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Tve been playing a lot of King of Toyko lately: There are six dice (at least at first), with sides labeled 1, 2, 3 and On each turn you get three dice rolls; each...

Question

Tve been playing a lot of King of Toyko lately: There are six dice (at least at first), with sides labeled 1, 2, 3 and On each turn you get three dice rolls; each time; you keep the ones you like and can reroll any of the others_ Now , am currently in Tokyo; so does me nO good (you have to leave Tokyo to heal). Im aiming for many as possible: If I end up with at least three of one number, though will get some victory points (YOu get the value of the die if you have three; and one more for each

Tve been playing a lot of King of Toyko lately: There are six dice (at least at first), with sides labeled 1, 2, 3 and On each turn you get three dice rolls; each time; you keep the ones you like and can reroll any of the others_ Now , am currently in Tokyo; so does me nO good (you have to leave Tokyo to heal). Im aiming for many as possible: If I end up with at least three of one number, though will get some victory points (YOu get the value of the die if you have three; and one more for each additional occurance of that number) So here's my strategy: on each roll _ keep all the get , and keep all the 3 s get, but [ reroll everything else. What is the probability that end up with three and three victory points (meaning the other three dice are all 3s)? The following parts break down the computation Fix 0 < jk < 3 and find the conditional probability that end up with three and three 3*s, given that start my third roll with j k3s, and 6 - j - k other things. Now fix p and q with 0 < p < j and 0 < q < k and find the conditional probability that I start my third roll with k3*s, and 6 - j - k other things, given that I start my second roll with p q 3*s, and 6 q other things. Use the previous three parts to calculate the desired probability: Please explain you reasoning and simplify YOur answer to the extent possible: your final answer should involve only one multinomial coefficient BONUS: actually get a numerical aswer



Answers

Craps. In the game of craps, a player rolls two balanced dice. Thirty-six equally likely outcomes are possible, as shown in Fig. 4.1 on page $147 .$ Let $A=$ event the sum of the dice is 7 $B=$ event the sum of the dice is 11 $C=$ event the sum of the dice is 2 $D=$ event the sum of the dice is 3 $E=$ event the sum of the dice is 12 $F=$ event the sum of the dice is $8,$ and $G=$ event doubles are rolled. a. Compute the probability of each of the seven events. b. The player wins on the first roll if the sum of the dice is 7 or $11 .$ Find the probability of that event by using the special addition rule and your answers from part (a). c. The player loses on the first roll if the sum of the dice is 2,3 , or $12 .$ Determine the probability of that event by using the special addition rule and your answers from part (a). d. Compute the probability that either the sum of the dice is 8 or doubles are rolled, without using the general addition rule. e. Compute the probability that either the sum of the dice is 8 or doubles are rolled by using the general addition rule, and compare your answer to the one you obtained in part (d).

No one in question, one of one were given information. Similar games played for Chinese New Year in Vietnamese New Year. As it explains, the game's explains how the each game looks. It tells us how the rules of the game work were you better dollar and values. And the overall interest of this game is in the number of matches that you get in your game because that determines how much money you win. You get to a inwards to find the random variable X In this case, because X is gonna represent the number of Matthew's, we're just simply going to say X is the number of matches. The list of values that X may take on this game involves house role in three dice, and we're trying to determine the number of matches in the three dice. We could have zero matches, one match to match or all three matches so X could be any value 012 or three. See give the distribution of X because we can consider a match being a success and a non match being a failure. We have a set number, um, of roles here. We can say that X is about no meal distribution with N B and three, and the probability of a match is 16 D list. The vase that why may take on, then construct one pdf table that includes both X and why? And they're probabilities. Well, in this case, why is it being the profit per game? So why you could end up being a loss of $1 $1 $2 for $3 for the pdf table, I'm going to go to the side and created table in which the exes wise and the probabilities are all displayed. So form A pdf. I'm gonna write out my ex values, which was zero one, two and three. I'm also gonna write up my wife values, which were negative one one, 23 because both my ex and my wife values had the same probabilities instead of writing p of X appeal. Why? I'm just simply gonna right crop, as in probability, indicating that's the same probability for either the X or the why and the probabilities for each of these events. The probability, the X zero or that why is negative one ends of Ian 0.5 seven. So and I got this value because we do have a binomial distribution as we figured out in court. See, because we have a binomial distribution calculated this value using the information, find me a pdf in which I had a total of three rolls, the probability of 16 and upload in zero as my ex by you, this gave me 0.5787 I used the same approach with 12 and three to fill in the rest of my table. Because I do this, I get 0.3472 get 0.6 94 0.46 This creates a PdF table for both maxes and wives with their probabilities. E calculate the average expected matches over the long run of playing this game for the player because it's asking about the number of matches. I'm looking at the X values here, so I'm gonna use the formula in Times P. There's three total rolls. Probability is 16 So I would expect 0.5 matches calculate the average expected earnings over the long run playing this game with a player. In order to do this, we're gonna figure out our average winnings for why, which is calculated by the summation of taking every single why value and multiplying by the probability of his wife values. In this case, we have negative one. Multiply about probability 0.5787 We're gonna add this to the next Y value, which is one in its probability. 10.34 72 plus next, wise to times of ability 0.694 then what are added to three times point 00 for six. All this gives you an expected earnings to be negative 0.8 Or you would expect to lose about eight cents of the long run per player. G determine who has the advantage, the player or the house. The house has the advantage big time in this case because the expected winnings for the player would be negative. So House expected winnings for the player negative. That indicates that the house is making some money

Christian 61. We're looking at both of it have to be to be a fair game. So really asking three times to six. Last. What would be are they are times for six. What? He pulls the rope. So, three, This would be one plus gonna reduce 46 to 2/3 X equal. Zero. I'm going to subtract the one for both sides. So I have 2/3 X equals negative one. You know, this was this was a negative negative one. This here is a negative. So this is it. And I'm gonna multiply both sides by negative three halves so that I get excited by itself. So X equals positive three halves, which is the same as 1.5 part B. I have three times. I'm gonna reduced to 6 to 1/3 plus negative one times to third and again, I reduced for 6 to 2/3. That's one blasts, which is the same as one plus negative to third, which equals 1/3

So we have first part that a knot is equal to zero and a five is equal to one because Alan can't win without money. And here's one if he has all the $5. So this is party for part B. By the law of total probability, we have that. Ate, too, is equal to the property that the first flip is H times it probably that Alan wins with $3 plus a property. That first flip is tails times to probably that Alan wins with $1. So it's this 0.5 a three plus 0.5 a one, and the same logic yields as four equations for part C. So we have a one is equal to 10.582 plus 0.5 a knot, and a two is equal 2.5, 83 plus 0.5 a one a 30.584 plus 0.582 and a 40.5 a five plus 50.583 So then, if we use initial conditions from Part A, which are a not equals zero and a five equals one, then we can solve this system and we end up. Getting that anyone is 1/5 into his 2/5 83 is 3/5 and a four is forfeits. So we follow the pattern in part, see, which we just worked up. Then in party, we have that the chance that Alan beats Beth with their initial fortunes being $8 and B dollars. So the chances that Allenby it's Beth is going to be given by a over a plus B or equivalently one minus B over a plus B.

Problem 101. We have a board game between a player and the house. A player versus the house. The board of the sport game has sex objects. For example, From 1 to 6, 1 2 3456, the player bets on one object. The player puts one. Yeah, for one object, for example, the player select this three as his pit or her. But then the house or the player rolls three fair dies. One boys contains the same objects for the board. Rwanda is for example, here is from 1- six. Then the rules are if one, there is no dice that matches the selected object. Mm The player gets zero profit and if one dies here is zero. If only one day's matches the selected object, the player goods his bit plus one profit. And if only two matches, the player gets one Plus two as a prophet. And finally, if the three days matches the object, the player gets one plus three as a prophet. Let's continue to see what is the questions of this problem for party. We want to define the random variable X. X. Is the number of matches between the dice and the selected object object over the border for about three. We want to list the values that X may take on. We can see that X can make 012 or three. Then X may take one 0, 1, 2 and three. Well, but see we want to give the distribution of X. X follows a binomial distribution with a success rate, one divided by six. Because the probability for one device To match the object is one divided by six because it's a fair dice. And we have six objects on it. Then it's one body by six with three trials. Sorry. These three with three trials. Because the player or the house roles. Three paradise. Then the success rate is one divided by sex. And we have three tribes. This is the distribution for the random variable X. We want to list the values that why make taken knowing that. Why represents the proof? Let's see here. What is the profit for each value of X. If we define random for a boy that has the following values. Here is the income, income is zero. Here is the income is two. There is the income is three, is the income is four. But we should notice that the player puts one bit before rolling the dice. Then there is a cost and the prophet is the income minus the cost. This means we have here here zero minus one And here we have 2 -1, 3 -1 for -1. Then for barney the values for Y is minus one, one, two and three for birth. But before we switched to party we want to construct one pdf table that includes X and Y. Let's make this table let's make this table in the next bitch. We have six. Made a con 0123. Why May take on -11 two three. And the probability for X. Or for white we can get it using the binomial distribution formula. The probability X equals and holly of X. Oy. Or the probability of I equals N. I don't employ boy. Success probability which is B. It was about I was employed by one minus beep to the board of n minus or for example the probability For x equals three equals 10 33 deployed by the success rate which is one divided by sex with about of three. The blind by one minus P which is five divided by six is a bar of 3 -3 is you? This gives The probability of 4.046. And applying In the same formula with all equals two. We get 4.694 and Applying by I equals one We get 4.34 72 And putting our equal zero we get 4.5 787 for birth. We want to calculate the average expected matches which means you want to get the expected value for the random variable X. We can get bye but deploying X submission of X. I are deployed by its probability equals zero multiplied by the first probability Which gives zero then 1 Depoted by 4.3472 Plus two. multiplied by 4.0694 Plus three. multiplied by 0.046. These gifts almost oh boy and foot. Then the expected value for the number of matches is half. Let's move to board. Yeah. We want to calculate the average expected earnings or we want to get the expected value of Boy which equals submission of why I multiplied by its superb ability then equals -1. multiplied by the first probability here. 4.5787 plus one. multiplied by the second probability plus to multiply it Boy a certain probability Plus three multiplied by 0.46. This gives there is seven. This gives -4.7. 4.0787. This means there is no expected profit. It's a loss which answered the board. G who has the advantage? The player with the house. This means the house has the advantage because this is a negative. Which means the player blues by repeating this game along and along.


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