coin brainteaser..please help

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dblock21

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

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

DashRiprock said:
so not as bad a game as before but still not great.
Click to expand...

That depends how many times you play it :p
 
mb325 said:
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?
Click to expand...

That's st petersburg paradox no?
 
How much money do you have to lose? Would you allow Kelly betting?
 
arabianights said:
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
Click to expand...

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
 
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...
 
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
 
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