Can this bet have +EV

[MHoydilla]

I was in a local vegas casino and saw this prop bet at a blackjack table 15-1 if you bet the next hand you get BJ (A + T). This is a six deck game with 5 decks being dealt. I can count the Tens and Aces at what point would this be a profitable bet.

[crazy canuck]

I'll assume a one deck game for simplicity.

A=number of aces
T=number of ten valued cards
N=number of decks in the card

Probability of getting a BJ right of the deck =

2*(A/N)*(T/N-1)=1:20.7

Now suppose all the cards that are dealt in the beginning are not aces and not tens, so we want determine how many cards do we need to be left so the bet is +EV.

So we need

2*(A/N)*(T/N-1) > 1:15 for the bet to be +EV.


or

2*(4/N)*(16/N-1) > 1:15 => N=45

To summarize if the first 7 cards that come off the deck are non aces or tens the bet is +EV!

This is promosing so far, but we need to consider how often this opportunity arises.

The prob. of the first 7 cards that come off the deck are not aces or tens is given by:

(32/52)*(31/51)*(30/50)*(29/49)*(28/48)*(27/47)*(26/46)=1:40.


So one time in 40 we will have a +EV situation right off the deck. In general I think we'd have to do a simulation to see how often the situation is +EV. I mean, what happens if there is half or one third of the deck is left. Not too hard of a code to write tho.

However, suppose the first 10 cards that come off are not aces or tens. The prob of this happening is:

(32/52)*(31/51)*(30/50)*(29/49)*(28/48)*(27/47)*(26/46)* (25/45)*(24/44)*(23/43)=1:240.

and lets see what return our bet produces.

The odds of getting a BJ now is

2*(4/42)*(16/41)=13.5

so the return is 15/13.5= 1.11 for a low 11% return.

So even considering such a relatively rare opportunity our investment only produces a 11% return. So it is not very likely that the bet is exploitable if you consider losing money at the table while waiting for the opportunity to come up and including variance and time value. So don't quit your day job /images/graemlins/smile.gif.