Lonely's Interesting Tale: Single Seat vs Multiple Seat Betting

It's late, poor Lonely's wife recently left him, he's got a snootful of scotch and he's in a reckless mood. After unsuccessfully hitting on the dealer (he's married, so he's rusty), he settles for playing cards. He's counting a shoe and hits a really good count, let's say + 4 TC (or as close to + 4 as all that scotch allows).

At + 4, Lonely's normal bankroll rules dictate a $100 play, which also happens to be the table max. But if he wants, he can play all 7 spots. Cursing God and the harlot who left him, he plunks down $100 on all seven spots.

How do you view Lonely's actions? To your mind has he bet on 7 different hands, and therefore not violated his bankroll rules, since he (but not the dealer) is about to be dealt 7 different totals? Or do you take the opposite view, that he's definitely violated his bankroll rules because, despite getting 7 different hands himself, all 7 hands will be played against just one dealer's hand?

If, like me, you think betting 7 times the recommended level against one dealer's hand violates at least the spirit of bankroll rules, might you still do it if we add in the very reasonable assumption that the + 4 advantage will be short lived due to an upcoming reshuffle? That would sorely tempt me, given the fact that the multi bet approach nets $700 bet at + 4 for 8 hands' worth of cards dealt versus the one spot approach which only nets $400 bet for 8 hands' worth of cards.

PS: Lonely's story has a happy ending. The day after the divorce was finalized, he bought a winning Lotto ticket and his ex-wife was eaten by rabid grizzlies. Today, he owns his own Caribbean island. I'm there right now; come on down.

first of all, is this a true story, specifically? Do you remember how the 7 hands played out?

If your bankroll kept your max bet to 100, losing upwards of 500 on this hand would have definitely hurt. This could easily happen if dealer had an Ace showing and a 9 underneath. Do you make the insurance bet 7 times against that ace?

But the fact that you get more penetration is a viable advantage. Perhaps 7 bets of $75, or 4 bets of $100 would have taken away the risk of losing 4 to 7 max bets- which is a lot.

Quote: surrender88s

first of all, is this a true story, specifically?

There's one exaggeration: the rabid bears that ate his wife were black bears, not grizzlies.

Just kiddin'. No, it's not a true story.

I'm with you on solving the problem, though: I think I'd have gone with maybe 3 spots of $100's.

It would also matter to me how imminent the reshuffle was. If it was so close that this was definitely going to be the last hand, I would probably bet more spots than otherwise.

But the thread's original author badly dropped the ball on that (typical!).

Quote: hmmm23

But the thread's original author badly dropped the ball on that (typical!).

Enjoyed the story, only to find out it was a farce! Grizzlies have to eat, too, ya know! lol

I like your sense of humor though.

Counters like to size their bets in proportion to the ratio of their expected value to their variance. Refer to this page from the Wizard of Odds. Let var = the variance of one hand and cov = the covariance between the results of two hands of blackjack. We see on the Wizard of Odds page that the variance of several hands of blackjack played at the same time is n*var +n*(n-1)*cov. Think of the total money bet on one round of n hands as one bet with 1/n bet on each spot. Then the variance of this total bet is (n*var +n*(n-1)*cov)/(n^2) which equals var/n + (n-1)*cov/n. If var = 1.32 and cov = 0.48, then this variance is 0.60. If Lonely should bet $100 on one hand, then he should bet a total of $100*(1.32)/(0.60) = $220 on seven hands. That's about $31 on each hand.

Quote: camapl

Enjoyed the story, only to find out it was a farce! Grizzlies have to eat, too, ya know! lol

I like your sense of humor though.

Thanks. I really like your posts, as well.

Quote: ChesterDog

Counters like to size their bets in proportion to the ratio of their expected value to their variance. Refer to this page from the Wizard of Odds. Let var = the variance of one hand and cov = the covariance between the results of two hands of blackjack. We see on the Wizard of Odds page that the variance of several hands of blackjack played at the same time is n*var +n*(n-1)*cov. Think of the total money bet on one round of n hands as one bet with 1/n bet on each spot. Then the variance of this total bet is (n*var +n*(n-1)*cov)/(n^2) which equals var/n + (n-1)*cov/n. If var = 1.32 and cov = 0.48, then this variance is 0.60. If Lonely should bet $100 on one hand, then he should bet a total of $100*(1.32)/(0.60) = $220 on seven hands. That's about $31 on each hand.

Thanks for the work you put into that post. I suck at math, so it may take me a while to parse through the equations, but I'll get out my fingers and toes and start working at it.

In the meantime, am I right in guessing this equation doesn't take into account the idea of how many cards are burned, and instead assumes an infinite amount of times Lonely's going to be in that situation?

If so, and I think that's probably right because $31 per hand seems pretty "wasteful" on the cards side, is there any way we can import the idea of limited opportunities (or maybe incorporate some kind of cost Lonely has to pay to get to all those later possibilities, those being his HE losses while playing long enough to get to all of those good, $100 level counts)?