Five 4's - rare? Wizard?
Too bad you only pushed.
Quote: paisielloWell there's 24 of them in a six deck shoe and you got five of them. Assuming an inifinite deck then a good approximation would be (1/13)^4 = 0.0035% or about 35 chances in 1 million (approx.). Pretty rare indeed.
Too bad you only pushed.
Yes, condolences on pushing, where is the 5-card charlie rule...lol
And paisiello, you could be more accurate here without too much trouble (the infinite deck assumption is a little off here), start at 311 cards left because of dealer upcard is always >6 here, also I am calculating just 5 4s to start with (you calculated 5 of any rank).
(24/311)*(23/310)*(22/309)*(21/308)*(20/307) = 1.8106787e-6 = 1 in 552k
But...OP asks the same question if he also gets 5 Aces (7/8ths of the time 7-K up...1/8th of the time Ace up: Prob. is 1 in 698k w/Ace up), 2s, and 3s.
Given all possible scenarios, 5 Aces, 2s, 3s, or 4s, the probability of any of these happening is roughly four times more likely or 1 in 139k.
Quote: BozAny issue with snapping the picture?
I am very friendly with all the staff and they let me "check a text message" wink wink nod nod while holding the phone up so I could see due to a "glare"
Quote: s2dbakerI think the fact that you pushed with that hand is a crime!
Yeah the cliché of "you worked hard that hand" was heard immediately upon the dealer flipping over her hole card
Long time ago AC used to pay for Four 5's.
Quote: tringlomane
(24/311)*(23/310)*(22/309)*(21/308)*(20/307) = 1.8106787e-6 = 1 in 552k
But...OP asks the same question if he also gets 5 Aces (7/8ths of the time 7-K up...1/8th of the time Ace up: Prob. is 1 in 698k w/Ace up), 2s, and 3s.
Given all possible scenarios, 5 Aces, 2s, 3s, or 4s, the probability of any of these happening is roughly four times more likely or 1 in 139k.
The original poster asked about ANY card, not just 4's. So my estimate is 0.0035% or 1 in 28,500.
If you want an exact number then the answer would be:
(23/311)(22/310)(21/309)(20/308) = 0.0023% = 1 in 43,000.
This is off from your estimate of 1 in 139,000. I didn't follow what your reasoning was there.
Quote: paisielloThe original poster asked about ANY card, not just 4's. So my estimate is 0.0035% or 1 in 28,500.
If you want an exact number then the answer would be:
(23/311)(22/310)(21/309)(20/308) = 0.0023% = 1 in 43,000.
This is off from your estimate of 1 in 139,000. I didn't follow what your reasoning there.
Reasoning is based on blackjack rules. If a player gets 3 8's for example, he won't be able to ask the dealer to hit him again. Sorry for not clarifying that earlier. Yeah if you are allowed to hit without restriction, then 1 in 43k would be correct.
Quote: bbvk05Fun, but you shouldn't be hitting a 16 with four 4's in it vs. a ten.
He shouldn't be playing with 66% penetration either.
Quote: bbvk05Fun, but you shouldn't be hitting a 16 with four 4's in it vs. a ten.
Sure I should, no 5's had been out in the shoe yet.
Quote: 1BBHe shouldn't be playing with 66% penetration either.
and where are you finding better than that?
Quote: 1BBBorgatta, Foxwoods, Pennsylvania, pit 1 at Revel.
As I have not played at Borgata, Foxwoods or PA, I can only attest that I have never seen better than 2/3 penetration at Pit 1 Revel - do you have a specific dealer? I have discussed this with floor managers and pit bosses that they shuffle check and look at the penetration - you must have caught a great situation at a time when nobody bothered to shuffle check
Quote: 1BBHe shouldn't be playing with 66% penetration either.
You're right. He should be playing with a 50% penetration, or on a CSM. Or playing dice or UTH.
24/312*23/311*22/310*21/309*20/308 is correct for "off-the-top" calculations per rank, which is used here regularly.
Since its possible that A-2-3-4 are the only possibilities of 5-in-a-rowthe formula is multiplied by 4. This answer is about 1 in 140318.
If one counts the burn card one MUST include the chance a A-2-3-4 gets burned, complicating the math quite a bit.
Quote: PaigowdanYou're right. He should be playing with a 50% penetration, or on a CSM. Or playing dice or UTH.
CSM makes sense for the casino. 50% pen is flushing money down the toilet.
Quote: aceofspadesAs I have not played at Borgata, Foxwoods or PA, I can only attest that I have never seen better than 2/3 penetration at Pit 1 Revel - do you have a specific dealer? I have discussed this with floor managers and pit bosses that they shuffle check and look at the penetration - you must have caught a great situation at a time when nobody bothered to shuffle check
I have observed pit 1 over a dozen times since April 2nd and found the pen to be dealer dependent. I was able to find close to 75% all but a couple of times the last being Columbus Day weekend. I also played twice with a very weak dealer who cut off just over 5 out of 6 decks as well as making other mistakes. I have not seen that dealer since early June.
Actually I was watching Ace when this happened. I was charting the dealer. Has Ace done so, he would have know to split those
4's !
Quote: PaigowdanYou're right. He should be playing with a 50% penetration, or on a CSM. Or playing dice or UTH.
At least you didn't say Keno or Big Six. I imagine that not all casinos want the extra shuffling that 50% pen brings or the fees for the CSMs.
Quote: 98Clubs
If one counts the burn card one MUST include the chance a A-2-3-4 gets burned, complicating the math quite a bit.
1/13th chance the burn card is the card we need:
(1/13)*4*(23/311)*(22/310)*(21/309)*(20/308)*(19/307) + (12/13)*4*(24/311)*(23/310)*(22/309)*(21/308)*(20/307) = 1 in 140318 for 5 Aces, Twos, Threes, or Fours ignoring dealer's upcard.
So, no it doesn't matter. Think of it this way, is the probability of cards 1 thru 5 being 5 4's the same for cards 2 thru 6, or cards 207 thru 211, etc?
