Probability of a 20 losing to Blackjack - is this correct?

Hello, I am curious about the odds of getting a 20 and losing to blackjack.

4 standard decks. 208 cards

Total Hands: 208 × 207 = 43056 hands

Player Hard 20: 64 x 63 = 4032 hands
Player Soft 20: 32 x 19 = 512 hands
Player any 20: (4032 + 512) / 43056 = 284/2691 [ ≈ 10.5537% ]

Dealer 21(P hard 20): 62 x 16 = 992 hands
Dealer 21(P soft 20): 64 x 15 = 960 hands
Dealer any 21: (992 + 960) / 43056 = 122/2691 [ ≈ 4.5336% ]

Probability of both events happening:
(284 x 122) / (2691 x 2691) = 34,648 / 7,241,481 ≈ 0.4785% chance

This feels low to me so I'm not sure if I made a mistake somewhere along the way.

Can anyone verify that the work is correct or point out my error(s)? Thank you!

Let me consider an infinite deck game. The probability of a dealer blackjack is

8/169=0.0473.

The probability of a player two-card twenty is

18/169=0.107.

The probability of both events is

(8/169)x((18/169)=0.00504 =0.504%.

However, the probability of a known player 20 of any number of cards losing to a dealer Blackjack is 4.73%, but that hand can lose to a dealer 21 of multiple cards too.

I used to want to buy insurance with a 20 against a dealer Ace.

The odds of losing with a twenty seem to change depending on the size of the bet one makes. Nothing attracts a dealer BJ like a fish making a particularly large bet.

I didn't budget for losing on the dealer drawing to a 21 without a Black Jack.

I bought in for 20units ($200) and lost it all without winning a hand playing basic strat. (6 or 8decks)
I didn't lose 20 hands... more like ~17.
Some were either splits or doubles

edit:
doh.. misread title of thread

Quote: pbM9000

Hello, I am curious about the odds of getting a 20 and losing to blackjack.

4 standard decks. 208 cards

Total Hands: 208 × 207 = 43056 hands

Player Hard 20: 64 x 63 = 4032 hands
Player Soft 20: 32 x 19 = 512 hands
Player any 20: (4032 + 512) / 43056 = 284/2691 [ ≈ 10.5537% ]

Dealer 21(P hard 20): 62 x 16 = 992 hands
Dealer 21(P soft 20): 64 x 15 = 960 hands
Dealer any 21: (992 + 960) / 43056 = 122/2691 [ ≈ 4.5336% ]

Probability of both events happening:
(284 x 122) / (2691 x 2691) = 34,648 / 7,241,481 ≈ 0.4785% chance

This feels low to me so I'm not sure if I made a mistake somewhere along the way.

Can anyone verify that the work is correct or point out my error(s)? Thank you!
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Using your work as a template, here's how I would do it:
Total Hands Player: 208 × 207 = 43056 hands
Player Hard 20: 64 × 63 = 4032 hands
Player Soft 20: 2 × 16 × 16 = 512 hands

Total Hands Dealer: 206 × 205 = 42230 hands
Dealer 21(P hard 20): 2 × 62 ×16 = 1984 hands
Dealer 21(P soft 20): 2 × 64 × 15 = 1920 hands

Probability of both events happening:
4032 × 1984 / 43056 / 42230 + 512 × 1920 / 43056 / 42230 ≈ 0.4940% chance

But if I only take insurance with a 20, I'll only win the insurance bet 4/13ths of the time.