Blackjack question for math junkies


Aug 3, 2002
Las Vegas
I deposited $100 for a $110 bonus at a casino. I played 5 $10 hands of BJ at a time. It ended up being 17 total hands for 75 dealt hands to me. I got blackjack one time, the dealer got blackjack 5 times. Given the probability of being dealt blackjack is 3.4% (yes i understand these arent alot of hands to show a fair game or not that really isnt my point) what was the probability a dealer could get 5 blackjacks in 17 hands? My one blackjack in 75 hands is only off by one so im not so curious about that. I dropped that $210 playing basic strategy is 75 hands One hand was only a $5 bet. About 7 times the expected loss in my math. Is this right? I just do not know how to figure out the math on a dealer getting roughly ten times the blackjacks expected.

Does this make sense? I do not intend on stating the casino I am just curious about the math probabilities.
I don't know that you can really use the 75 hand figure in this case, since you were playing 5 hands at once. I'd think for statistical reasons you'd need to count one dealer hand and 5 player hands as one hand since all 5 are playing against that one dealer hand. Of course saying that now makes it seem even MORE lopsided! Hopefully clayman or grandmaster will have a more detailed analysis.
pokeraddict said:
Thats what I meant. It is really 5 for 17 and my hands are not really relevent. My question is simply what are the probabilities of getting blackjack 5 of 17 times.

I'll give it a shot but don't believe anything I say.

First of all the probability of getting a BJ goes from 4.8265% in SD down to 4.7451% in 8 deck games. Approximately once in every 21 dealt hands. Where did you get 3.4% from?

If you played against 17 dealer upcards, I'm guessing 98 hands were dealt (16*6 + your last $5 hand and dealer's last hand.) So one would expect about 4.7 BJ's and 6 actually occurred. I doubt if this is very unusual.

Given that 6 BJ's occurred and assuming they are equally likely to be dealt to any of the 6 spots, the question becomes "What are the odds that one person out of 6 people will get 5 of 6 BJ's?" I'm thinking 11%.

If 5 BJ's occurred in just 17 hands, that would be an unusual event but I don't think that that is what happened here.

Were you really betting $50/round on a $210 bankroll? Even at multi-hand, I'm guessing that's a pretty high risk of ruin. Hopefully you realized that going in and aren't too disappointed you busted.

Perhaps posters more knowledgeable than I will add their comments too.
(Please correct me if this is wrong, I just started reading up on probability last week. Here is a
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if anyone's interested).

The odds of k successes in n trials where the probability of success is p can be calculated using the
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b(n,p,k) = COMBIN(n,k) * p^k * (1-p)^(n-k)

Using a probability of a blackjack of 0.0474895 for a six-deck game calculated by
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, the odds of 5 successes in 17 trials are:

COMBIN(17,5) * (0.0474895^5) * (0.9525105^12) = 0.000834

or almost 1 in 1000, so not especially unlikely. I had a bitchin' run with 10 blackjacks in 45 hands today so I was interested to see how unlikely that was (3 in 10,000).

Clayman, I think your method would also work but the 11% estimate is off, for example the chance of the dealer getting 5 out of 5 blackjacks would be (1/6)^5 = 1.3 in 10,000 so the chance of getting 5 out of 6 is only a little higher.
OK my math was off on the BJ%. Thank you all for the math here. 1 in 1000 cerainly isnt impossible. About the same as rolling a 2 twice in a row in craps which ive seen many times. Thank you all. I just wanted to know what the probability was.
This is nuts, I just played at this same casino. I won 43 1/2 bets playing $50 one hand vs the dealer. I got dealt one single BJ. I got a bunch of double, splits or both. In this time the dealer got 11 BJ's. Makes me shake my head at how this could be but if I win 43 1/2 bets in 114 hands it must be legit right? The dealer busted a whole bunch, just about every double I hit a rag on he busted.

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