How to calculate math on a new bet

I know it can be complicated or it can sometimes be fairly simple but either way it is out of the realm of knowledge for most of us. I am talking about the math. It is a necessary evil that must be done but is a daunting task for most. Could anyone possibly let me know how to calculate the math on a very simple blackjack bet that doesn't exist just yet. Let's say that I am playing a traditional game of blackjack and I am dealt a winning blackjack with a Ace and a Ten and am paid 3 to 2. Now what if the casino would let me make one additional bet that allows me to draw on additional card. If that card is another 10 I will win a second blackjack. If it is not a 10 the house would win. So basically I would win 2 blackjacks off of just one ace. How would I calculate the house edge on that second bet?

lehmorr,

Welcome to the forum! I think there are just a couple of things needed to give you an answer.

1. Verify the 2nd Blackjack pays 3:2 on your bet, same as the first one.
2. How many decks are you using?

In general, I would say that if the 2nd bet only pays 3:2, that it's a very high house edge, (around 55%) because there are 9 cards that lose your bet for every 4 that win, and you've already used up one of the 10's. I think your bet would have to pay 3:1 or thereabouts to be even in the neighborhood of a good bet (8% or so?). But once you answer the 2 questions above, we can probably give you an answer. Disclaimer: by "we", I mean the math guys, though I'll give it a shot. I'm not in their league.

I cant do the math either. Card game math is a pain in the ass because you have to consider card removal. And, for BJ, a variety of multiple deck sizes. So, for simple math, we think about the infinite deck, which is why Babs said 9/13 of the time it looses.


If, as Babs believes, it only pays if you draw a BJ, then it's a very high edge like she said.

However, if it plays and pays like a normal/second hand, then it has the same edge as a Free Ace coupon, which I believe is a player advantage coupon.


Food for thought.

Quote: lehmorr13

I know it can be complicated or it can sometimes be fairly simple but either way it is out of the realm of knowledge for most of us. I am talking about the math. It is a necessary evil that must be done but is a daunting task for most. Could anyone possibly let me know how to calculate the math on a very simple blackjack bet that doesn't exist just yet. Let's say that I am playing a traditional game of blackjack and I am dealt a winning blackjack with a Ace and a Ten and am paid 3 to 2. Now what if the casino would let me make one additional bet that allows me to draw on additional card. If that card is another 10 I will win a second blackjack. If it is not a 10 the house would win. So basically I would win 2 blackjacks off of just one ace. How would I calculate the house edge on that second bet?

This bet does exist as Super 31 Black Jack,and the house edge is high. (t*1.5-x)/(f+x) where t are the number of tens and face cards left and x is the number of non ten and non face cards left gives the player EV. -0.2338709677 for 6 decks without considering the dealer's card.

Quote: miplet

This bet does exist as Super 31 Black Jack,and the house edge is high. (t*1.5-x)/(f+x) where t are the number of tens and face cards left and x is the number of non ten and non face cards left gives the player EV. -0.2338709677 for 6 decks without considering the dealer's card.

What's that bit about First Ace trying to say-- a side bet that you can place if your first card is an ace, betting that you'll get two 10s in a row?

Absent card removal, this is just like insurance, except it pays worse.

Given an infinite deck, the return to the player is 16/52 * 1.5 - 36/52 * 1 = -23.077%.

Quote: miplet

This bet does exist as Super 31 Black Jack,and the house edge is high. (t*1.5-x)/(f+x) where t are the number of tens and face cards left and x is the number of non ten and non face cards left gives the player EV. -0.2338709677 for 6 decks without considering the dealer's card.

Cool that you found the Super31 bet, miplet! If I'm reading that correctly, you're rolling over the base bet for an additional 10 card and wagering the entire amount rather than a separate sidebet to get that EV, right? If I'm understanding the OP, his sidebet is, after a BJ, you can wager the sidebet that the next card will be another 10, but your BJ gets paid separately. Or is that what you're actually calculating, without regard to the base bet that won 3:2?

Quote: CrystalMath

Absent card removal, this is just like insurance, except it pays worse.

Given an infinite deck, the return to the player is 16/52 * 1.5 - 36/52 * 1 = -23.077%.

So the more than 2 to 1 chance of a loss is mitigated by the 3:2 payout for a win to this amount. Ok, I can go with that, thanks, Crystal Math!

If the second bet paid 2:1 (4:2), the HE would be -7.692. A bit fairer.

" Given an infinite deck," Walmart here in Grand junction doesn't carry them.

Quote: Buzzard

" Given an infinite deck," Walmart here in Grand junction doesn't carry them.

Of course it carries them.
Just buy any single deck any brand, each time you draw a card write down the card, put the card back to the deck, shuffle the deck and draw the next card.
Repeat procedure for second card and so on.
For infinite amount of hands repeat procedure infinite amount of times

As to the original question it depends on number of decks. But let's assume, for simplicity, it's one deck.

You have Ace and Picture, that leaves 50 cards, 15 are pictures (you've used 1 of 16) and the other 35 aren't. There are now two ways of thinking about it.
(a) Ignore the dealer's upcard and what is the next card out. Then the chances of being a Picture are 15/50 = 30%.
(b) Consider the dealer's upcard, then either it's a Picture (in which case the numbers are 49,14,35) or not (49,15,34).
(i) p (dealer=Picture) = 15/50, p (player=Picture) = 14/49
(ii) p (dealer~=Picture) = 35/50, p (player=Picture) = 15/49.
Total = (15*14+35*15)/(50*49) = (210+525)/(50*49)= (15*49)/(50*49) = 15/50.

The actual odds tend towards 4/13 as the number of decks increase and the effect of already having a Picture diminishes.