Greasy John: A very rare blackjack hand

[Greasy John]

Here?s something you?ll probably never see: a single-deck 5-card 21 comprising all four 5s. I?ve thought about this hand through the years and started to make some observations about it. This hand can only be made by starting with two 5s, hitting and getting another 5, and then hitting and catching an ace followed by the case 5. The ace can fall in no other position otherwise the player would stand.

Now in order to hit 5,5 the dealer would usually have to have a 10 or ace up, otherwise the player would probably double. And if the dealer had an A or 10 up he could not have a blackjack or the hand would be over. Also, with a 10 or ace up the counting player would stand a good portion of the time with 5,5,5,A (he would not hit the sixteen).

This situation is more rare than a royal flush, if I may use this analogy, for several reasons. First, a royal can be in one of 4 suits and can also be made on the draw. Also, there are 20 cards available to work with initially. Getting a blackjack hand with all four 5s which includes an ace as the second hit card, I would think, would be about 4 times as hard to achieve as being dealt a royal flush in any of four suits. (Yes, you would have a chance at any of four aces ( three if the dealer?s up card is an ace) to complete the 5-card 21, but it must be in the second hit-card position only.)

My math may be faulty, but I calculated that being DEALT a royal flush is one chance in 649,739. Being dealt one in a specific suit is about 1 chance in 2,598,956. And since the player would normally hit a 5,5 about one-third of the time ( against a 10 or ace where the dealer has no blackjack) the odds of the 5,5,5,A,5 hand are about one in 7,796,868. And even then the player has only about a 53% chance that he would hit a 5,5,5,A (sixteen) because of the count. So, the chances are about 1 in 14,711,071 of this scenario playing out.

If a player played 70 hands an hour he would have to play continuously for about 24 years for this scenario to happen.

Greasy John

[nightstalker]

Num hands required for getting 5,5,5,5,A on top of single deck =
inverse (4/52 * 3/51 *2/50 *1/49 *4/48 )

= 13*17*25*49*12
= 3,248,700

assuming player hit hit hit hit

[Greasy John]

I think you overlooked a mathematical component of the scenario. That the ace MUST be in the fourth position
( the second hit card). Your mathematical analysis implies the ace could appear anywhere. I think you have to multiply your result by 5.

GJ

> Num hands required for getting 5,5,5,5,A on top of
> single deck =
> inverse (4/52 * 3/51 *2/50 *1/49 *4/48 )

> = 13*17*25*49*12
> = 3,248,700

> assuming player hit hit hit hit

[Don Schlesinger]

> I think you overlooked a mathematical component of the
> scenario.

Nope.

> That the ace MUST be in the fourth position
> (the second hit card).

That's why he DIDN'T include a factor of 5 in his calculation. He put each card in one specific spot only, which is correct.

> Your mathematical analysis
> implies the ace could appear anywhere.

No it doesn't.

>I think you have to multiply your result by 5.

Nope.

Don

[Greasy John]

#1, the sequence is not 5,5,5,5,A. It?s 5,5,5,A,5. But relating to the math is seems wrong to me. You can place the four fives and an ace in any order and come up with the same 3,248,700 :1 that nightstalker arrived at.

You could say that in video poker a royal flush, of say A,K,Q,J,10 in spades ( left to right) has the same chance of arriving (left to right) as an 10,A,Q,J,K of diamonds. But if you MUST receive the sequential, suited royal flush the chances are about 18,000,000:1. Whereas, any royal flush ( JoB with correct BS) is about 40,000:1. Wouldn?t a specific order increase the odds substantially? Maybe I?m missing something.

Greasy John

[Don Schlesinger]

"Maybe I?m missing something."

I explained what you're missing. His method of calculating the event was, in fact, for one specific order. Had it been for any order, the ace could have been in any one of the five spots, and he would have multiplied by five, which he didn't.

Don

[Greasy John]

Then nightstalker would have divided by 5 not multiplied. It's not 5 times harder to make the sequence in ANY particular order, it's 5 times easier. The point isn't: How many ways can you make the hand? The point is that making it in one specific order is harder than making it in any particular order.

Greasy John

> "Maybe I?m missing something."

> I explained what you're missing. His method of
> calculating the event was, in fact, for one specific
> order. Had it been for any order, the ace could
> have been in any one of the five spots, and he would
> have multiplied by five, which he didn't.

> Don

[Don Schlesinger]

> Then nightstalker would have divided by 5 not
> multiplied.

Maybe you missed "inverse" in his explanation.

> It's not 5 times harder to make the
> sequence in ANY particular order, it's 5 times easier.

Don't you think you're being a little patronizing "explaining" that to me?

He already calculated it for the single, particular order that you require. You keep missing the point and don't understand.

> The point isn't: How many ways can you make the hand?
> The point is that making it in one specific order is
> harder than making it in any particular order.

Last time, then I can't help you anymore. The probability of receiving the single, specific order of 5,5,5,A,5 in single deck is:

(4/52)(3/51)(2/50)(4/49)(1/48) = 1/3,248,700.

That is the correct answer that nightstalker gave. If you were trying to calculate it for any order, you would have to multiply the fraction above by five, making the value larger, and the hand easier to achieve.

If you don't understand this, or don't want to understand this, you're on your own at this point.

Sorry.

Don

[NightStalker]

> #1, the sequence is not 5,5,5,5,A. It’s 5,5,5,A,5. But
> relating to the math is seems wrong to me. You can
> place the four fives and an ace in any order and come
> up with the same 3,248,700 :1 that nightstalker
> arrived at.
> Greasy John

I got your point... I apologize for not being clear.
Yes, you are correct. I calculated the probability for getting 5,5,5,5,A (din't read your post completely)

But mathematically, probability of getting
5,5,5,5,A == 5,5,5,A,5

Num hands required for getting 5,5,5,5,A on top of single deck = inverse (4/52 * 3/51 *2/50 *1/49 *4/48 )
Num hands required for getting 5,5,5,A,5 on top of single deck = inverse (4/52 * 3/51 *2/50 *4/49 *1/48 )

As you can see, order MUST be specified. This is the probability of picking four 5's and 1A from a single deck of cards in SPECIFIED order irrelevant of order.

Comparing it with a specific example:
Number of hands to get a blackjack = 21
num of hands to get a suited blackjack = 84
num of hands to get a suited blackjack in hearts = 336
num of hands to get a suited blackjack in spades= 336
num of hands to get a suited blackjack in clubs= 336
num of hands to get a suited blackjack in diamonds= 336

I've given
number of hands required to pick either of following combinations:
5,5,5,5,A or 5,5,5,A,5 or 5,5,A,5,5 or 5,A,5,5,5 or A,5,5,5,5

If your question is to find the num of hands required for picking 5 cards(4 5's and an ace) out of 52 where order is irrelevant. Then you can divide number by 5 as probability will increase by 5. This is equivalennt to asking: Num of hands required for suited blackjack.

In short, my answer is same for any specific order.
5,5,5,5,A or 5,5,5,A,5 or 5,5,A,5,5 or 5,A,5,5,5 or A,5,5,5,5

Good luck..