Odds of this strange rule

Picture coming but basically 7 handed video blackjack says if all 52 cards are gone then automatically win- I'm not sure this is mathematically possible with one deck- does anyone know for sure?

With lots of splitting. Whats the min and whats the max?

Are you sure it's not even more rare the 18 yos.
Split aces draw one card. Split up to 3 hands
If you average put all the cards on the table and subtract 20 for the dealer having two tens- that's an average of 320 point value on the table - even among 11 hands you would have more then 21 on each hand

Pretty sure the machine in question is this one:
https://www.ballytech.com/games/class3/video-slots/multi-play-blackjack-1116.html
https://www.youtube.com/watch?v=cBBABu7QKjM

This unit's a good way to get your head kicked in, in my experience.

Was hoping it was some new multi player VBJ. $1 to $500. bet $500++ in seat 7 everyone else bets $1 and intentionally eats up as many cards as possible.

Quote: Wizardofnothing

Picture coming but basically 7 handed video blackjack says if all 52 cards are gone then automatically win- I'm not sure this is mathematically possible with one deck- does anyone know for sure?

It could happen with as few as five players, each splitting to three hands.

Quote: Wizardofnothing

Are you sure it's not even more rare the 18 yos.
Split aces draw one card. Split up to 3 hands
If you average put all the cards on the table and subtract 20 for the dealer having two tens- that's an average of 320 point value on the table - even among 11 hands you would have more then 21 on each hand

Remember, it's real easy to go over 21 on 95% of all blackjack hands

I just did a quick proof-of-concept at my desk. I dealt out 7 spots which all split to three hands. I was able to make 21 hands that didn't bust and the dealer has 15. So it is mathematically possible but I have no clue of the odds!

But wait, is it mathematically possible using only one total desk? I'm missing something/ all cards would total 340 I can figure out how to spread that among 7 hands

And if it is possible any idea how astronomical the odds are?

Yeah I did it with one standard 52-card deck. I split 2,2,2; 3,3,3; 4,4,4; 5,5,5; 6,6,6; 7,7,7 and then dealt a random card to each of the 21 hands. Then after some trial and error and I was able to arrange the remaining cards such that no hand busted and the dealer needed to draw again. I feel like starting with split hands of 10,10,10 may be an easier way to construct this scenario.

Quote: TomG

It could happen with as few as five players, each splitting to three hands.



Remember, it's real easy to go over 21 on 95% of all blackjack hands

It's unlikely that going over 21 is going to help you --- except if it's the 6th and 7th hands when you know you're close to running out of cards. (You WOULD be counting down from 52 to see how many cards are left, of course. It's an unusual method of card counting.) It's probably only live hands that are paid.

But splitting 10s --- especially non-matching 10s if that's allowed --- up to 4 times each could make it happen. If the order at one point were T J Q K T J Q K that would eat up a count of 80 on only one hand. This could happen twice (not very likely --- but theoretically possible). Also if you split and resplit EVERY pair --- even 5s against a 10 which is about as bad a play as there is --- it doesn't strike me as impossible to run out of cards with 7 players. Also, if it's close and you have a number of "bad" results (like 16, 17, 14, 18, 18, 15 against a 10), you might want to hit naturals. You only want to risk busting if it will use up all the cards and promote otherwise not very good hands into winners.

What you have to remember is to think outside the box. You definitely don't want to be using basic strategy. Assuming you can't track the exact ranks of 52 cards in your head, you could have printed forms with all 52 cards on it --- such as A A A A 2 2 2 2 3 3 3 3 . . . . K K K K) and cross off cards as you see them. This will help you know how many and which cards are left. If most or all of the 10-cards are gone, you're more likely to hit 12s and 13s against any up card.

I don't know all of the rules (or any of them, for that matter) but with a modified strategy it strikes me as beatable.

I believe this unit also automatically stands on some (if not all) 21s, which'll also cut down on the number of cards you can take.

https://imgur.com/a/nJDtM


So basically I think this is near impossible

I'm not sure why you would want to do it unless as I said before you could bet very small on 6 spots and bet a bunch on the last hand for a guaranteed winner on all the hands. IIRC the VBJ machines you are talking about force to make equal bets and it's even money BJ.

Quote: Wizardofnothing

But wait, is it mathematically possible using only one total desk? I'm missing something/ all cards would total 340 I can figure out how to spread that among 7 hands

Again: it could happen with as few as four players if they all split to three hands. Obviously astronomical odds. But with seven players it would only require two to split to three hands to make 12 hands total, with 11 of them drawing until they go over 21. With those parameters, it starts to become fairly reasonable. Aren't the odds of at least three players being dealt a pair close to 2.5%? (and that should really be much higher given the ability to split 10-j or q-k, et cetera)

Quote: AxelWolf

I'm not sure why you would want to do it unless as I said before you could bet very small on 6 spots and bet a bunch on the last hand for a guaranteed winner on all the hands. IIRC the VBJ machines you are talking about force to make equal bets and it's even money BJ.

If not for even money on blackjack it should conceivably be a good play:

If fewer than three players are dealt pairs, play basic strategy and lose less than $5 (<1% on a $506 bet; plus earn points). If the three or more players are dealt pairs, everyone in the first six seats splits and hits burning as many cards as they can. When it's the seventh players turn, take as many hits as possible if there is a greater than 50% chance of using up all the cards. If he does use up all 52 cards, win $510 (or more accurately about $260 in EV since he could still win playing basic strategy). If there is only a very small chance of using up all the cards play basic strategy and only lose about $15

I saw a unit like this in a Vegas strip casino recently. The rules were awful. You couldn't even depth charge the game if you wanted to basically. I think the odds of dealing out all 52 cards is going to be exceedingly low. I forget, but I think the average number of cards a player hand takes is like 3.1 I want to say. Hell, let's round up to 4 for giggles... Okay, 7x4 = 28, plus the dealers hand (let's say another 4) then you're looking at 32 cards on average. You'd need to burn another 20 cards across your 7 hands, or possibly the dealer hand I suppose. Not saying it couldn't happen, but it would be rare, and in the mean time you'd get pounded on the machine.

Maybe there are better ones... I just know the one I saw was awful.

It is giving out a very high point return- so I'm trying to figure out what burning all the cards is worth- I'm already getting back almost 3 percent benefits

Quote: Wizardofnothing

It is giving out a very high point return- so I'm trying to figure out what burning all the cards is worth- I'm already getting back almost 3 percent benefits

FYI. ]the machines are rare nowadays. Someone could easily narrow down the field.

Quote: Wizardofnothing

It is giving out a very high point return- so I'm trying to figure out what burning all the cards is worth- I'm already getting back almost 3 percent benefits

Not enough in my opinion. Even money on blackjack.