Easy math puzzles

We talk about risk of ruin only when the edge is greater than zero.

Quote: lilredrooster

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Blackjack Puzzle

"You go to Vegas over a long weekend to play a single-deck blackjack game with a 0% edge ($500 minimum). You decide to play 1,200 hands but you will quit if you bust or double your initial bankroll before reaching 1200 hands.

What size bankroll should you bring to give yourself a 1/3 chance of busting, 1/3 chance of doubling and 1/3 chance of finishing the 1200 hands without busting or doubling?

Assume a standard deviation of 1.1547, flat betting $500 one hand at a time and perfect basic strategy to realize the 0% edge"


we all know that the house edge against basic strategy is not 0 - the puzzle is set that way intentionally

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blackjack Hall of Fame member and author Don Schlesinger solved the puzzle in another forum

this is what he wrote:

" I'm going to give you a very elegant solution that doesn't rely on calculators With SD = 1.1547, the total SD for 1,200 hands is 1.1547 * 1200^0.5 = 40 units exactly. To have 33.33% probability as the END POINT solution (doubling or going broke), using my "bumping into the barrier concept" (see BJA, p. 124, second full paragraph) you need exactly half of that value or 16.67% probability of busting or going broke IF you're guaranteed to get to play all 1,200 hands. If you look at cumulative normal tables, you'll see that, to lose enough to go broke, we have to lose 0.97 standard deviations, which has probability 16.6%. So, we're looking for B/40 = 0.97, whence B = 38.8 units, or $19,400. QED. No fancy math, no calculators!"

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What did I say above?

Quote: aceside

What did I say above?

you implied the puzzle was not worthwhile because the edge was set at zero
Don found the puzzle interesting and decided to solve it just as it was setup (with the edge at zero)
it's a mathematical puzzle - it wasn't set up to reflect the real world of blackjack - it was set up to challenge a person's math skills
so Don obviously didn't agree with your comment

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Don has dismissed me many times. He probably thinks I am assside. What matters is your math derivation.

Quote: aceside

We talk about risk of ruin only when the edge is greater than zero.
link to original post

Wrong. There is a RoR in any scenario with variance, including scenarios with a player advantage.

Also worth noting that games with <0.5% house edge (the only games worth playing) are close enough to 0% to use the RoR calculations for 0%

Maybe you are right. I only found two paragraphs on wiki to describe risk of ruin, so I guess it is a loosely defined financial terminology. Also, I notice we refer to different edges. I meant to say a player’s positive edge in gambling, because gamblers tend to invest on a game with a positive edge.

Risk of ruin is not “loosely defined”.

Very simple: It is the chance you lose a specified starting capital base before a specified event happens.

About 99.9999% of gamblers play against a house advantage. The most common RoR calculation for them is: if I bring X bankroll to the casino and plan to play Y game for Z hours, what is the probability that I bust X bankroll before playing for Z hours. No idea what you mean by “gamblers tend to invest on a game with a positive edge”

Let me give you a numerical example. I invented a fair coin flip game for players to bet on heads or tails. The payout is 1.01 for a head, but 1.00 for a tail. If a player has a bankroll of $1000 to bet $1 at a time on heads to play an infinite amount of times, what is his RoR?

There is no end goal so that’s technically not a RoR problem. You could be up a million dollars and then get 1,001,000 consecutive tails and bust your bankroll. Highly unlikely but mathematically possible. There are formulas for this kind of problem (chance of never busting problem) but I believe they only work for even money payouts. Probably have to use a Markov chain to calculate this case

An RoR calculation would be for: what is the chance you triple your $1000 bankroll before busting? Or what is the chance you play 5,000 flips without busting?

Actually, I think the other way around. I believe RoR applies only to infinite numbers of events. I am self taught in statistics, so I’m not so sure though.

Regarding the math problem I raised above, could you please solve it? I want to solve it myself too, but need somebody to check.

The most common use of RoR is probably in investing/trading. There will be a defined profit goal; the RoR won’t be stated in terms of making specific trades “forever” since that’s impossible. Even for a gambler that has a 1% advantage counting cards in blackjack, that advantage won’t last forever. “Forever” is theoretical but RoR is used for real life scenarios

I actually use RoR to size my bankroll for Las Vegas trips. I’m not an advantage player but I don’t play anything with a house edge >0.5%

The solution to your problem is easy with a Markov chain but I am without my computer at the moment

Quote: aceside

Let me give you a numerical example. I invented a fair coin flip game for players to bet on heads or tails. The payout is 1.01 for a head, but 1.00 for a tail. If a player has a bankroll of $1000 to bet $1 at a time on heads to play an infinite amount of times, what is his RoR?
link to original post

you've stated you think AI is wrong but most of the time I believe it is correct - I acknowledge that sometimes it is wrong
I put your question to google AI
hopefully Ace2 will confirm that the answer it gives is correct (or incorrect)



"AI overview

Determine Bankroll Sizes for Near-Safety

Because (0.99015)B will approach, but never exactly equal, zero for any finite B, you can never reach a absolute 0% risk. However, you can make the risk so infinitesimally small that it is practically impossible to bust:

With a bankroll of $1,000 and bets of $1.00 your risk of ruin is approximately 0.005%, which translates to about a 1-in-20,000 chance of going bust.

For a 1% risk of ruin: You need a bankroll of 465 units ($465 if betting $1 at a time).

For a 0.1% risk of ruin: You need a bankroll of 697 units.

For a 0.0001% risk of ruin (1 in a million): You need a bankroll of 1,400 units.

Final Answer:

To never go bust with flat betting, you would theoretically need an infinite bankroll. For practical safety, a bankroll of 1,400 times your bet size reduces your risk of ruin to less than one-in-a-million.


how the problem was solved:


Calculate the Risk of Ruin

For an asymmetric random walk with a positive edge, the probability of going bust (Risk of Ruin, R) from an initial bankroll of B units (assuming $1 flat bets) is calculated using the characteristic root equation:

Where:

p = 0.5 (probability of winning)

q = 0.5 (probability of losing)

W = 1.01 (units won)

L = 1.00 (units lost)

Solving for the root r (where 0 < r < 1) gives approximately 0.99015. The probability of going bust is then: R=rB=(0.99015)B"

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The exact answer can only be obtained via Markov chain. I’m on the road without my computer so I’m unable to do that now.

However, the game is statistically quite similar to using a weighted coin with heads coming up 50.25% of the time and paying even money. For that case, the chance of ever busting is ((1-.5025)/.5025)^1000 =0.0000454 or about 1 in 22,000. With a starting bankroll of 1,000 units, the chance of ever busting is obviously going to very very low, even with a smallish advantage of 1/2 percent