A question for math-minded folks...
If the prior probabilities of a given card landing on the first box, second box, third box are 40%, 15%, 10%, respectively,
what is the posterior probabilities of the first box and second box when the 40% event did not happen??
All help appreciated !!
chuckybaby
In a normally cutted game, the probability of receiving a single given card will never be as high as 40%.
The probability of receiving a 10,J,Q,K may reach 40% and will go up if the next card is a non-ten.
So if the initial probability of receiving a ten value card is 40 of 100 cards in the first box, it will become 40/99 in the second box.
What am I missing in your question? Is it related to shuffle tracked segments?
Last edited by Secretariat; 06-25-2024 at 08:03 AM.
So, a total of 65%.Originally Posted by chuckybaby
Did you mean this? Doesn't make sense. Did you mean to say "of the second and third box when the 40% event did not happen?"? If you did, then you have 65% left for those last two boxes, and you're saying that box two has a 50% greater probability of hitting than does box three. The new probabilities would then be 39% for box two and 26% for box three.
Does that make more sense to you?
Don
Here's example to show relationship:
Single deck, S17, SPL1, SPLA1, NDAS
Overall EV best strategy: +.0248%
EV given up card A: -36.03%
EV given up card 2: +9.995%
EV given up card 3: +13.59%
EV given up card 4: +18.19%
EV given up card 5: +23.35%
EV given up card 6: +23.94%
EV given up card 7: +14,53%
EV given up card 8: +5.441%
EV given up card 9: -4.383%
EV given up card T: -17.08%
Suppose you want EV given up card is not T
EV given up card is not T = (overall EV - prob(T) * (EV given up card is T)) / (1 - prob(T))
EV given up card is not T = (.0248% - 4/13*(-17.08%))/ (1 - 4/13) = +5.29%
Note that prob(A)+prob(2)+.....+prob(T) = 1
This needs to be true for calculation to be valid.
Hope this helps.
k_c
"Perfection is the enemy of success."
-Elon Musk-
But he wasn't asking for EVs. He was asking for the probability of a card landing in one of the first three boxes and then the probability after one sees that box 1 isn't successful that it will land in box 2 or box 3. Do you have a different answer to that specific quustion than the one I offered?
Don
Here's how I would look at it.
Successes of card landing on spot 1, spot 2, spot 3, eventually somewhere
success(1) = .4
success(2) = .15
success(3) = .1
success(>3) = .35
if trial 1 fails
success(1) = 0
success(2) = .15/(1-.4) = .25
success(3) = .1/(1-.4) = .16666
success(>3) = .35/(1-.4) = .58333
Sum of probabilities need to equal 1.
k_c
"Perfection is the enemy of success."
-Elon Musk-
Yep, I believe this is the correct approach. Therefore, if the Ace didn't land in the first spot, the probabilities increase for the following spots:
25% (instead of 15%) for the second spot, 16.7% (instead of 10%) for the third spot, and 58.3% for any other spot.
Sincerely,
Cac
Luck is what happens when preparation meets opportunity.
At the risk of disagreeing with all of the above, I'll defend my point of view. Suppose this is a shuffle-tracking play consisting of a single round. Nothing beyond the one round matters. And suppose I assess my ability to call the correct box exactly as the OP stated, namely 40%, 15%, and 10%. If I fail in my attempt to predict with those probabilities, I have made a gross error in calculation, and, therefore, there is no remaining probability that it appears eventually somewhere else in this round. I don't care about "eventually"! It's probability of appearing THIS ROUND in box 4, 5, 6, or, 7 is zero. I have completely failed in my attempt to locate the card in question.
And so, the probabilities are exactly as I presented them early. 39% for the second spot, 26% for the third, and ZERO for the others in this round.
Don
Sorry..I think I should clarify...it isn't about "missing" a card.
Let's take a step backward...
The 40%, 15% 10% probabilities are to do with the card displacement due to a certain shuffle...
When I say 40% it means there is zero displacement between the key and the target
15% means that 1 random card has appeared before the target
10% means that 2 random cards have appeared before the target
I need to know the change in probability of Boxes 1 and 2 after I've seen from the previous hand how many random cards have appeared after the key
in order to know how much to bet on Boxes1 and 2 to catch the target
[40%, 15%, 10%], where did these figures come from?
Usually, these figures come from actual table records, computer simulations, mimicking the dealer's shuffle at home, or from other people's studies.
Why 65% totally instead of 100%?
We record data this way: when the key card comes out, we take a record of the target card appearing for the first card, second card, third card, or elsewhere. In this case, we get 40%, 15%, 10%, and 35%. A sequencer knows this well, sometimes the target card doesn't appear in the first three cards, but shows up on the 4th, 5th, 6th, 7th, or even not at all!
A) 40% 1st,x,x,x
B) 15% x,2nd,x,x
C) 10% x,x,3rd,x
D) 35% x,x,x,else
(assuming one deck game, so no more than one target card in the deck)
If key card is the last card, the following probabilities apply: A,B,C,D
If key card + 1*non target card came out,.. B,C,D
If key card + 2*non target card came out,.. C,D
If key card + 3*non target card came out,.. D
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