December 28th, 2018 at 6:04:08 AM
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Quote:mustangsally

see

Sally, I was just making an Ask the Wizard question out of this and was thinking that with 120 total hands, what we should solve for is T^119. That will show us the probability of being in each state 119 hands after the first hand.

That said, I'm changing my answer to 0.0001402719. Do you agree?

It's not whether you win or lose; it's whether or not you had a good bet.

December 29th, 2018 at 12:31:27 PM
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I disagree.Quote:WizardSally, I was just making an Ask the Wizard question out of this and was thinking that with 120 total hands, what we should solve for is T^119. That will show us the probability of being in each state 119 hands after the first hand.

That said, I'm changing my answer to 0.0001402719. Do you agree?

because your transition matrix does actually start with a 0 state.

(I think I have been corrected by those that know on this enough to know)

instead of starting with a "start" state (they are the same as per my understanding)

even though P and P^1 are the same. that is why I think this works at 120.

Let us see with my transition matrix

matrix = P

some photos (orange cell is what power the matrix is raised to)

P^1 is after the first trial

P^2 is after the 2nd trial

P^3 is after the 3rd trial

shows we can get 3 in a row after 3 trials

17*221*221

from that transition matrix starting at state 0

we get the correct answer for 3 in a row over 3 trials

now, WITH A START STATE

P^120 is after the 120th trial

looks like the same result be starting at

state 0 or

state start

Happy New Year!

Sally

I Heart Vi Hart

December 29th, 2018 at 5:13:20 PM
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I also think it should be 120 iterations (for a probability of 1 in ~7,069) and I don’t understand the reasoning behind using 119.

On the very first hand you have a 1/17 chance of going to the state of 1 pair or 16/17 of state of no pair. And the 120th hand also counts in this case (it wouldn’t if the problem involved achieving success on that hand).

Just can’t see why you’d deduct 1 from 120.

On the very first hand you have a 1/17 chance of going to the state of 1 pair or 16/17 of state of no pair. And the 120th hand also counts in this case (it wouldn’t if the problem involved achieving success on that hand).

Just can’t see why you’d deduct 1 from 120.

It’s all about making that GTA

December 29th, 2018 at 6:01:23 PM
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I think it comes from this thread where the Wizard made a transition matrix that already started after trial 1Quote:Ace2Just can’t see why you’d deduct 1 from 120.

https://wizardofvegas.com/forum/questions-and-answers/math/32095-last-open-number-roulette/2/#post689723

then the explanation in the post below that by my Uncle

I agree on both those posts

Sally

Happy NEw Year!

(Mom wants to be in Connecticut instead of Miami for her Christmas present)

I Heart Vi Hart

December 29th, 2018 at 6:36:27 PM
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Thank you Sally and Ace for your corrections. I guess I was confusing the state at the beginning of the hand and the end of the hand.

This problem appears in Ask the Wizard column #311.

This problem appears in Ask the Wizard column #311.

It's not whether you win or lose; it's whether or not you had a good bet.