coin brainteaser..please help

[dblock21]

We each flip three fair coins. I offer to pay you $1 if we do not get the same amount of heads, if you agree to pay me $2 if we do (get the same amount of heads). Will you agree to play this game?

I haven't had prob. in a couple years and can not get my head around this question..any help would be great

 

[arabianights]

well... probability of each getting the same number is 5/16 so I expect to lose $1/16 each time I play so no...

Not to mention it's a really **** game so I would say no anyway

 

[DashRiprock]

I have E(X) = (11/16 * $1) - (5/16 * $2) = +$1/16

so not as bad a game as before but still not great.

 

[mb325]

Seems like a **** version of this game:

The pot starts at £1.
A fair coin is tossed.
For every head, the pot is doubled.
The game ends when a tail appears - you keep everything in the pot.

How much would you be willing to pay to play this game?

 

[jassaal123]

Thanks for given these info to all members dear friends,

 

[mb325]

I make it 5/16 chance to get the same amount of heads, so 11/16 a different amount.

E(X) = 1(11/16) - 2(5/16) = +$1/16 expectancy.

so not as bad a game as before but still not great.

That depends how many times you play it

 

[arabianights]

Seems like a **** version of this game:

The pot starts at £1.
A fair coin is tossed.
For every head, the pot is doubled.
The game ends when a tail appears - you keep everything in the pot.

How much would you be willing to pay to play this game?

That's st petersburg paradox no?

 

[wackypete2]

REAL brain teaser...
How much head do I get for my $?

Peter

 

[6am]

How much money do you have to lose? Would you allow Kelly betting?

 

[barjon]

well... probability of each getting the same number is 5/16 so I expect to lose $1/16 each time I play so no...

Not to mention it's a really **** game so I would say no anyway

mmm, I begin to see why I've always lost money playing these things in the pub How come probability of the same is 5/16 - why not 4/16?

mug punter, me

jon

 

[arabianights]

cause there are four ways of each having the same no of heads... you can have no heads each, one head each, two heads each or three heads each

for each of these you need to add up the probabilities... i.e. p(all the same) = p(me no heads) X p (you no heads) + p (me 1 head) x p (you 1 head) +...

which is 1/8 * 1/8 + 3/8 * 3/8... = 20/64 = 5/16

or use binomial distribution but don't need that...

 

[DashRiprock]

P(both 0 heads) = P(0h) * P(0h) = 1/8^2 = 1/64
P(both 1 head) = P(1h) * P(1h) = 3/8 * 3/8 = 9/64
P(both 2 heads) = P(2h) * P(2h) = 3/8 * 3/8 = 9/64
P(both all heads) = P(3h) * P(3h) = 1/8^2 = 1/64

SUM = 5/16

 

[barjon]

thanks, both

 

[shortsell]

if it's 5/16 why wouldn't you play arabian? +$1/16 expectancy no?

 

[arabianights]

cause hungover and thought 16-5 was 9

I'm still not playiung such a **** game!

 

[Hotch]

cause hungover and thought 16-5 was 9

nice

 

[crude_lover]

can't i just go down the casino instead ?