course Mth 152 I had a hard time doing any of the 11 assignments... I'm going back and restudying that section. ???????????assignment #010?€?~?????S??i?`??Liberal Arts Mathematics II
......!!!!!!!!...................................
09:05:37 Query 12.5.6 fair dice game pays $3 for 6, $2 for 5, $1 for 4. What is a fair price to pay for playing this game?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
09:27:03 Query 12.5.6 fair dice game pays $3 for 6, $2 for 5, $1 for 4. What is a fair price to pay for playing this game?
......!!!!!!!!...................................
RESPONSE --> As it is a single fair die with 6 sides and three payouts, there is a good chance (50% in fact) of winning something. Actually such a game is unlikely as it puts the ""house"" at even odds with the better and is thus unprofitable... however, lets look at the problem.... according to the text a 6 pays $3, a 5 pays $2, and 4 pays $1. If one was to have to roll the die all six times (in ref to the total faces on the die) in order to get the payout side for each (aka on the last roll), an even payout versus an even amount of betting value would break down like this.... (with the probability being 1/6 of getting a winning roll of a singular, particular number) $3 (a six) by 1/6= .50 cents $2(a five) by 1/6= .33 cents (roughly) $1(a four) by 1/6=.16 cents (roughly) as you can see, rolling a six has a much better payout than rolling a four, thus requiring a higher gambling bet to equal out... however, rolling a five or a four changes that drastically. Therefore, in order to find the best mean I shall add all of the winnings up and divide by 6 (taken from 1/6). first to add the winnings.... $3+$2+$1= $6. Now to divide this by 6 we would get an even $1. Therefore, though I may be wrong, I would have to say that an even $1 would be a mutually fair bet.
.................................................
......!!!!!!!!...................................
09:27:10 ** A 1 in 6 chance of getting $3 is worth 1/6 * $3 = $.50 . A 1 in 6 chance of getting $2 is worth 1/6 * $2 = $.33 1/3 . A 1 in 6 chance of getting $1 is worth 1/6 * $1 = $.16 2/3 . The total expectation is $1.00 * 1/6 + $2.00 * 1/6 + $3.00 * 1/6 = $1.00 So a fair price to pay is $1.00 **
......!!!!!!!!...................................
RESPONSE --> Okay, so I did have it right
.................................................
......!!!!!!!!...................................
09:37:37 Query 12.5.10 expectation Roulette $1 bet 18 red, 18 black one zero What is the expected net value of a bet on red?
......!!!!!!!!...................................
RESPONSE --> Okay, we have a wheel with 18 red, 18 black, and 1 zero slot. We are betting $1 and we want to find the net value of a red bet. First we must look at the average gain... the average gain is a pretty good chance, 18 in 37 (total of slots is 37)... which equals out to .486 (or .49 or .48 (cents), depending on whether you round up/off or round down) However, as you can see, the chances of winning are less than 50%, so if you keep betting and betting without a stop on your dollar the chances are that you will start losing and lose consistantly more. The average loss is a slightly greater chance, 19 to 37, which equals out to .514 (or 51 cents). Therefore we will subtract this from our win... .486-.514= -0.028 -0.028 is the net value
.................................................
......!!!!!!!!...................................
09:38:06 ** If your net gain is $1 for a win and -$1 for a loss the expected value is 18/37 * (+1) + 19/37 * (-1) = -$.027. **
......!!!!!!!!...................................
RESPONSE --> Okay, I was off by one fraction of a digit... it may have been a result of rounding off.
.................................................
......!!!!!!!!...................................
09:48:20 Query 12.5.20 exp sum of 2 of 5 cards 1-5. What is the expected sum of the numbers on the two cards drawn?
......!!!!!!!!...................................
RESPONSE --> I had some trouble with this one because I found it to be too dependent on the exact cards drawn... I could find an average expected sum... but a singular, exact sum I could not. Here's how I found an average.... the greatest possible sum would be the five card and the four card... or 9 the lowest would be the one card an the two card... or 3 add these together (12) and divide by 2 (for an even mean) and I got a 6. But I don't think this is right.
.................................................
......!!!!!!!!...................................
09:48:55 ** You can't get a sum of 1 on two cards. There is also no way to get a sum of two, since the lowest total possible is 1 + 2 = 3. There are 2 ways to get total 3. You can get 1 on the first and 2 on the second, or vice versa. There are 2 ways to get total 4. You can get 1 on the first and 3 on the second, or vice versa. There are 4 ways to get total 5. You can get 1 on the first and 4 on the second, or vice versa, or 2 on the first and 3 on the second, or vice versa. There are 4 ways to get total 6. You can get 1 on the first and 5 on the second, or vice versa, or 2 on the first and 4 on the second, or vice versa. There are 4 ways to get total 7. You can get 2 on the first and 5 on the second, or vice versa, or 4 on the first and 3 on the second, or vice versa. There are 2 ways to get total 8. You can get 3 on the first and 5 on the second, or vice versa. There are 2 ways to get total 9. You can get 4 on the first and 5 on the second, or vice versa. You can't get more than 9. There are 2+2+4+4+4+2+2 = 20 possibilities, so the probabilities are 2/20, 4/20, 5/20, etc.. The expected sum is therefore 2/20 * 3 + 2/20 * 4 + 4/20 * 5 + 4/20 * 6 + 4/20 * 7 + 2/20 * 8 + 2/20 * 9. This gives 120 / 20 = 6. **
......!!!!!!!!...................................
RESPONSE --> Ah, I see... though I did this completely different and did not go into the same detail. I see the mechanics of it now though.
.................................................
......!!!!!!!!...................................
09:49:11 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> I had a bit of trouble on that last one, but the others I felt fine with.
.................................................
??????????^??assignment #011 ?€?~?????S??i?`??Liberal Arts Mathematics II 07-04-2006
......!!!!!!!!...................................
09:59:43 **** Query 12.6.6 rnd # table to simulate 50 one-and-one foul shooting opportunities if 70% prob of success; 2 shots Give the results of your tally. How does your empirical probability compare with the theoretical probability?
......!!!!!!!!...................................
RESPONSE --> I found this interesting as I have done programming with random number generation and as a game designer (my side job outside the mental hospital) of paper and dice role playing games I deal alot with random principles. However, I'm not sure how to work that side chart (though it would be interesting to know) to get a percentage chance of hitting baskets....
.................................................
......!!!!!!!!...................................
10:00:11 ** In 1-and-1 shooting you only get a second shot if you make the first. So there are 3 possibilities: miss the first, don't get another shot make the first, get another shot and make it make the first, get another shot and miss it prob of 0 = prob of miss on first shot = .3 prob of 1 = prob of hit on first and miss on 2d = .3 * .7 = .21 prob of 2 = prob ot hit * prob of hit = .49. 'Hits' happen with 70% or .7 probability, misses with probability 30% or 3. The theoretical probability of 2 misses is probability of miss * probability of miss = .30 * .30 = .09. The theoretical probability of 2 miss and 1 hit is probability of miss * probability of hit + probability of miss *hit probability of miss = .30 * .70 + .70 * .30 = .21 + .21 = .42. The theoretical probability of 2 hits is probability of hit * probability of hit = .70 * .70 = .49. Note that these probabilities add up to .09 + .42 + .49 = 1, as they must since these three events cover all possibilities.
......!!!!!!!!...................................
RESPONSE --> Okay, I understand this much, but I'm still unsure as to how we use the chart.... I will try to work on that again before the test.
.................................................
......!!!!!!!!...................................
10:00:45 Query 12.6.12 rnd walk start N then right, left or straight with prob 1/2, 1/6, 1/3; 1 st 2 columns of table
......!!!!!!!!...................................
RESPONSE --> Oh goodness, that again.... I wish I could be allowed to use some of my dice for this but then you would need to see my work.
.................................................
......!!!!!!!!...................................
10:01:26 ** Your probabilities are given as 1/2, 1/6 and 1/3. These can all be expressed in terms of the common denominator 6: 1/2 = 3/6, 1/6 = 1/6, 1/3 = 2/6. So a move to the right has 3 chances out of 6, a move to the left has 1 chance out of 6 and a move straight has 2 chances out of 6.
......!!!!!!!!...................................
RESPONSE --> Okay, I see what you mean.... oh, I thought we had to use the chart... sorry... I can work something like this without it because it uses the basic principles of probability that we learned back in the first section.
.................................................
......!!!!!!!!...................................
10:02:38 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> Sorry about this... I guess I've ""goofed"" this one assignment up. I did find your explanations useful, however, and I am wondering if I could have skipped the chart to have done that last question. If I can figure the chart out I may go back and try to rework my problems in this section as there was very little that I could do.
.................................................
"