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This cache is a member of The Ten Pieces Of Pie Series
What are the odds? Ask any professional gambler and he will tell you that knowing the odds makes a crucial difference between making a good living or being broke. But calculating the actual probabilities is not always easy and often counter intuitive. For instance imagine you are on a game show and they show you three doors. Behind one of the doors is a large cash prize and there is nothing behind the other two. So let’s say you pick door number 1 but instead of showing you door number one the host opens one of the other doors revealing it to be empty. Now he says you can either keep your original choice or you can switch and take the other door. Is it best to keep your door, change to the other door, or doesn’t it make any difference. Most people will see that now they have a choice between two doors. One has the cash and on does not so the odds are 50% of winning no matter which door you choose and since we are told that you should not second guess yourself most probably keep their door. However the actual odds are that you have twice as much chance of winning the cash by switching than keeping your original door. Not only does the math show this but running hundreds of thousands of trial with computers has shown that if you keep your door you will win on average one third of the time but if you switch doors you will win two thirds of the time. It has to do with that when you chose your door it had one chance to win and the other two doors had two chances to win between them. When they show you an empty door nothing has changed. The remaining door still represent the two times your original door is empty and actually has two chances to win.
Here is one you can make money on. Take a regular deck of cards with no jokers (52 cards) and make sure it is well shuffled. Then ask someone “What do you think the odds are that if you pick two cards not counting suite like a ten and a five or a queen and a seven that one of those cards will be in the top three cards or if not they will be back to back somewhere in the deck?” Almost always they say “Not very likely” and offer me odds of anywhere from three to one up to ten or more to one against it happening. Then I say “Well if we play for low stakes I will give it a try at two to one odds. I will pay you a dime every time it does not happen and you pay me twenty cents every time it does happen”. They almost always agree to that. Not only am I collecting double every time I win but the actual odds are in my favor. On average I will win seven out of ten times. Approximately three times in ten one of the cards will be in the top three and approximately four times in ten they will be back to back somewhere in the deck. It is always good if they happen to win the first couple of times so that when you start winning you can say “Wow I must be on a hot streak.”
Here is one more example. If you are in a room with 22 other people what are the odds that at least two of you will have the same birthday (not counting year)? Again most people will say this is unlikely. If you were looking for someone in the room that has the same birthday as you it would indeed be very unlikely but the odds of at least two people in a group of 23 people having the same birthday is 50% and thus will happen half of the time. Any more than 23 people and it will happen more often than not and by the time you have 75 people in the room the odds are up to 99% that at least two of them share a birthday.
You can check your solution below but the coordinates must be entered in Degrees and Minutes in the normal format like the posted coordinates. I will add a hint after there is a First To Find.
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P1= The probability in decimal form of drawing four yellow marbles in the first four draws from a bag originally containing seven yellow marbles and four blue marbles.
P2= The probability in decimal form of getting at least one five by rolling five dice once.
N= 425.8687714 multiplied by P1
W= 155.8378000 multiplied by P2
I am giving this a difficulty of 2.5 because based on your background solving it could be anywhere from very easy to very hard.
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