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giorgio
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Expectation in coin flipping

August 1st, 2004, 2:14 pm

You flip a fair coin winning $2 for a head, and losing $1 for a tail. Game A: You flip 100 times and you receive the payoff or pay the balance after the 100th flip.Game B: You start with $50 and settle after each flip. You can not go under. You must stop at 0.a) How do the expectations of the two games compare? Are they equal, almost equal, quite a bit apart?b) More precisely, do they differ by i) more than 20% ii) btw 1% and 20% iii) btw 5 bp and 1% iv) btw .1 bp and 5 bp v) less than .1 bp?c) Is the expected value of game B representable in closed form? (Finite sums do not qualify!!)
 
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giorgio
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Expectation in coin flipping

August 3rd, 2004, 11:12 am

Dear all,An 'obvious' correction in my statement of the problem. The question compares theexpected winnings; or you can think of starting Game A also with $50. Otherwise,first two parts are trivial.
 
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Aaron
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Expectation in coin flipping

August 8th, 2004, 8:57 pm

I think you mean that you have to stop the second game after 50 tails. If you could pay for tails with your winnings, game 2 has an infinite expectation.The expectation of the first game is easy. You get $3 times the number of heads minus $100. The average number of heads is 50, so the average profit is $50.In the second game, your profit is $2 times the number of heads minus $50. The average number of heads is still 50, so the average profit is the same $50. The argument here is simple. Suppose you flipped a coin putting $1 in one pile for heads and $1 in another pile for tails. Clearly the expected amount in each pile is the same, regardless of your stopping rule. Therefore, if your stopping rule is stop when the tail pile reaches $50, the expected amount in the head pile must be $50.
 
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ScilabGuru
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Expectation in coin flipping

August 9th, 2004, 3:19 pm

Game A: is 100* mean( 1 trilal) =100*1.5 = 150 - I don't understand what Aaron is calculating here. GAme B is also not so difficult too: Mean profit_B = mean profit_A * (1- Probability to get zero during 100 trials) - 50* ( Probability to get zero during 100 trials) == 150 - 200 * (Probability to get zero during 100 trials )Let P_n^k = Probability to get 0 FIRST TIME during n trials (starting from k) The only problem is to solve this difference equation now And again
Last edited by ScilabGuru on August 8th, 2004, 10:00 pm, edited 1 time in total.
 
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Aaron
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Expectation in coin flipping

August 11th, 2004, 9:14 pm

In game A, you pay, not receive, $1 for each tail. So the expectation per flip is $0.50, not $1.50. After that, our caculations agree.In game B, you are interpreting it as playing for 100 flips, or stopping sooner if you get to $0. I agree with your solution to that problem, except k - 1 and k + 2 should be k + 1 and k - 2 in the first clause of your definition of P. Making that change, I get 7.8*10^(-12) for P^50_100 so the second game is a very small amount worse than the first.I interpreted game B as you keep playing until you get 50 tails, regardless of how many heads or flips.
 
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Candle
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Joined: October 21st, 2003, 2:55 am

Expectation in coin flipping

August 12th, 2004, 12:54 am

Wait a minute. I thought the rule of game B says we cannot go below $0. This means we can, say, get one head ($52) and then get 51 tails (down to $1) and still remain in the game. I think game B is a random walk with an absorbing barrier at 0, and we are calculating the expected position after 100 steps.
 
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Aaron
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Expectation in coin flipping

August 12th, 2004, 12:02 pm

That is how ScalibGuru interpreted it as well. Since there is nothing in the statement of the problem about stopping after 100 flips, I assumed you were no allowed to apply your winnings to paying losses (otherwise the expectation is infinite). So you and ScalibGuru compute the expected value after 100 flips given an absorbing barrier at zero, in which case the answer is that the expectation is not significantly different from the barrier-free solution. I computed the expected value after 50 losses.
 
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giorgio
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Joined: May 25th, 2004, 5:43 pm

Expectation in coin flipping

August 25th, 2004, 3:14 pm

Dear Aaron,Thanks for the comments and solution. You are right I should be more precise and say that game 2 stops at $0 or after 100 flips (whichever comes first).Any chance that the recursion is solvable in closed form? I arrived at my solution by direct computationbut it is a nice finite sum I can't get in closed form.Cheers,Georgios
 
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giorgio
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Expectation in coin flipping

August 25th, 2004, 3:15 pm

Candle,You are right on the absorbing barrier at $0 and stopping when n=100 (flips)Any chance for the recursion proposed in closed form?
 
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Aaron
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Expectation in coin flipping

August 25th, 2004, 6:46 pm

There is no known closed-form exact answer. Would you settle for a Normal approximation?
 
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RR
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Expectation in coin flipping

September 1st, 2004, 4:25 pm

candle has right solution...
 
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greenleaf
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Joined: March 10th, 2004, 6:02 pm

Expectation in coin flipping

September 1st, 2004, 5:20 pm

Simple barrier options have closed-form solution. What could prevent this problem from having one? I might try viewing the total amount of money as some stock price diffusing normally with a constant drift of 50 cents and a volatility of order $1. A BS-like difussion eqn thus follows (the "interest rate" set equal to 50 cents as well) and then a barrier is imposed.Perhaps the trouble with Giorgio's original problem is its discreteness in time. Well, I'll try.
 
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Aaron
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Expectation in coin flipping

September 2nd, 2004, 12:51 pm

The barrier option formula is a Normal approximation.You could write down a very long formula for all the binomial terms, since the problem is bounded (it must end after 100 flips). You could call this a closed-form solution. What I meant is there is no known general closed form for N flips.
 
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giorgio
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Joined: May 25th, 2004, 5:43 pm

Expectation in coin flipping

September 2nd, 2004, 4:08 pm

I am afraid Aaron, is exactly right. The discreteness moves the problem from a simple barrier optionto the finite world where formulas are, generally, tougher to obtain.I do have a simple formula (finite sum, ie) of simple combinatorial coefficients but it doesn'tseem I can push that forward.Any concrete ideas here? The book 'Concrete Math' by Knuth could potentially help, but I slightly doubt it; they have a 'big bag' of tricks that apply to similar tasks. Any other thoughts?Cheers,Georgios
 
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Aaron
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Expectation in coin flipping

September 3rd, 2004, 3:25 pm

Why do you want a closed-form solution? You can write a computer program to give exact answers (subject to machine accuracy) for N up to a few million; the Normal approximation gives very accurate answers for N > 30.
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