## Theory problem update

It’s been a few days since I first posted my puzzle. If you missed it, you can find it here. Aside from the comments there, there is also some discussion about it going on over at Poker VT which you can see here (no account needed). People have also talked to me in private about the problem.

In my history of asking this problem, there are some common solutions that people come up with:

**Bet $1 because you have less risk of busting. **

When I did bonuses like this years ago, betting $1 is exactly the method I used precisely for the reason that I was risk-averse and had a limited bankroll. Such reasoning doesn’t apply to this situation, though, where we don’t care about variance, only EV.

**Bet whatever you want because it doesn’t matter. **

Some people reason that it doesn’t matter what bet we place, since a $100 bet has the same expectation as a series of 100 $1 bets. This is true, however despite the truth of this statement it’s not a valid reason as a solution to the problem.

**Martingale.**

Some people come up with the martingaling strategy, doubling bets every time you lose and such. Many people believe this is not only the best strategy, but one in which you can’t lose. This, also, is not right. Plenty of good information on the subject is available here.

A note about martingaling, and here’s a hint about the correct solution: martingaling has no worse EV than betting $1 repeatedly. Would it be possible for me to convince you that martingaling has a higher EV than betting $1 at a time?

The solution will be posted in a couple of days.

## rouliroul said:

I’m in the “It doesn’t matter camp” I don’t see how its not a valid solution to the problem. The problem is: “How much should we bet to maximize EV, and why? Does it even matter? If so, why? If not, why not?”

The answer is the size of the bet won’t change the EV, it will only increase or decrease variance. So it doesn’t matter.

## EngineerSean said:

It’s obviously a percentage of your current bankroll, which is a Kelly problem. Unfortunately the Kelly criterion states that you only bet if you have an edge. In $2500 of play, and EV of half a cent lost per dollar of play, your EV is -$12.50. With your $100 bonus, however, you expect to win $88.50 over $2500 of play, which means that you now have a 3.5% edge.

Kelly criterion states that you bet f = (bp – q)/b, where f = percentage of bankroll you bet, b = odds if you win, p = probability of winning, and q = probability of losing

b = 1.035 (Our expected value is 3.5% so we expect to get $1.035 back for every dollar we bet)

p = 0.5

q = 0.5

bp = 0.5015

bp – q = .0015

(bp – q) / b = ~0.0014 = f

So you would want to bet 14 cents for every hundred dollars in your current bankroll, or $1, whichever is greater. This assumes you want to make the full Kelly bet. Betting any more than that increases risk of ruin without increasing your chance of clearing the $100 bonus.

## EngineerSean said:

Also it seems clear that if you are claiming that Martingaling has the same EV as betting $1 every time (a claim that I just verified and it’s true) then Martingaling is certainly not the correct answer, since your chance of going to zero is far greater and we’re only concerned with doing $2500 worth of volume, even if the median value of all possible bankrolls is higher by Martingaling than by just betting $1 every time. Also I made a mistake above, I meant $87.50 in profit, not $88.50, but the numbers don’t significantly change.

If you haven’t Martingaled for breakfast in Vegas I highly recommend it. Free breakfast buffet never tasted so good.

## Derk’s blog » Blog Archive » Gambling theory problem/puzzle solution said:

[...] originally posted a problem about a week ago here and an update with a hint here. If you haven’t read those, please do and give it a try before reading the [...]