I will be playing 100 hands for $10.00 using basic strategy. What is my expected loss in a game with a house advantage of 0.4%?
Does the same figure apply if playing 2 hands to a total of 100?
Bill k,
Flat-betting $10/hand for 100 hands at a house edge of 0.4% will cost you, on average, ($10/hand)*(100 hands)*(0.4/100) = $4.
If you play 2 hands per round, then your expected loss will double, to $8.
Hope this helps!
Dog Hand
...you could lose hundreds of dollars, or win hundreds of dollars in that short a time period flat betting, basic strategy. This game can be vicious at times.
Good luck,
Cobbson
OldCootFromVA,
When Bill k wrote "Does the same figure apply if playing 2 hands to a total of 100?", I read that to mean "total of 100 ROUNDS", but I now see you point, that he might have meant "total of 100 INITIAL HANDS", which would of course be 50 rounds.
Bill k, if you did indeed mean 100 INITIAL HANDS, then your expected loss will, of course, be exactly the same as the original problem: $4.
What WILL change if you play 50 rounds of 2x$10 as opposed to 100 rounds of 1x$10 will be the DISTRIBUTION of your results, due to the "covariance" of playing simultaneous hands.
Here's a simple illustration of what I mean. Say you're playing an 8-deck shoe game, and rather than play 100 rounds of 1x$10, you play ONE round, with 100 spots in play... did I mention that this is a VERY LARGE TABLE? ;-)
Now if the dealer has a BJ, you won't win ANY of your hands (since you're playing B.S., you would NOT take "even money" on any of your naturals). You'll probably have about 5 naturals yourself, so on this round you'll lose 95x$10 = $950.
On the other hand, if the dealer has a "6" showing, you'll be doing lots of splits and double downs, and you won't take the chance to bust any of your hands (since you're playing B.S.). In all, you'll have close to 200 bets riding on the felt. If the dealer subsequently busts, you'll win ALL of your hands, and thus you'll win probably around $2000.
Now, the first case here will occur around one time in 21 tries (actually, the probability is 2*32*128/(416*415) = 0.04745... or once in 21.074... tries), while the second will occur roughly one time in 31 tries (the dealer will have a "6" 1/13th of the time, and will subsequently bust about 42% of the time, depending on the house rules for Soft 17). Therefore, if you play 100x$10, you'll often have results of -$950 and +$2000. On the other hand, if you play 100 rounds at 1x$10, your chances of losing EVERY unnatural hand are vanishingly small, as are your chances of winning EVERY hand. Thus, playing 100 rounds at 1x$10, you'll essentially NEVER end up at -$950 or at +$2000.
Hope this helps!
Dog Hand
...if the reason for the question was to meet the playing requirement for some kind of bonus hustling.
