Toasty's Math Puzzel 2

[Toasty]

Scenario :

You are invited to play a number drawing game from 1 to 100 integers only. You are told if the number drawn is below 52 you win 1:1 on any money you wager.

You start with a bankroll of $1,000.

The twist is you can only bet a % of your bankroll and the game is then played 10,000 times using this percentage (i.e. 10% would go $100 if you won the next bet would be $1,100*10%=$110). If you go bust during this time the game stops (If you reach less than $1 you are considered bust).

Ok, would you play in this game? Do you have an edge and if so how much? What % of your BR would you wager?, why? and is there an optimal amount to bet?

[Fnord]

Ok, would you play in this game?

Yes

Do you have an edge and if so how much?

51:49 or 1%

What % of your BR would you wager?, why? and is there an optimal amount to bet?

It comes down to variance. In the course of 10,000 trials, how many can you expect to lose in a row? If I had to wager a guess, I would go with 5-10%. Certainly, I would expect the correct answer to be a percentage of your bankroll.

[Toasty]

Going along the right lines, although your edge is 2% 51 numbers will win 49 number will lose (51/100)-(49/100)=2%.

I won't comment on the betting percentages yet but I will mention that you may need to make a simulation or be very good with mathematics to arrive at the answer.

I'll accept the answer to 0.5% up or down.

You are also on the right lines when you mention variance.

[xbones]

I'd have thought 50% of your bankroll would be better, assuming that you 50% of your bankroll at the time and not your starting bankroll.

[xbones]

(you can't drop below $1,000)

Do you mean you can't drop more than $1,000?

[Toasty]

I've updated the question.

[xbones]

Surely if you bet a percentage each time you could never go into debt.

Not even sure why I'm bothering to ask, Stats always was the weakest part of my maths

[Toasty]

Surely if you bet a percentage each time you could never go into debt.

Not even sure why I'm bothering to ask, Stats always was the weakest part of my maths

True but if you start with $1,000 and drop down to 5c, its pretty safe to say that you lost all of your money. Hehe I should write a list of caveats when I do these. You are right though you can't go into debt, i'll change the question.

[michael1123]

I used trial and error on my calculator to find the right equation.

x = starting amount for each bet
y = number of turn (out of the 10,000)

x * 1.1^y = amount bet on a specific turn

So, on the 10,000th turn, you'd be betting about 8.45 times your original betting amount.

Now here's where I get completely random. Lets say, accounting for variance, and to be reasonably sure you don't go bust, you want your final bet to be $100.

x*8.45=$100
$100 / 8.45 = 11.834

So my guess is $11.83 or if you want to be really cautious (since there's bound to be a lot of variance in 10,000 bets) and have your final bet be $50, then I'd start with $5.92.

edit: BAH! ... I misread the original question. BUT! If your next question says that you choose a starting betting amount, and say that each time the amount you bet increases by 10% ... then I win ...

Bah. Maybe I'll take another look at doing it right later, if it hasn't been answered by then ... all that work for nothing ...

[Toasty]

I take it you mean 5.92% and continue from there?

I've ran it through my simulator and the mean average for 50 plays is a loss of ($999). Although 4 of those times you would make a considerable profit. That is if you waged $1000 generally you would expect to lose $999 but there would be a small chance you would win big.

[michael1123]

Read this part.

edit: BAH! ... I misread the original question. BUT! If your next question says that you choose a starting betting amount, and say that each time the amount you bet increases by 10% ... then I win ...

Bah. Maybe I'll take another look at doing it right later, if it hasn't been answered by then ... all that work for nothing ...

:-/

[Toasty]

I must have replied while you were editing your post.

I'll point you all towards the answer and say "Kelly's Criterion".

[koolmoe]

This is clearly related to the rate of ruin problem, though that assumes a standard size bet every time. For that situation, I think the answer is 4.08%, but I'm not 100% sure.

I think betting 100% gives you the highest expectation, but that is skewed by the result that you win all 10k bets and make 1000*2^(10000). This happens with frequency 0.51^10000, so the expected value is 1000*(2*0.51)^10000 or 1x10^87.

The difference equation is:

B[k+1] = B[0]prod{0,k,(1 – r + rx[i])}

where B[k] is bankroll after bet [k], r is your chosen betting rate, and x[k] is a stochastic variable with the following distribution:

Win, x[k] = 2 probability =0.51
Lose, x[k] = 0 probability = 0.49

Thus, B[10000] = 1000*(1-r+2*r*p)^10000

To answer the question, you need to know with what frequency you're willing to go broke (defined by B[10000] < number that equals bust in your mind). Then you just compute the variance of B[10000] in terms of r, and find the biggest r that satisfies the constraint based on the normal distribution (B[10000] has a nearly normal distribution by the central limit theorem). I'll try to give an example in a bit.

[Toasty]

A theorom covers the answer its called Kelly's criterion.

The gist of it is, if you bet a higher % of your bankroll then the actual edge you have, you will eventually go broke. Not straight away but eventually.

The best % of your BR to wager was a % that was just below the % of your edge, In this case 1.5% to 1.75% would be optimal. This would build your bankroll at the fastest rate without going bust.

Betting more than twice your edge is where you really increase your chance of going broke even though you are repeatedly making +EV bets (If you bet the same amount each time i.e. $10 and not the same % of your BR you would make money). Betting just over your edge would also build your BR but on avg not as fast.

This therom us used by quite a few BJ players.

[koolmoe]

From Kelly's original paper:

Now the gambler could bet his entire capital each time, and in fact this would maximize the expected value of his capital.

And a bit further down...

One might still argue that the gambler should bet all his money in order to maximize his expected win after N times. It is surely true that if the game were to be stopped after N bets the answer to this question would depend on the relative values (to the gambler) of being broke or possessing a fortune. If we compare the fate of two gamblers, however, playing a nonterminating game,the one which uses the value [equal to the player's edge in the game] will, with probability 1, eventually get ahead and stay ahead of one using any other [value].

http://www.racing.saratoga.ny.us/kelly.pdf

Note that, after 10000 bets, there is a finite probability of being (close to) broke using almost any betting rate that is nonzero. As the betting rate increases, the expectation does as well, but the distribution of results becomes more skewed. To illustrate, I ran 1000 trials using a rate of 2% and a rate of 4%. At 2%, the player never went broke (which you defined as less than $1) although there were 18 finishes below $100. At 4%, the player went broke 42 times but on average finshed with over 10x as much money (over 700K vs. 64K). Of further interest: At 4%, there were 507 finishes where the player's bankroll decreased compared to 151 at 2%. At 4%, there were 42 finishes above $1,000,000 compared to 7 at 2%.

What's optimal depends on your perspective. A professional gambler would never accept the 4% results, while a recreation gambler would likely much prefer the 4% results.

Fun problem, Toasty!

[johnnyawe]

heh, another good question.

This reminds me of a roulette strategy. Always bet on black. Start out by betting $5. If you lose, bet $10 on the next spin. If you lose that, bet $20 on the next spin. Keep increasing like this until black hits. When black does hit, you will have +$5 profit. Put that profit in your pocket, and repeat the process from the beginning by betting $5 on black. Repeat this process for the rest of the night.

I believe the problem with this is that any casino's roullette table will enforce a "maximum bet" that is sufficiently low enough to foil this strategy.

[fishstick]

heh, another good question.

This reminds me of a roulette strategy. Always bet on black. Start out by betting $5. If you lose, bet $10 on the next spin. If you lose that, bet $20 on the next spin. Keep increasing like this until black hits. When black does hit, you will have +$5 profit. Put that profit in your pocket, and repeat the process from the beginning by betting $5 on black. Repeat this process for the rest of the night.

I believe the problem with this is that any casino's roullette table will enforce a "maximum bet" that is sufficiently low enough to foil this strategy.

it's called the progression method, and is advocated by some people for black jack as well. and as you said, the max bet makes this strategy fail.

e.g. a blackjack table with a $5 min and $500 max
bet 5 - lose
bet 10 - lose
bet 20 - lose
bet 40 - lose
bet 80 - lose
bet 160 - lose
bet 320 - lose

so - if you lose 7 hands in a row, your "system" just fell apart.

[Humphrind]

heh, another good question.

This reminds me of a roulette strategy. Always bet on black. Start out by betting $5. If you lose, bet $10 on the next spin. If you lose that, bet $20 on the next spin. Keep increasing like this until black hits. When black does hit, you will have +$5 profit. Put that profit in your pocket, and repeat the process from the beginning by betting $5 on black. Repeat this process for the rest of the night.

I believe the problem with this is that any casino's roullette table will enforce a "maximum bet" that is sufficiently low enough to foil this strategy.

it's called the progression method, and is advocated by some people for black jack as well. and as you said, the max bet makes this strategy fail.

e.g. a blackjack table with a $5 min and $500 max
bet 5 - lose
bet 10 - lose
bet 20 - lose
bet 40 - lose
bet 80 - lose
bet 160 - lose
bet 320 - lose

so - if you lose 7 hands in a row, your "system" just fell apart.

A friend of mine dropped $600 in one night at the casino using negative progression. It's iffy, but if you get yourself into a BJ tournament, it' a great move. (after he put $150 on 1 hand and lost he just gave up. He barely spoke the rest of the night)

In a tournament, you need to easily re-coup your loses, and wining and losing don't matter as much as placing.

I think negative progression is too risky for real money, I'd rather just grind. (I go to have fun, making money is a side-note) but it is a crutual to know if you are in a BJ tournament.

[Toasty]

I went to the casino once and I'm not sure why, but they were allowing £1 bets on columns, made a tidy sum that night, waiting for 5 misses and then betting and of course its 2:1 so your stake twice to break even.

[HaD]

Kelly Criterion:

f=fraction of ur bankroll to maximize ur bankroll increasing rate
prize odds= v:1
p=probability to win

f=( (v+1)*p% - 100% ) /v

so in ur example:

v=1 (u win 1 to 1)
p=51/100

so:
f=( (1+1)*51%-100%)/1= 102%-100%=2%

noticed that u bet exactly the edge here. its because ur prize odds is 1:1

[gabe]

Posted: Thu, 01 Jul 2004, 3:16am

...

Posted: Mon, 25 Apr 2005, 6:40am

Quite a difference. Nice job I guess.

[teddosan]

Toasty, what simulator are you using?

[stevedonel]

heh, another good question.

This reminds me of a roulette strategy. Always bet on black. Start out by betting $5. If you lose, bet $10 on the next spin. If you lose that, bet $20 on the next spin. Keep increasing like this until black hits. When black does hit, you will have +$5 profit. Put that profit in your pocket, and repeat the process from the beginning by betting $5 on black. Repeat this process for the rest of the night.

I believe the problem with this is that any casino's roullette table will enforce a "maximum bet" that is sufficiently low enough to foil this strategy.

Martingale systems work decent in the short run, but the house advantage will eventually catch up and take away whatever edge you can build up. If you want to go with a progression system, try adding 1 bet to your double. So....

1st loss - $5 ($5 profit, if win)
2nd loss - $11 ($6 profit, if win)
3rd loss - $23 ($7 profit, if win)
4th loss - $47 ($8 profit, if win)
5th loss - $95 ($9 profit, if win)
6th loss - $191 ($10 profit, if win)
7th loss - $383 ($11 profit, if win)

This increases your risk slightly, but has a larger increase in long term return, because you make an extra few bets on each streak.

[Greedo017]

not sure how familiar any of you are with biology, but genetics can kindof lead you toward the answer also, but not so precisely. Without going into great detail, in a small population, randomness causes everyone's genes to be the same after a while. Every allele (different gene, aka brown hair = one allele, black hair = one allele, etc.) will either be 100% or 0%. So basically, you're going basically broke or having a fortune most of the time, unless you're betting tiny. This doesn't explain the betting less than your edge idea, but i thought it might be interesting to some of you.