Probability game?

Question

I am selling tickets to a game, wherein I toss a coin. If it's heads, I give u one rupee; tails, the game stops. The game can have infinite chances, provided heads keep coming. What will u be willing to pay for a ticket, or what should be the fair value of a ticket?
Have wracked my brains but just can't seem to get this one!

Answer 1

garypopkin's answer is correct, but as an addendum, this infinite series sums to 2. So the expected value of winnings from the game is $2, making this the fair ticket price.

Answer 2

The answer is 2 rupees

Fair Amount=(1/2)*(1)+(1/2)^2 *(2) + (1/2)^3 *(3) +........ (1/2)^infinity * (infinity)

I dont remember exactly how to arrive at 2 using binomial theorem

But you can neglect the the larger values of n as they will result in very small values.

Take n=9

Fair amount=(256+256+192+128+90+48
+28+16+9)/512
=(512+501)/512
which approximately equals 2

If you take larger values of n, the first part in the above numerator will be equal to the denominator and the second part will tend towards the same denominator

Thus for n=infinity,the second part will also equal denominator

Therefore, fair amount= ((denominator value) + (denominator value)) / denominator
= 2 rupees

Answer 3

half a rupee is what i am willing to pay because then both players have a probility of 0.5 in losing the first term of the game and both will loose 0.5 rupee.

If I was the one who sells tickets I would ask much more than 0.5 rupee because I risk to loose an infinite amount of money while the other only risks a loss of x rupees, I would ask at least 4 rupees then because of the intuitive unlikelyness of the event that 4 head apear in a row

I will NOT pay 2 rupees because then I will have a chance 0.5 of loosing 2 rupees in the very first turn, while the other will have a chance of loosing two rupees only after 4 times head which is very unlikely

I bet that if you ask 2 rupee and play this game long enough you will be left with nothing. ( you is the one who pays the one rupee )

Answer 4

The expected value of the payoff is

1 * 1/2 + 2 * 1/4 + 3 * 1/8 + 4 * 1/16 + 5 * 1/32 + ...

or

1/2 + 1/2 + 3/8 + 1/4 + 5/32 + ...

In general, the nth term of this series is

n / 2^n

and we want the sum of the terms as n goes to infinity. You can see that the value of the terms decreases rapidly as n increases, since the denominator increases exponentially.

Answer 5

The payoff is Rs1 each time I win.
To arrive at the right price, we can sum up the multiples of prob and payoffs

i.e.

first time win, expected payoff= (1/2)*1
second time win, expected payoff= (1/2^2)*(1+1)
third time win, expected payoff = (1/2^3)*(1 + 1 + 1)
and so on...

Sum of payoffs = 1/2 + 2/2^2 + 3/2^3 + ... infinity


This is an AG series to infinity and sum is
Say, S = 1/2+ 2/2^2 + 3/2^3 + ... infinity

Let 1/2 = x
Then,

S = x + 2x^2 + 3x^3 + .... infinity
This is AG series and its sum is = x/ (1-x)^2

So, sum of this series is = 1/2 / (1/2^2) = 2

So, I would be ready to pay Rs 2

Answer 6

The potential dividends are random. You can't put an exact price on something like that. The amount you are willing to pay will be different for each person based on their willingness to risk the purchase price of the ticket.