Parimutuel market: maximize your EV

[techmologist]

Your coworkers are competitive people who like to gamble. They have set up a bet pool on the longest run of heads (call this R) that occurs in a sequence of 100 fair coin flips. In this pool you are allowed to bet on any combination of three mutually exclusive outcomes:

1) R < 6
2) 6 <= R < 10
3) R >= 10

Everyone has placed their bets and they are waiting for you to do the same. So far the money in the pool breaks down like this:

$666 on outcome 1
$333 on outcome 2
$1 on outcome 3

for a total of $1000 in the bet pool. Assume your goal is to maximize your total expected value (as opposed to rate of return, or longterm growth rate). You have $200 in cash on you. Do you bet it all? Do you need to run to the ATM to get more? How much do you bet on each outcome?

There is no "take" like at the race track. The whole $(1000 + your wager) is paid back to the winning bettors. Also, it doesn't matter how you arrive at the probabilities for the three outcomes, as long as they are close. It is only important that it is possible to calculate them as accurately as you like. This is not the case with a horse race. You can use .472, .484, and .044 as approximate values for outcomes 1, 2, and 3.

 

[ImaLooser]

Your coworkers are competitive people who like to gamble. They have set up a bet pool on the longest run of heads (call this R) that occurs in a sequence of 100 fair coin flips. In this pool you are allowed to bet on any combination of three mutually exclusive outcomes:

1) R < 6
2) 6 <= R < 10
3) R >= 10

Everyone has placed their bets and they are waiting for you to do the same. So far the money in the pool breaks down like this:

$666 on outcome 1
$333 on outcome 2
$1 on outcome 3

for a total of $1000 in the bet pool. Assume your goal is to maximize your total expected value (as opposed to rate of return, or longterm growth rate). You have $200 in cash on you. Do you bet it all? Do you need to run to the ATM to get more? How much do you bet on each outcome?

There is no "take" like at the race track. The whole $(1000 + your wager) is paid back to the winning bettors. Also, it doesn't matter how you arrive at the probabilities for the three outcomes, as long as they are close. It is only important that it is possible to calculate them as accurately as you like. This is not the case with a horse race. You can use .472, .484, and .044 as approximate values for outcomes 1, 2, and 3.

Well, the more money I bet the greater the expected value. So I would bet every cent I had to maximize the expected value. If I have a million bucks then I bet 666,000 on 1), 333,000 on 2) and 1,000 on 3). I'm certain to win $999.00 and the losers split a dollar. They beat me up in the parking lot afterward and I pay a $10,000 hospital bill.

 

[techmologist]

Well, the more money I bet the greater the expected value. So I would bet every cent I had to maximize the expected value. If I have a million bucks then I bet 666,000 on 1), 333,000 on 2) and 1,000 on 3). I'm certain to win $999.00 and the losers split a dollar. They beat me up in the parking lot afterward and I pay a $10,000 hospital bill.

I don't see how you are guaranteed $999. If you bet in proportion to the other bets, won't that guarantee that you break even?

You are right to point out that this model (as in the other thread from a couple weeks ago) does not factor in many things, such as hospital bills, the double cheeseburger you had to eat for extra brain power to solve the problem, etc. Assume that your coworkers are good sports and enjoy gambling for the fun of it. They wear cowboy hats and drink bourbon and call you "partner". They will not beat you up for winning their money.

 

[techmologist]

Since the pot is relatively small (compared to the stock market, say) and your bet on any particular outcome tends to move the odds in an unfavorable direction, simply betting the largest amount possible will not maximize the EV as it usually does. Neither does betting a tiny amount on the most favorable outcome. There must be some way to squeeze the most out of this inefficient market.

 

[BWV]

given that there is about at least an 8.4% chance of getting a string of 10 heads, bet it all on outcome 3 (assuming abstractly that you don't move the line). so a bet with odds of 8.4% pays 1000-1, that is an an $84 expected value for $1

 

[BWV]

Here are the approximate odds:

R<6 = 5%
6 <= R < 10 = 78%
R >= 10 =16%

so accounting for moving the line, then you can either bet on 3 up to a limit of expected value of (16%*999) = 160 or bet on 2 and 3 to the point where a 95% chance of winning the $666 bet on 1 has a fair value of 633

 

[techmologist]

Here are the approximate odds:

R<6 = 5%
6 <= R < 10 = 78%
R >= 10 =16%

so accounting for moving the line, then you can either bet on 3 up to a limit of expected value of (16%*999) = 160 or bet on 2 and 3 to the point where a 95% chance of winning the $666 bet on 1 has a fair value of 633

Thanks for the replies. We definitely must account for moving the line with our bet. That is what makes this problem tricky. I got values very different from yours for the probabilities of the three outcomes

1) ~47%
2)~48%
3)~4-5%

this website calculates streak probabilities:
http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html

and it gives answers similar to the ones I got by an okay approximation. I see where you are going, I think. The current odds on outcome 3 are so good that you bet a small amount of money on 3 for a chance at winning a fraction of that $999. The more you bet, the larger your fraction. So if you only bet $1 and won, you would split the $999 with the other guy who bet a dollar on outcome 3. If you bet $9 on outcome 3 and won, you would get 9/10 of that $999, about $900. That's still pretty good, whether outcome 3 happens 5% of the time or 16% of the time as you say. But you are also putting more money at risk, and the fact is you will lose your $9 the great majority of the time. So that brings the expected value back down.

 

[BWV]

Does not look right to me, the odds of a 6 head streak are 1.6% = (.5)^6, so with 94 chances to hit a 6 run streak out of 100 tosses the odds of getting one are about 77%: 1-(1-.016)^94

 

[techmologist]

You don't have quite 94 chances, usually. You may get a run of 4 or 5 heads and then get a tail, messing up the potential run of 6. In this one "attempt" at running 6 heads, you have used up 5 or 6 of those 94 chances. See what I mean? Those 94 chances to run 6 overlap too much. So effectively you have fewer chances. Still, you can use your reasoning to find the expected number of runs of 6 or more in 100 flips. Then use the Poisson distribution to calculate the probability that there are no runs of 6 or more. That is exactly how I got my numbers, which I think are within a few percent of the exact values.

But getting the exact probabilities isn't the main thing here. Just that it is theoretically possible.

 

[BWV]

Ah, OK I see my mistake