# Flipping Coins

• Jun 20th 2006, 04:08 PM
Quick
Flipping Coins
There is a gambling game that goes like this, you place a $5 bet to begin with. A coin is then flipped until it flips tails (or the house maximum of$\displaystyle x$flips). Each time it lands on heads the house pays an amount equal to$1 times the number of consecutive flips, for example, land three heads in a row gets you three dollars($3), which is then added to the money you got from landing two flips in a row($2), and the one beginning flip on heads ($1), in other words, 3 heads in a row will give you$6.

a)What would be the expected house winnings if 16 people played?
b)32 people?
c)64?

a)54
b)103
c)200

Is there one equation to model all the answers?
• Jul 6th 2006, 07:16 AM
CaptainBlack
Quote:

Originally Posted by Quick
There is a game that goes like this, you pay a $5 fee to begin with. A coin is then flipped until it flips tails (or the house maximum of$\displaystyle x$flips). Each time it lands on heads the house wll give you an amount equal to$1 times the number of consecutive flips, for example, land three heads in a row gets you three dollars($3), which is then added to the money you got from landing two flips in a row($2), and the one beginning flip on heads ($1), in other words, 3 heads in a row will give you$6.

a)What would be the expected house winnings if 16 people played?
b)32 people?
c)64?

a)54
b)103
c)200

Is there one equation to model all the answers?

Am I missing something here? Why are the expected house winnings
not just proportional to the number of players?

RonL
• Jul 18th 2006, 05:19 PM
Quick
Quote:

Originally Posted by CaptainBlack
Am I missing something here? Why are the expected house winnings
not just proportional to the number of players?

RonL

well, I didn't realize when I asked this question that the expected house winnings could be found by just multiplying by the odds, what I did to find those answers was just to say...

for 16 people:
8 will win the first flip because there is a 1/2 chance
4 will win the second flip because there is a 1/4 chance
2 will win the third flip because there is a 1/8 chance
1 will win the fourth flip because there is a 1/16 chance

and then I figured out how much money that would be, still this method seems somewhat more accurate...
• Jul 18th 2006, 09:11 PM
CaptainBlack
Quote:

Originally Posted by Quick
well, I didn't realize when I asked this question that the expected house winnings could be found by just multiplying by the odds, what I did to find those answers was just to say...

for 16 people:
8 will win the first flip because there is a 1/2 chance
4 will win the second flip because there is a 1/4 chance
2 will win the third flip because there is a 1/8 chance
1 will win the fourth flip because there is a 1/16 chance

and then I figured out how much money that would be, still this method seems somewhat more accurate...

The expected house winnings should be:

$\displaystyle E(\mbox{winnings with N players})=$$\displaystyle \mathrm{N} E(\mbox{winnings with 1 player})$

RonL
• Dec 21st 2006, 12:41 PM
F.A.P
Quote:

Originally Posted by Quick
There is a gambling game that goes like this, you place a $5 bet to begin............... Is there one equation to model all the answers? Let Y be the number of head flips before a tail, and W the house winnings with one player. Y then has a probability mass function$\displaystyle P(Y=k)=\left(\frac{1}{2}\right)^k\cdot\left(1-\frac{1}{2}\right)= \left(\frac{1}{2}\right)^{k+1} $Now the expected win for the house given Y = k flips is$\displaystyle E(W|Y=k)=5-\sum_{n=0}^{k}n$Using the law of total expectation we get an expected winning of$\displaystyle E(W)=\sum_{k=0}^{x}E(W|Y=k)P(Y=k)=\sum_{k=0}^{x}\l eft(5-\sum_{n=0}^{k}n\right)\left(\frac{1}{2}\right)^{k+ 1} $where x is the maximum allowed flips by the house. Now if many persons are playing, either one at a time or all at once, the expected house winnings is$\displaystyle NE(W)$or equivalently$\displaystyle E[W_1 + W_2 +...+W_N]$where Wi is the house winnings for person i. As pointed out by CaptainBlack. This can be verified by calculating the two quantities using the above approach. • Sep 4th 2007, 12:31 PM Obsidantion Quote: Originally Posted by Quick for 16 people: 8 will win the first flip because there is a 1/2 chance 4 will win the second flip because there is a 1/4 chance 2 will win the third flip because there is a 1/8 chance 1 will win the fourth flip because there is a 1/16 chance Well you're on the right track but it's still not as accurate as it could be, plus I think you must have miscalculated your results with this method because they're quite a bit off. CaptainBlack is right that the amount of winnings is directly proportional to the number of players because each player plays the same game which will leave the house with the same amount of expectable gain. So, the way I would do it would be to concentrate on the expected profit for the house in only one game: There's a 1/2 chance of the player loosing on the first flip, in which case the house would get$5. So, on average they would gain $(5*1/2) or$2.5 on the first flip. There's a 1/4 chance of the player loosing on the second flip, in which the house would get $4 and so they would gain an average of$(4*1/4) or $1 on the second flip ($(2.5+1) or $3.5 average gain in the first 2 flips). I would then continue this process for every flip, adding up the average gains, until the amounts become negligible (depending on your desired degree of accuracy). The figure I would be left with would be an accurate approximation of how much the house would gain on average for every game. I work this out to be$4.000015259 (precise to at least 8 decimal places). By multiplying this amount by the number of players in your questions, I receive the answers:
a. $64 approx b.$128 approx
c. \$256 approx

I think that F.A.P gave us a more accurate and certainly faster method but it's difficult to understand for those who aren't familiar with the notation and concepts he uses.