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Consider this gambling game. The player flips fair coin repeatedly until the player gets head, counting the number of flips required to get the head. If a head is a...

Question

Consider this gambling game. The player flips fair coin repeatedly until the player gets head, counting the number of flips required to get the head. If a head is achieved on the first flip, the player wins $2. If a head is achieved on the second flip (the Mlips went tails, heads), the player wins S4_ Ila head is achieved on the third flip (the flips went tails, tails, heads), the player wins $8_ The winnings continue the following this pallern. That is, if the first head is archived on the nth

Consider this gambling game. The player flips fair coin repeatedly until the player gets head, counting the number of flips required to get the head. If a head is achieved on the first flip, the player wins $2. If a head is achieved on the second flip (the Mlips went tails, heads), the player wins S4_ Ila head is achieved on the third flip (the flips went tails, tails, heads), the player wins $8_ The winnings continue the following this pallern. That is, if the first head is archived on the nth flip, then the player wins S2" What would you be willing t0 pay tO play this game? How many possible Outcomes are there t0 this game? Label the oulcomes 01,0z, 0= Whal are the values ol each outcome? We will now consider the probability of each outcome_ What is the probability that the firsi flip is a head? This IS the probability of the first oulcome Whal is the probability thal the player flips the coin (wice and gets tails first, then heads? This is the probability of the second oulcome Whal is the probability thal the player flips the coin three times and gets tails Gicsl; then tails again, then heads? This is the probability of the third outcome. Whal is the probability of each utcome? Whal the expecled value of this game? Now what would you be willing spend to play this game? What if the game is modified so that the first Outcome has value zero, now what is the expected value? What if the game is modified so that the firsl ten outcomes have value zero , now what is the expected value? Whal if the game is modified so that the first hundred oulcomes have value zero, now whal is the expecled value? What if the casino running this game has enough resources t0 make one thousand dollar payout, what is the actual expected value of the game? What if the casino running this game has enough resources to make one million dollar payout, what is the aclual expecled value of the game? What if the casino running this game has enough resources to make one billion dollar payout; what is the actual expected value of the gane?



Answers

Gambler's Ruin. Allan and Beth currently have $\$ 2$ and $\$ 3,$ respectively. A fair coin is tossed. If
the result of the toss is heads, Allan wins $\$ 1$ from Beth, whereas if the coin toss results in tails, then Beth wins $\$ 1$ from Allan. This process is then repeated, with a coin toss followed by the exchange of $\$ 1,$ until one of the two players goes broke (one of the two gamblers is ruined). We wish to determine $a_{2}=P($ Allan is the winner $|$ he starts with $\$ 2) .$ To do so, let's also consider
$a_{i}=P($ Allan wins $|$ he starts with $\$ i)$ for $i=0,1,3,4,$ and $5 .$
(a) What are the values of $a_{0}$ and $a_{5} ?$
(b) Use the Law of Total Probability to obtain an equation relating $a_{2}$ to $a_{1}$ and $a_{3} .$ [Hint:
Condition on the result of the first coin toss, realizing that if it is heads, then from that point Allan starts with $\$ 3 . ]$
(c) Using the logic described in (b), develop a system of equations relating $a_{i}(i=1,2,3,4)$ to $a_{i-1}$ and $a_{i+1} .$ Then solve these equations. [Hint: Write each equation so that $a_{i}-a_{i-1}$ is on the left hand side. Then use the result of the first equation to express each other $a_{i}-a_{i-1}$ as a function of $a_{1},$ and add together all four of these expressions $(i=2,3,4,5) . ]$
(d) Generalize the result to the situation in which Allan's initial fortune is $\$ a$ and Beth's is $\$ b$ .
[Note: The solution is a bit more complicated if $p=P($ Allan wins $\$ 1) \neq .5 .$ We'll explore Gambler's Ruin again in Chap. 6.1

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

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.

Remember that odds are calculated and expressed a little differently than probability. Probability is based on sample space. Odds are expressed as the odds for or against something on. So, for instance, rolling a die getting a too well. There's one outcome out of six to roll a die and get a two and five outcomes out of six that are not getting it, too. And so if we express that as a fraction of a fraction weaken, express it then as a ratio of one to five in favour and five to one against rolling a guy and getting an even number, that would be, uh, half of the outcomes 24 and six or even half the outcomes or not. And so we can ah, express that as one, 21 in favor and one to one against drawing a spade from the deck of cards. There are 1/4 of the cards in a deck or spades. That means 3/4 are not. And so we can express that as a ratio of one to three in favor and three to one against a red card. Half the cards air red so similarly to, uh, rolling a dining, getting an even number. If half the cards are red, it's gonna be 1 to 1 in favor, 1 to 1 against drawing a queen from the deck, a deck of cards. There are four queens out of 52 that's one out of 13 in favor and 12 out of 13 against. So we could express that as one 2 12 in favor 12 21 against and then tossing a coin and getting two tails. So if we have, we're tossing Teoh two coins getting two tails. We could get head, tail heads and tails. We could get tails and heads. We could get heads and heads, and we could get tales and tails. That's 1/4 of the outcomes are getting two tails. That means 3/4 or not. Therefore, we can express the ratio is 1 to 3 in favor, 3 to 1 against tossing two coins and getting exactly one tail. Well, there's two outcomes where you have one tail that's two out of the four, so that's 1/2 versus 1/2. So once again, that's 1 to 1 in favor, 1 to 1 against


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