A mind game

[tropicalknight]

Optimally, if one was only concerned about the expected value, then the answer would be to bet 100% every time.

If one was concerned about the geometric mean, assuming completely average luck, the answer would be to bet 66.6% every time.

When one factors in diminishing marginal utility of money, the answer would shift lower by some amount.

For me, I would bet 60% for no good reason besides it seems about right for my risk tolerance.

 

[peilthetraveler]

Quote from Josef K:

No, that is just the mathematical expectation, not a mathematical law. You could also have six losses in six years. Hence, you need to practice risk management which takes this into account.

So you take a much much smaller payday because you are afraid of a 6 losses in a row (which has a 1 in 46,000+ chance of happening)

Steven Wynn builds billion dollar casinos because he has around a 53% chance to win and a 47% chance to lose. This scenario has an 83.3% chance to win and a 16.6% chance to lose.

To give you an idea of the odds of 1 in 46,000, your odds of being murdered are 1 in 18,000. The odds you will die on the job this year is 1 in 27,000. The odds are 1 out of 23,483 that you will be wrongly declared dead by the Social security administration this year.

So why do you need risk management in this situation. Even if you were right, even with risk management betting only 10% every year, if you had 6 straight losses, it would take you 7 years to recover and by then 13 years have passed since you first started. You are trying to protect yourself against something that is very very unlikely to happen.

 

[Josef K]

Quote from peilthetraveler:

So you take a much much smaller payday because you are afraid of a 6 losses in a row (which has a 1 in 46,000+ chance of happening)

Steven Wynn builds billion dollar casinos because he has around a 53% chance to win and a 47% chance to lose. This scenario has an 83.3% chance to win and a 16.6% chance to lose.

First of all, I assume that Steve Wynn's odds that you're referring to are the odds that his casino will win on any one play of a game in the house. Given that, then your comparison is completely inappropriate. Games are played over and over again all day long in a casino, while our game is played only once a year. Our game is supposed to be analogous to "that special once-a-year market scenario where you get that maximum-conviction, 'perfect setup' type of trading opportunity", not high-frequency trading.

Second, the six losses in a row scenario is extreme and has very little chance of happening, but the odds of losing at least two years out of the six are much greater. What are those odds? There are 15 different combinations (not permutations) of two years selected randomly out of the six. For each of those two year combinations, say a six is thrown. For the other four years, we don't care what's thrown, and there are 6^4 possible permutations for those other four years. This gives us 15 * 6^4 ways in which at least two losing years could occur in the first six. Since the total possible permutations of results is 6^6 = 46,656, then the odds of losing at least two times in the first six years are

(15 * 6^4)/(6^6) = 15/(6^2) = 15/36 = 41.67%.

Now, if you have exactly two losing years during the first six and you bet only 40% each time, then you'd end up with $691,488 at the end of six years. But if you bet 70% each time, you'd end up with only $375,844.50, so things are riskier than they seem.

 

[FreakofNature]

Odds of losing 7 in a row is so damn rare, I would space the bets accordingly and Martingale it for life taking into consideration and re-investing the new gained capital but always allowing room for the 7th bet.

 

[Josef K]

One more thing: the odds of having at least two losing years during the first nine are ([(9*8)/2] * 6^7) / (6^9) = 36/(6^2) = 36/36 = 100%.

If you have exactly two losing years during the first nine and bet 40%, you end up with $1,897,443.07. If you bet 70%, then you end up with $1,846,524.03.

 

[Pension_Admin]

Just run a simulation on MS Excel with Rand() and IF formulas. The resulting chart over your remaining life time would give you the answer.

 

[Josef K]

That 100% doesn't seem right, does it? There's probably a mistake in my calculations which I'll have to fix later when I'm more awake.

 

[Josef K]

I realize what I did wrong: my method of counting the different permutations ended up in counting some of them more than once. Ignoring the what happens in the other years besides ones set for losing is what caused the problem.

The probability of having at least two losses in the first six years is actually only 26.32% and the probability of at least two losses in the first nine years is about 45.7%.

 

[peilthetraveler]

Quote from Josef K:

One more thing: the odds of having at least two losing years during the first nine are ([(9*8)/2] * 6^7) / (6^9) = 36/(6^2) = 36/36 = 100%.

If you have exactly two losing years during the first nine and bet 40%, you end up with $1,897,443.07. If you bet 70%, then you end up with $1,846,524.03.

100% chance to have 2 losing rolls in 9 years in a game of chance? Come on...does that really make sense? Lets disprove it. Get a dice, roll it 9 times and see how many times 6 comes up. If you do that over and over and 6 always comes up within 9 rolls, I will concede.

I'm not saying that there isnt a chance to have 2 losing rolls in 9 years or even in 6 years, I'm just saying its much less likely. And sure there are scenarios where betting a lower wager will return more money than betting a 70% wager, but when you do that you are playing AGAINST the odds.

Now in your scenario of 2 rolls ending up 6 in 9 years, you betting 40% and me betting 70% we end up within $50k dollars of each other. Nothing to cry about. But if we roll 9 times and we only get 1 six over 9 rolls, then you get $4,427,000 and I get $10,463,000. A difference of 6 million dollars.

In your scenario, I lost $50k by not betting your way. In my scenario, you lost $6 million by not betting my way. Which is worse?

 

[Josef K]

Quote from peilthetraveler:

100% chance to have 2 losing rolls in 9 years in a game of chance? Come on...does that really make sense? Lets disprove it. Get a dice, roll it 9 times and see how many times 6 comes up. If you do that over and over and 6 always comes up within 9 rolls, I will concede.

I'm not saying that there isnt a chance to have 2 losing rolls in 9 years or even in 6 years, I'm just saying its much less likely. And sure there are scenarios where betting a lower wager will return more money than betting a 70% wager, but when you do that you are playing AGAINST the odds.

Now in your scenario of 2 rolls ending up 6 in 9 years, you betting 40% and me betting 70% we end up within $50k dollars of each other. Nothing to cry about. But if we roll 9 times and we only get 1 six over 9 rolls, then you get $4,427,000 and I get $10,463,000. A difference of 6 million dollars.

In your scenario, I lost $50k by not betting your way. In my scenario, you lost $6 million by not betting my way. Which is worse?

You're right, I made a mistake. I've corrected it in my subsequent posts.