[Guest]

Amir and Nabil are at a bar, bored. So they decide to do a little gambling.

Amir proposes:

"You put up $100 and I'll put up $105. Then you can pick either Heads or Tails and I'll flip a coin 100 times.

If your choice appears more than the other, you win the whole lot, otherwise, I do."

Should Nabil take the bet?

[Guest]

Amir and Nabil are at a bar, bored. So they decide to do a little gambling.

Amir proposes:

"You put up $100 and I'll put up $105. Then you can pick either Heads or Tails and I'll flip a coin 100 times.

If your choice appears more than the other, you win the whole lot, otherwise, I do."

Should Nabil take the bet?

Not take the bet.

Since there are 100 flips (even number), 3 things could happen:

1) more heads than tails

2) more tails than heads

3) same number of each

1 and 2 are equally likely, but Amir has the edge because he also wins in case 3

[Guest]

Not take the bet.

Since there are 100 flips (even number), 3 things could happen:

1) more heads than tails

2) more tails than heads

3) same number of each

1 and 2 are equally likely, but Amir has the edge because he also wins in case 3

There are 101 things that could happen:

0:100

1:99

2:98

3:97

..

50:50

..

97:3

98:2

99:1

100:0

And they are all equally likely. Meaning you have a 50/101 (49.5%) chance of winning. $5 is more than 0.5% of $200 (2.5%) so it's a good gamble in the long run.

[Guest]

Actually, I'd question whether they are all 'equally likely'. If gave you the top 26 (0:100 - 25:75) and the bottom 25 (75:25 - 100:0) and I kept the middle 50 (26:74 - 74:26) and put up $105 to your $100 - would you take the bet?

There are 101 things that could happen:

0:100

1:99

2:98

3:97

..

50:50

..

97:3

98:2

99:1

100:0

And they are all equally likely. Meaning you have a 50/101 (49.5%) chance of winning. $5 is more than 0.5% of $200 (2.5%) so it's a good gamble in the long run.

[Guest]

Actually, I'd question whether they are all 'equally likely'. If gave you the top 26 (0:100 - 25:75) and the bottom 25 (75:25 - 100:0) and I kept the middle 50 (26:74 - 74:26) and put up $105 to your $100 - would you take the bet?

Yes.

Flipping exactly 100 heads out of 100 flips is the same chance as flipping exactly 46 heads out of 100 flips is the same chance as flipping exactly 92 heads out of 100 flips.

Edited January 27, 2009 by Llam4

[Guest]

Here's a link about tossing the coin 7 times and you will see that the middle ones will occur more often than the outer ones.

http://answers.google.com/answers/threadview/id/526284.html

Heads (or tails) all 7 times occurs 1 in 128 times (not one in 7 times).

Here's a little bit from the page:

7H: 1 way

6H and 1T: 7 ways

5H and 2T: 21 ways

4H and 3T: 35 ways

3H and 4T: 35 ways

2H and 5T: 21 ways

1H and 6T: 7 ways

7T: 1 way

Total: 128 ways

Yes.

Flipping exactly 100 heads out of 100 flips is the same chance as flipping exactly 46 heads out of 100 flips is the same chance as flipping exactly 92 heads out of 100 flips.

[bonanova]

There are 2¹⁰⁰ equally likely outcomes for flipping a coin 100 times.

Of these, 1 outcome has no heads and 1 outcome has no tails.

And 100 outcomes have 1 head, and another 100 outcomes have one tail.

And so on.

Not every enumeration of outcomes is a set of equally likely ones.

The fraction of outcomes with 50 heads and 50 tails is quite large.

It swamps the 5% edge.

[Guest]

Amir and Nabil are at a bar, bored. So they decide to do a little gambling.

Amir proposes:

"You put up $100 and I'll put up $105. Then you can pick either Heads or Tails and I'll flip a coin 100 times.

If your choice appears more than the other, you win the whole lot, otherwise, I do."

Should Nabil take the bet?

If this is strictly a math problem, then 205/100=2.05 and 49/101=0.485, multiplied together is 0.995... not advantageous, but a fair bet among friends . Incidentally, I've seen evidence that a skilled person can substantially predict the outcome of his spinning a coin to land on heads or tails, but have read that manipulating a coin flip to cheat is much more difficult.

[Guest]

Ok here goes

The total number of possibilities is 2^100 ~= 1.27 E30 (let's call this quantity A)

The number of those resulting in 50H and 50T is C(50,50) ~= 1.009 E29 (let's call this quantity B)

Therefore, we have (A-B) / 2 chances of winning and (A+B) / 2 chances of losing.

(A-B)/2 ~= 5.83 E29 (probability ~ 46.0%)

(A+B)/2 ~= 6.84 E29 (probability ~ 54.0%)

So, for 100 bucks, you have an expected return of .46*205 (+ .54*0) , that is ~ 94.3 bucks.

Don't take the bet.

note that for it to be fair, our friend would have to put a bit less than 118 dollars on the table

[Guest]

Congrat's to hellmake for figuring out the formula for the probability of equal flips: For 2k flips, the odds of k heads and k tails are p(k,k) = Combination(2k,k)/2^(2k)

Thus, Nabil's EV = p(success)*(total $) = ((1-Combination(100,50)/2^(100))/2)*$205 = ~0.46*$205 = $94.30

As $94.30 < $100, he shouldn't take that bet.

But what if the coin was weighted (say for any flip: 80% Heads, 20% Tails). For 100 flips, what would be a fair bet for Tails > Heads? Heads > Tails?

Edited January 28, 2009 by vinays84

[Guest]

But what if the coin was weighted (say for any flip: 80% Heads, 20% Tails). For 100 flips, what would be a fair bet for Tails > Heads? Heads > Tails?

Ok, my take on this:

Love your puzzles btw...

If the coin lands Heads with a probability q (in this case q=0.8), the probability of having n Heads out of 100 flips is:

P(n) = (q^n) * (1-q)^(100-n) * C(n,100)

Now, the sum of P(n) for n = 0:50 (included) with q=0.8 is S(50) = 2.14 E-11.

Our expected return (hoping I didn't mess the numbers up) on 100 bucks is given by (1-S(50))*(100+X) where X is the opponent's wager.

For the bet to be fair, that value should be equal to 100, and therefore we get X = 100/(1-S(50)) - 100 = 2.14 E -9 (this means I'd play even if his wager was a cent).

Conversely, he'd have to put someting like 1.9E13 dollars on the table for me to accept the bet with tails.

[Guest]

Love your puzzles btw...

Thanks!

If the coin lands Heads with a probability q (in this case q=0.8), the probability of having n Heads out of 100 flips is:

P(n) = (q^n) * (1-q)^(100-n) * C(n,100)

Now, the sum of P(n) for n = 0:50 (included) with q=0.8 is S(50) = 2.14 E-11.

Our expected return (hoping I didn't mess the numbers up) on 100 bucks is given by (1-S(50))*(100+X) where X is the opponent's wager.

For the bet to be fair, that value should be equal to 100, and therefore we get X = 100/(1-S(50)) - 100 = 2.14 E -9 (this means I'd play even if his wager was a cent).

Conversely, he'd have to put someting like 1.9E13 dollars on the table for me to accept the bet with tails.

Sounds about right. Curious, though, did you calculate S(50) using a speadsheet (as I did) or can a formula be created for that as well (and by this I mean a directly calculatable [is this a word?] formula..w/o the summation)?

[Guest]

Thanks!

Sounds about right. Curious, though, did you calculate S(50) using a speadsheet (as I did) or can a formula be created for that as well (and by this I mean a directly calculatable [is this a word?] formula..w/o the summation)?

Hellmake has already given the formula as

Spoiler for formula:

P(n) = (q^n) * (1-q)^(100-n) * C(n,100)

To calculate this easily is provided by google.

Really, if you write in google search editbox as 0.2^50 google gives you 1.12 x 10^-35

And for c(100,50) we know that it's equal to 100!/50!x50!

if you write 100! to editbox, google gives 9.3 x 10¹⁵⁷

[Guest]

What he meant was "can we get a formula for S(50) = \sum{n=0->50) of P(n)" without using the summation notation. I've thought a bit about this, and I don't think there is any "simpler expression"... and yea i did use a spreadsheet

[Guest]

Hmm, now I'm curious as to if and how to find that formula w/o summation.

A direct example of its use would be using it to answer the question "What weight (probability) on heads would the coin need to be so that Amir's original proposal (100 flips, see if #tails > #heads) would be an even bet?"

I actually can find that answer by manually trying out multiple probabilities, using a method similar to the bisection method of numerical analysis, to find the probability. It came out to 0.4950167 (this is as far as excel will calculate, as it shows the probabilities with this weight to be exactly 0.5 to 0.5 (even though it probably isn't but just really close)).

[Guest]

diffidently take the bet! he should also choose tails, because there is a grater chance on the coin landing on tails. than, and if hes in a bar, he's probable drunk, so he wont have the concentration for a bunch of fancy math, just says ok

[Guest]

are pure math problems considered brain teasers? i don't understand...

[bonanova]

are pure math problems considered brain teasers? i don't understand...

They can be.

Here are some ways:

1. The question looks like it has a simple answer, but it's the wrong answer.
   
2. The question looks like it's not possible to solve, but with a little thought it's simple.
   

In other words, just doing a calculation is not a puzzle. Some thought has to be involved.

Often, questions dealing with probability become interesting puzzles.

That's because intuition about probability is often misleading.