A mathematician walks into a bar, sits down, and...

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Jason_C

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...challenges the drunk next to him to a mental game.

He says, "I've got this penny here - a legitimate, properly printed, balanced, U.S. coin with one side heads and one side tails - and I'll wager you one cent that if I flip it, it will land on heads". The drunk accepts the wager. The mathematician flips, and it is indeed heads, so the drunk pays a penny. "Now," says the mathematician, "I'll wager you two cents I can flip heads again. After that, I will double my wager each time that I shall continue to get heads". The drunk, not understanding the potential risks, agrees to continue this game indefinitely. Now let us suppose the coin is flipped a total of 99 times, and each time, the mathematician wins - that is, he get 99 heads in a row.

Finally, the mathematician proposes they make it an even 100. Again, the wager is of course doubled.

Now for some math questions.

1) What are the odds of all 100 flips in a row being heads?
2) If the 100th flip is in fact NOT heads, how much money with the drunk have made?

A man is on a game show. The host tells him he may pick any one of three doors. Behind two doors are goats, and behind the other door is a sports car. Suppose the man rather arbitrarily selects door number 2. The host now opens door number 1, revealing a goat. At this point, the host gives him an option. The man must now choose which door shall hold his prize. He may stick with his choice of door number 2, or he may change it to either door number 1 or 3. However, whatever he decides, his decision must be final - there shall not be another opportunity to see and change doors. Assuming the man wants the sports car, what should he do? Does it matter?

Explain your mathematical reasoning on all answers, please.
 
Jason_C said:
He says, "I've got this penny here - a legitimate, properly printed, balanced, U.S. coin with one side heads and one side tails - and I'll wager you one cent that if I flip it, it will land on heads". The drunk accepts the wager. The mathematician flips, and it is indeed heads, so the drunk pays a penny. "Now," says the mathematician, "I'll wager you two cents I can flip heads again. After that, I will double my wager each time that I shall continue to get heads". The drunk, not understanding the potential risks, agrees to continue this game indefinitely. Now let us suppose the coin is flipped a total of 99 times, and each time, the mathematician wins - that is, he get 99 heads in a row.

Finally, the mathematician proposes they make it an even 100. Again, the wager is of course doubled.

Now for some math questions.

1) What are the odds of all 100 flips in a row being heads?
2) If the 100th flip is in fact NOT heads, how much money with the drunk have made?
Click to expand...
1) Not very likely, that's for sure. 1267650600228229401496703205376 to 1 against, says my calculator. That's the reciprocal of 0.5^100.

2) The man gains 1 cent.

If he won the first flip, he would have gained 1 cent.
If he won the second flip, he would have gained 1 cent (2 cents - 1 cent that he already lost).
If he won the third flip, he would have gained 1 cent (4 cents - (1+2=3) cents that he already lost).
If he won the second flip, he would have gained 1 cent (8 cents - (1+2+4=7) cents that he already lost).
If he won the second flip, he would have gained 1 cent (16 cents - (1+2+4+8=15) cents that he already lost).
Et cetera.
Jason_C said:
A man is on a game show. The host tells him he may pick any one of three doors. Behind two doors are goats, and behind the other door is a sports car. Suppose the man rather arbitrarily selects door number 2. The host now opens door number 1, revealing a goat. At this point, the host gives him an option. The man must now choose which door shall hold his prize. He may stick with his choice of door number 2, or he may change it to either door number 1 or 3. However, whatever he decides, his decision must be final - there shall not be another opportunity to see and change doors. Assuming the man wants the sports car, what should he do? Does it matter?

Explain your mathematical reasoning on all answers, please.
Click to expand...
He should choose number 3.

There are 2 boxes left, and 1 of those has the prize. There is a 50% chance of picking the prize box.

However, when the man picked his box, there was only a 33% chance of picking the prize box. Even though there are now only 2 boxes, the probability that he initially chose the correct box is 33%.

Compare this with 1000 boxes, with one of them being a prize box. Pretty slim odds, yes? 0.1% chance of picking the correct one. Now if 998 of the boxes were eliminated, leaving only your box and the prize box, you'd want to change your mind, wouldn't you?
 
You were correct on all but the first one.

The answer to problem number 1 is "50%" or "1/2". That is, 99 of the flips have already been established as heads. At this point we ask what the odds are that all 100 are heads. The odds are, of course, 1 in 2. I know, that one was kind of dirty. :p

9800000.00 points donated to Maruno successfully!
 
Jason_C said:
You were correct on all but the first one.

The answer to problem number 1 is "50%" or "1/2". That is, 99 of the flips have already been established as heads. At this point we ask what the odds are that all 100 are heads. The odds are, of course, 1 in 2. I know, that one was kind of dirty. :p
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Especially since the question asks the probability of all 100 flips in a row being Heads, making it as such a different question to the one you meant to ask.
 
Maruno said:
Especially since the question asks the probability of all 100 flips in a row being Heads, making it as such a different question to the one you meant to ask.
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Hence why I gave you the points anyway.

I should try to come up with more ambiguous wording for that. I need a more effective way to get people on the wrong track without actually stating the wrong thing.

Suggestions?
 
A woman has a simple bracelet with seven golden links. She needs to spend some time in a hotel and the clerk has agreed to accept each link as payment for one night's stay. She isn't sure how long she'll be staying, so she only wants to pay one night at a time. Therefore, she must cut her bracelet into pieces. In the end, it turns out she decides to stay all seven nights. That is, she payed her entire bracelet. BUT she still only payed one night at a time.

What is the minimum number of cuts she would have made in the bracelet to pay one night at a time for seven nights?
 
These are fun.
Jason_C said:
A woman has a simple bracelet with seven golden links. She needs to spend some time in a hotel and the clerk has agreed to accept each link as payment for one night's stay. She isn't sure how long she'll be staying, so she only wants to pay one night at a time. Therefore, she must cut her braces into pieces. In the end, it turns out she decides to stay all seven nights. That is, she payed her entire bracelet. BUT she still only payed one night at a time.

What is the minimum number of cuts she would have made in the bracelet to pay one night at a time for seven nights?
Click to expand...
Umm.. *counts* Four.

- cut
O
- cut
O
- cut
O
O
- cut
O
O
O
back to top

She divided the bracelet into a 1-link, a 1-link, a 2-link and a 3-link. With these she could make different combinations for each number of days up to 7. She merely swapped groups of links each night when she paid.

1: 1-link
2: 2-link
3: 2-link + 1-link
4: 2-link + 1-link + 1-link
5: 3-link + 2-link
6: 3-link + 2-link + 1-link
7: 3-link + 2-link + 1-link + 1-link
 
Ah! Does she overpay? That is, does she give the entire bracelet at the start, and then would make 2 cuts to subtract any links she doesn't owe when she checks out?

Having stayed for 7 nights, thus owing the entire bracelet, she makes zero cuts.
 
Maruno said:
Ah! Does she overpay? That is, does she give the entire bracelet at the start, and then would make 2 cuts to subtract any links she doesn't owe when she checks out?

Having stayed for 7 nights, thus owing the entire bracelet, she makes zero cuts.
Click to expand...
She pays one link per night. One night at a time.
 
One cut, but down the middle of the entire chain, thus leaving her with lots of halves of links, all separate.

What kind of bracelet is it? I'm picturing a chain with 7 hoops connected together.

Ah again! Six of the links are attacked only to the seventh loop, and the seventh loop is worn as the bracelet itself with the other 6 dangling free. Therefore only 1 cut is made, in the seventh loop to let the other 6 free.
 
I'm picturing a chain with 7 hoops connected together.
Click to expand...
It is a chain with 7 links together, but what you're forgetting is that it's a BRACELET. Bracelets have to come off, and this a gold bracelet, not an elastic one. Gold bracelets have latches that allow you to take them off. So once it's removed, picture it not as a loop but simply as a chain.

OOOOOOO

Only two cuts are needed now.

O | OO | OOOO

Night 1: She pays the 1-link.
Night 2: She pays the 2-link and gets the 1-link back.
Night 3: She pays the 1-link again.
Night 4: She pays the 4-link and gets the 1-link and 2-link back.
Night 5: She pays the 1-link again.
Night 6: She pays the 2-link again and gets the 1-link back.
Night 7: She pays the 1-link again.
 
EmeraldDragon said:
You gave us the answer? Awww. You robbed us of our potential reward :p
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Maruno had already figured out how to pay the links and how many breaks in the bracelet were necessary. He was just forgetting one break already existed so there wouldn't have to be a cut there. As far as I'm concerned, he gets the reward.
 
Yah. For some reason I was thinking of a loose bracelet that slid off without any undoing. Such bracelets exist.

That's nearly 20 million points in a few hours. Not bad. Thanks!
 
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