It says odds of busting with 14 is 56% but 6/13 = 46%, why?

[LukizKing]

By the 6/13 + (6/13 * 1/13) + (6/13 * 1/13) calculation we get the odds of all the possible outcomes until we hit the soft 17, WHICH IS ~ 53%. Does that make sense?

In the combinatorics, '*' means AND, while '+' means OR. Having 6/13 as our starting probability, we add the probability of getting the A or 2, resulting in the ~ 53% for the player or the dealer to bust.

6/13 chance of hitting on 14 and busting.
(6/13 * 1/13) chance of hitting on 14, getting 15, hitting and busting.
(6/13 * 1/13) chance of hitting on 14, getting 16, hitting and busting.

Correct me if I'm wrong

[PinkChip]

6/13 chance of hitting on 14 and busting.
(6/13 * 1/13) chance of hitting on 14, getting 15, hitting and busting.
(6/13 * 1/13) chance of hitting on 14, getting 16, hitting and busting.

But in the second case (dealer gets an Ace, with probability of 1/13):
when dealer has 15, his probability of busting by the next card is not 6/13 anymore.
If he has Hard 15, there are now 7 cards which would him bust: 7,8,9,10,J,Q,K.
So you must compute 1/13 * 7/13 in the second case.

Analogously, 1/13 * 8/13 in the third case, since if the dealer has 16,
he busts with 8 cards: 6,7,8,9,10,J,Q,K.

It gets further complicated, since in the second case, if the dealer has 15,
he could draw another Ace (1/13) and then get into the third case.
In my opinion, the result of these overall four cases is

6/13 (start on Hard 14 and draw 8,9,10,J,Q,K) +
1/13 * 7/13 (start on Hard 14, then draw an Ace to Hard 15, then draw 7,8,9,10,J,Q,K) +
1/13 * 8/13 (start on Hard 14, then draw a Two to Hard 16, then draw 6,7,8,9,10,J,Q,K) +
1/13 * 1/13 * 8/13 (start on Hard 14, then draw two Aces to Hard 16, then draw 6,7,8,9,10,J,Q,K)

= 6/13 + 7/169 + 8/169 + 8/2197
= 1014/2197 + 91/2197 + 104/2197 + 8/2197
= 1217/2197
= 55.4 %

[LukizKing]

6/13 (start on Hard 14 and draw 8,9,10,J,Q,K) +
1/13 * 7/13 (start on Hard 14, then draw an Ace to Hard 15, then draw 7,8,9,10,J,Q,K) +
1/13 * 8/13 (start on Hard 14, then draw a Two to Hard 16, then draw 6,7,8,9,10,J,Q,K) +
1/13 * 1/13 * 8/13 (start on Hard 14, then draw two Aces to Hard 16, then draw 6,7,8,9,10,J,Q,K)

= 6/13 + 7/169 + 8/169 + 8/2197
= 1014/2197 + 91/2197 + 104/2197 + 8/2197
= 1217/2197
= 55.4 %

Thank you, mate I was thinking of it the same, then made the edit because I thought 6/13 stays the same if we start on 14 but the thing is, we've already counted that probability in by writing 6/13 and your calculations make sense to me because we can't assume the probability of bust remains static no matter what. In that case, the resources on the internet were correct about the 56% (excluding the second screenshot of mine).

Thank you very much PinkChip If the other people agree with the calculations, please let me know

[PinkChip]

Thank you, mate I was thinking of it the same, then made the edit because I thought 6/13 stays the same if we start on 14 but the thing is, we've already counted that probability in by writing 6/13 and your calculations make sense to me because we can't assume the probability of bust remains static no matter what. In that case, the resources on the internet were correct about the 56% (excluding the second screenshot of mine).

Thank you very much PinkChip If the other people agree with the calculations, please let me know

You are welcome. I also had to go through my calculations and correct them several times. Should there still be any error, Don probably will tell us :-)

[DSchles]

But in the second case (dealer gets an Ace, with probability of 1/13):
when dealer has 15, his probability of busting by the next card is not 6/13 anymore.
If he has Hard 15, there are now 7 cards which would him bust: 7,8,9,10,J,Q,K.
So you must compute 1/13 * 7/13 in the second case.

Analogously, 1/13 * 8/13 in the third case, since if the dealer has 16,
he busts with 8 cards: 6,7,8,9,10,J,Q,K.

It gets further complicated, since in the second case, if the dealer has 15,
he could draw another Ace (1/13) and then get into the third case.
In my opinion, the result of these overall four cases is

6/13 (start on Hard 14 and draw 8,9,10,J,Q,K) +
1/13 * 7/13 (start on Hard 14, then draw an Ace to Hard 15, then draw 7,8,9,10,J,Q,K) +
1/13 * 8/13 (start on Hard 14, then draw a Two to Hard 16, then draw 6,7,8,9,10,J,Q,K) +
1/13 * 1/13 * 8/13 (start on Hard 14, then draw two Aces to Hard 16, then draw 6,7,8,9,10,J,Q,K)

= 6/13 + 7/169 + 8/169 + 8/2197
= 1014/2197 + 91/2197 + 104/2197 + 8/2197
= 1217/2197
= 55.4 %

Nicely done. Obviously, this is an infinite-deck calculation, where you don't account for the two cards that comprise the original 14, nor the fact, in the last part, that, since you've already drawn one ace to get to 15, the probability of drawing the second ace to get to 16 is now less than 1/13.

Don

[PinkChip]

Nicely done. Obviously, this is an infinite-deck calculation, where you don't account for the two cards that comprise the original 14, nor the fact, in the last part, that, since you've already drawn one ace to get to 15, the probability of drawing the second ace to get to 16 is now less than 1/13.

Don

Thank you Don. Yes, I referred to the simplification of infinite decks from the OP. With changing deck composition, the combinatorial analysis gets much more complicated.

I once wrote a program which computes some probabilities like dealer bust rate, depending on number of decks (see the other thread on number of cards per hand). It doesn't take into account player hands, just the dealer hand. But even this requires a tree with a depth of 12 or 13 levels, since the dealer might draw that many low cards (Aces and Twos) before busting.

Of course these dealer hands are very rare. Mostly the dealer hand will comprise 2, 3 or 4 cards, so the path in these cases can be aborted because its subtree needs not be traversed. The program counts the number of pathes it had to traverse, and luckily it was not 10^12 or 10^13 but only about 50000 (for S17 rule) or 70000 (for H17 rule).

It guess it would be impossible to compute the play through a full 6 deck shoe with 312 cards in this manner, which could produce exact results, in contrast to a Monte Carlo simulation.