#### Debates Forum

1. 18 Jul '11 00:41
Isn't it always better to flip more than H coins, to maximize the odds of getting exactly H heads?

Say a casino gives a bettor a chance to bet a small amount to win a bigger prize. He will get to flip a number of coins and if exactly H coins come up heads, he wins. He has to choose in advance how many coins (N) to flip, and then he has to flip all N of them (but as soon as he gets to H+1 heads, the gamble is over and he loses.)

Let H = 5 in this example. Obviously he'll choose to flip at least 5, but is 5 also the most he should choose?

Without doing any math, what N would you choose? Do you think people will tend to underestimate N, and thus tilt the odds away from what the casino knows is best? What does the math say N should be, for H = 5?

Once upon a time I knew the formulas to use for this sort of thing.
2.  FMF
Main Poster
18 Jul '11 01:00
[If I may just say something off-topic: I was in a meeting the other day with 3 people of varying ages whom I had not met before and who knew each other only as work colleagues. Not sure exactly how it came up but it turned out that our four birth days were 24th June, 25th June, 26th June and 27th June, all in different years. Carry on, as you were.]
3.  AThousandYoung
West Coast Represent
18 Jul '11 01:01 / 2 edits
He should choose 9 or 10. 2N and 2N-1.

It will always be a bell curve with most probability at the halfway point; thus 2N coins gives most likely N
4. 18 Jul '11 01:03
Originally posted by JS357
Isn't it always better to flip more than H coins, to maximize the odds of getting exactly H heads?

Say a casino gives a bettor a chance to bet a small amount to win a bigger prize. He will get to flip a number of coins and if exactly H coins come up heads, he wins. He has to choose in advance how many coins (N) to flip, and then he has to flip all N of the ...[text shortened]... N should be, for H = 5?

Once upon a time I knew the formulas to use for this sort of thing.
If I understand your question, i would want to flip the coin as few times as possible and here is why:

If you flip a coin zero times the odds of getting zero head are 1. That's the most so that's what i would pick. But I don't think you are looking for that answer so if you flip a coin once. Half of the time you get 1 head, half of the time you get 0 heads. As you flip more the distribution spreads out. If you flip a coin twice. 1/4 of the time you get no heads 1/2 you get 1, 1/4 you get 2. If you flip 3 times. 1/8 you get no heads 3/8 you get 1 head, 3/8 you get 2 heads 1/8 you get three heads.
5.  Soothfast
0,1,1,2,3,5,8,13,21,
18 Jul '11 02:03 / 1 edit
Originally posted by JS357
Isn't it always better to flip more than H coins, to maximize the odds of getting exactly H heads?

Say a casino gives a bettor a chance to bet a small amount to win a bigger prize. He will get to flip a number of coins and if exactly H coins come up heads, he wins. He has to choose in advance how many coins (N) to flip, and then he has to flip all N of the N should be, for H = 5?

Once upon a time I knew the formulas to use for this sort of thing.
If you get to choose the number of coins to be tossed and the objective is to get 5 heads, then toss 9 or 10 coins (as ATY says). Probability of 5 heads: ~0.246 (i.e. 24.6% chance).

It's a binomial distribution type thingy.

If you only toss 5 coins you need 5 heads out of 5 tosses, which is far less likely. Probability is (1/2)^5 = 1/32 = 0.03125 (i.e. 3.125% chance)
6.  Wajoma
Die Cheeseburger
20 Jul '11 10:15
Originally posted by FMF
[If I may just say something off-topic: I was in a meeting the other day with 3 people of varying ages whom I had not met before and who knew each other only as work colleagues. Not sure exactly how it came up but it turned out that our four birth days were 24th June, 25th June, 26th June and 27th June, all in different years. Carry on, as you were.]
This is something like the poser that asks how many people in a room are needed before there's a better than 50/50 chance that two have the same birthday. And no, it is nowhere near 365.
7.  Soothfast
0,1,1,2,3,5,8,13,21,
20 Jul '11 23:29
Originally posted by Wajoma
This is something like the poser that asks how many people in a room are needed before there's a better than 50/50 chance that two have the same birthday. And no, it is nowhere near 365.
I think it's only twenty-something individuals, though I don't remember for sure.
8. 20 Jul '11 23:38
Originally posted by JS357
Isn't it always better to flip more than H coins, to maximize the odds of getting exactly H heads?

Say a casino gives a bettor a chance to bet a small amount to win a bigger prize. He will get to flip a number of coins and if exactly H coins come up heads, he wins. He has to choose in advance how many coins (N) to flip, and then he has to flip all N of the ...[text shortened]... N should be, for H = 5?

Once upon a time I knew the formulas to use for this sort of thing.
Sorry, posted to wrong forum.
9.  FMF
Main Poster
21 Jul '11 00:08
Originally posted by JS357
Sorry, posted to wrong forum.
Yeah. I was wondering what you wanted us to fight and insult each other about?
10.  Soothfast
0,1,1,2,3,5,8,13,21,
21 Jul '11 02:10
Originally posted by FMF
Yeah. I was wondering what you wanted us to fight and insult each other about?
Who owns the coin? The proletariat or the bourgeoisie?
11. 27 Jul '11 17:32
Originally posted by JS357
Isn't it always better to flip more than H coins, to maximize the odds of getting exactly H heads?

Say a casino gives a bettor a chance to bet a small amount to win a bigger prize. He will get to flip a number of coins and if exactly H coins come up heads, he wins. He has to choose in advance how many coins (N) to flip, and then he has to flip all N of the ...[text shortened]... N should be, for H = 5?

Once upon a time I knew the formulas to use for this sort of thing.
What year/denomination is the coin?
12.  Palynka
Upward Spiral
27 Jul '11 18:11
Prob of N tosses getting H heads = 0.5^H*0.5^(N-H) * N!/((N-H)!*H!)