 = 1 + 1 + 6 = 2 + 3 + 3 = 4 + 3 + 1 = 1 + 2 + 5 = 2 + 2 + 4 9 = 6 + 2 + 1 = 4 + 3 + 2 = 3 + 3 + 3 = 2 + 2 + 5 = 1 + 3 + 5 = 1 + 4 + 4
10 = 6 + 3 + 1 = 6 + 2 + 2 = 5 + 3 + 2 = 4 + 4 + 2 = 4 + 3 + 3 = 1 + 4 + 5
11 = 6 + 4 + 1 = 1 + 5 + 5 = 5 + 4 + 2 = 3 + 3 + 5 = 4 + 3 + 4 = 6 + 3 + 2
12 = 6 + 5 + 1 = 4 + 3 + 5 = 4 + 4 + 4 = 5 + 2 + 5 = 6 + 4 + 2 = 6 + 3 + 3 13 = 6 + 6 + 1 = 5 + 4 + 4 = 3 + 4 + 6 = 6 + 5 + 2 = 5 + 5 + 3 14 = 6 + 6 + 2 = 5 + 5 + 4 = 4 + 4 + 6 = 6 + 5 + 3 15 = 6 + 6 + 3 = 6 + 5 + 4 = 5 + 5 + 5 16 = 6 + 6 + 4 = 5 + 5 + 6 17 = 6 + 6 + 5 18 = 6 + 6 + 6 3: 1/216 = 0.5% 4: 3/216 = 1.4% 5: 6/216 = 2.8% 6: 10/216 = 4.6% 7: 15/216 = 7.0% 8: 21/216 = 9.7% 9: 25/216 = 11.6% 10: 27/216 = 12.5% 11: 27/216 = 12.5% 12: 25/216 = 11.6% 13: 21/216 = 9.7% 14: 15/216 = 7.0% 15: 10/216 = 4.6% 16: 6/216 = 2.8% 17: 3/216 = 1.4% 18: 1/216 = 0.5%
 
Why are you looking at the total of the three dice?
You need to find the probability that you get exactly 1 six when 3 dice are rolled. Then exactly 2 sixes, then exactly 3 sixes.
Totals don't come into the question.
 
After calculating what you said if 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 calculated the RTP = 0,81. Is this right?
 
Look at this expansion of [imath]{\left( {\sum\limits_{k = 1}^6 {{x^k}} } \right)^3}[/imath] You will see the term [imath]\large15x^{14}[/imath]
That tells us that there are fifteen ways to toss a sum of fourteen with three dice.
 
It might be helpful to set it up this way.
Let G be the random variable that represents the player's gain, and N be the number of 6 from the 3 die rolls, then we have the following:
[math]G =\begin{cases} 10 & Pr(N=1) \\ 50 & Pr(N=2) \\ 250 & Pr(N=3) \end{cases}[/math]Can you calculate the Pr(N=1), Pr(N=2),Pr(N=3)? Hint: Use binomial distribution
This will complete the distribution for G, and you can calculate the expected value of G, E[G]
The average profit, i.e E[Profit] = E[Gain] - E[Fee]
 
BigBeachBananas said:
It might be helpful to set it up this way.
Let G be the random variable that represents the player's gain, and N be the number of 6 from the 3 die rolls, then we have the following:
[math]G =\begin{cases} 10 & Pr(N=1) \\ 50 & Pr(N=2) \\ 250 & Pr(N=3) \end{cases}[/math]Can you calculate the Pr(N=1), Pr(N=2),Pr(N=3)? Hint: Use binomial distribution
This will complete the distribution for G, and you can calculate the expected value of G, E[G]
The average profit, i.e E[Profit] = E[Gain] - E[Fee]
Click to expand...
[EDIT] I didn't consider the case where Gain = 0 in the previous post.
Let G be the random variable that represents the player's gain, and N be the number of 6 from the 3 die rolls, then we have the following:
[math]G =\begin{cases} 0 & Pr(N=0) \\10 & Pr(N=1) \\ 50 & Pr(N=2) \\ 250 & Pr(N=3) \end{cases}[/math]Can you calculate the Pr(N=0), Pr(N=1), Pr(N=2),Pr(N=3)? Hint: Use binomial distribution
This will complete the distribution for G, and you can calculate the expected value of G, E[G]
The average profit, i.e E[Profit] = E[Gain] - E[Fee]
 
You posted this in three forums. As a simple everyday logic, you cannot have a gambling payout rule that doesn’t cover 100% of the probability. 125/216+25/216+5/216+1/216 is not 100%. I answered this in the other forum.

this is the same as a three dial passcode lock each dial has 1 to 6. For each dial to be affixed at 6 there are 25 outcomes that the other two dials form which neither dial get rolled to a 6. So for three dials, there are 3*25=75 formations you can have which one dial shows up 6 and the others shows two numbers that doesn’t have 6. That corresponds to rolling three dice and getting exactly one six in a roll, and the probability is not 25/216 but 75/216.
 
When you have dial 1 and 2 affixed at 6, or 2 and 3 affixed at 6, or 1 and 3 affixed at 6, for each case there are 5 numbers of the remaining dial to be rolled to if you don’t roll the remaining dial to 6. So three cases times 5 = 15. This corresponds to rolling three dice and getting exactly two six in a roll, and the probability is not 5/216 but 15/216.

So your payout is -10 * 125/216 + 0 * 75/216 + 40 * 15/216 + 240 * 1/216.
 
To simplify the problem, instead of rolling three different dice, think you only have one dice, but you roll it three times as they are identical.
Let N be the number of times you roll a 6, then N~Binominal(n=3, p =1/6)
[math]\text{Pr(N=0)}={3 \choose 0}(\frac{1}{6})^0(\frac{5}{6})^3 = \frac{125}{216}\\ \text{Pr(N=1)}={3 \choose 1}(\frac{1}{6})^1(\frac{5}{6})^2 = \frac{25}{72}\\ \text{Pr(N=2)}={3 \choose 2}(\frac{1}{6})^2(\frac{5}{6})^1= \frac{5}{72}\\ \text{Pr(N=3)}={3 \choose 3}(\frac{1}{6})^3(\frac{5}{6})^0 = \frac{1}{216}\\[/math]Then your expected gain, [math]E[Gain]= 0*\frac{125}{216} + 10*\frac{25}{72}+50*\frac{5}{72}+250*\frac{1}{216}=\frac{875}{108}\approx8.10185\\ E[Profit]= 8.10185 - 10 \approx -1.89815[/math]
 
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