how %dealler and players bust same time?
- Thread starter KOLAN
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Blue Efficacy
Well-Known Member
If you're playing heads up, the % is zero.
If you're wondering how often you bust first, and then the dealer goes on to break afterwards, page 50 of Don Schlesinger's Blackjack Attack III has data from which you can derive the answer.
It shows that the dealer's bust percentage is 28.2%, while a basic strategy player's bust percentage is 15.9%. In general, I believe you can just multiply the two (.159 x .282) to find that you will have busted your own hand and lost first, only to find that the dealer broke anyway on 4.5% of all your hands.
This of course, is across the entire spectrum of player totals and dealer up-cards. You could get more accurate info on specific cases by multiplying your own chance to bust your total x the dealer's chance to bust her up-card. When you have 16 against a 7 for example, if you hit, you'll bust 61 times out of 100. Of those 61 busts, the dealer will go on to break 26%, or 16 of those times. So with that hand, there's a 16% chance that you'll both end up breaking.
KOLAN
Well-Known Member
thanks i gonna check in book :1st::band:
KOLAN
Well-Known Member
i belive if reduce when players bust it is big+
jack.jackson
Well-Known Member
lemme know how liked it. i was thinkin about purchase :1st::2nd::3rd:
That assumes the two probabilities are independent which is not a good assumption to make because the two hands are very much correlated. For example, a basic strategy player will never bust against a dealer 4,5, or 6 which is precisely when the dealer is most likely to bust. Without doing the math, I'd expect the probability of both the player and dealer busting to be much lower than 4.5% because in general basic strategy avoids busting when the dealer is more likely to bust and risks busting when the dealer is less likely to bust.
You raise a provocative question. Even so, it may still average out very close to correct by just multiplying the two bust frequencies together. Notice that the vast majority of simultaneous busts will occur when the player hits a stiff against the dealer's 7 thru Ace up, and "bust-bust" happens roughly 12% of those times. You'll face one of these "decisions" (could contain more than two cards) about 31% of the time (according to hand decision distribution charts). So it looks like you'll get "bust-bust" on about 3.7% of all your hands from "stiff vs. big up-card" alone. The few other potential "bust-bust" situations would be 9 vs. 2 -- 12 vs. 2 or 3 -- 9 vs. 7 thru Ace -- 10 vs. 10 or Ace, and 11 vs. Ace.
QFIT
Well-Known Member
Sorry, makes no sense to me. As Blue Efficacy said
"If you're playing heads up, the % is zero."
KOLAN
Well-Known Member
whot about wining % when player take card 12,13,14,15,16, vs 7,8,9,10,A
i think it is les dem 3,7%
I'm assuming Kolan was not playing heads up, and somebody else at the table has finally stood, forcing the dealer to play out her hand.
QFIT
Well-Known Member
You aren't going to bust unless the dealer has a lower than usual probability of busting.
To Guynoire and Norm: I see your points now. Just multiplying out the player's and dealer's overall respective bust percentages will lead to an errant answer due to the hands with which the player will never hit or bust.
Perhaps only ironically, the real answer may actually fall close that 4+% figure because of the high frequency (around 31%) of those relatively few hands where "bust-bust" will occur 10% to 16% of the time. Those may have been the hands Kolan was plagued by and focused on
To Kolan: I'm sorry, but I guess an accurate answer would have to involve multiplying the individual appearance frequency of each player's hitting hand x its bust frequency x the appearance frequency of the associated dealer's up-card x that bust frequency -- and then adding them all together. It's more work than I care to go thru, but if anybody wants to supply it, I'd be interested in the answer.
Perhaps only ironically, the real answer may actually fall close that 4+% figure because of the high frequency (around 31%) of those relatively few hands where "bust-bust" will occur 10% to 16% of the time. Those may have been the hands Kolan was plagued by and focused on
To Kolan: I'm sorry, but I guess an accurate answer would have to involve multiplying the individual appearance frequency of each player's hitting hand x its bust frequency x the appearance frequency of the associated dealer's up-card x that bust frequency -- and then adding them all together. It's more work than I care to go thru, but if anybody wants to supply it, I'd be interested in the answer.
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