Lecture Scene

[zengrifter]

This is a very old math "trick". It's better to switch cause the chance you made the right choise the first time is only 33.333%. After you switch it'll become 66.666%. It's hard to explain why but I always use the following example:

If there were 1000 doors and you would have to choose one the change of hitting the winning door is only 1/1000. Now, the quizmaster opens all other doors exept one (which he basically does in the 3-door quiz too only there's only one other door he can open). Now, he asks you if you'd like to switch. Of course the answer is YES. The chance of winning if you switched, now increased to 99.9%!

Hope this helps you a bit .

Yes, alright already, the dummy (me) gets it - just a little slow, you know! zg

 

[sagefr0g]

Yes, alright already, the dummy (me) gets it - just a little slow, you know! zg

so is it that your first choice was probably wrong? making it a good idea to eliminate it? so knowing that was probably a bad choice and monty showed the rest of the bad choices so you go with the other door left? so you might be right whatever percent of the time by switching but it's a higher pecentage of the time than your first choice woulda been?

 

[zengrifter]

so is it that your first choice was probably wrong? making it a good idea to eliminate it? so knowing that was probably a bad choice and monty showed the rest of the bad choices so you go with the other door left? so you might be right whatever percent of the time by switching but it's a higher pecentage of the time than your first choice woulda been?

Yes - now go to the head of the class and give us an example
of how this knowledge can be applied profitably in a prop bet. zg

 

[sagefr0g]

Yes - now go to the head of the class and give us an example
of how this knowledge can be applied profitably in a prop bet. zg

think of the dealer's hole card as an unopened door. the dealt cards as opened doors.
ok by the true count theorem http://www.bjmath.com/bjmath/counting/tcproof.htm (Archive copy) .
we know that "the expected value of the true count after a card is revealed and removed from any deck composition is exactly the same as before the card was removed, for any balanced count, provided you do not run out of cards."
we know that the dealer's down card is probably going to work out for her one way or the other cause that's how it goes.
so let's say we are playing heads up.
the true count is one card shy of going either up a tc or down a tc.
the dealers down card has not yet been shown. it's our turn to make a play decision.
we know the true count. we can expect that the dealers down card will not change the true count. it's like knowing what the dealers hole card is most likely to be.

ok am i
teachers pet now?........or class clown?

 

[mdlbj]

I think the goat is + EV.

 

[gibsonlp33stl]

Simple way to think about it.

For anyone still not getting the Monty Hall example...here is the easiest way to understand that switching is truly the correct choice.

Assume there is door A, door B, and door C. For this example lets assume doors A and B have goats, and door C holds the prize.

For the first scenario, lets assume that you always stick with the first door you choose. So choosing randomly a person will choose A 33.3% of the time, B 33.3% of the time, and C 33.3% of the time...so that person will get the prize exactly 33.33% of the times.



For the second, more complicated scenario, lets assume the contestant always switches to the other door.
So 33.3% of the time the person guess A, the host will show that B is a goat, and the person will switch his door to C and win.
33.3% of the time the person guesses B, the host will show that door A is a goat, and the person will switch to door C and win.
33.3% of the time the person will guess door C, the host will show that either A or B is a goat, the person will switch to either A or B and lose.
Therefore by switching the person will win 66.66% of the times she plays.


Another way to think about it...if you never switch, you have to guess the correct door first (33.3% chance) to win. If you always switch, you have to guess one of the wrong doors (66.6% chance) to win.

Therefore, always switch.

 

[ccibball50]

21

Same thing goes with Who wants to be a millionaire. If you get to the last suitcase, ALWAYS SWITCH.

The purpose of the 3 door equation in 21 is its correlation with the fact that percentages and probabilities change due to removal or alteration of elements involved in the game.

 

[KenSmith]

Same thing goes with Who wants to be a millionaire. If you get to the last suitcase, ALWAYS SWITCH.

I assume you mean Deal or No Deal, not WWTBAM.
Switching briefcases at DoND doesn't help. No information affected which briefcase you initially picked, so the probability at the end is 50/50 that you hold the correct case already. Switching neither hurts nor helps your chances.

If the producer of DoND is out there somewhere reading this, consider what it would be like to have a pro gambler on the show. It might give an interesting spin to the public's take on risk and reward. They could learn all about EV and Certainty Equivalent. Call me. :grin:

 

[ccibball50]

I assume you mean Deal or No Deal, not WWTBAM.
Switching briefcases at DoND doesn't help. No information affected which briefcase you initially picked, so the probability at the end is 50/50 that you hold the correct case already. Switching neither hurts nor helps your chances.

If the producer of DoND is out there somewhere reading this, consider what it would be like to have a pro gambler on the show. It might give an interesting spin to the public's take on risk and reward. They could learn all about EV and Certainty Equivalent. Call me. :grin:

First of all, you are correct, it is not who wants to be a millionair, but rather Deal or No Deal. However, if there are 25 cases, and you chose one, you have a 1 in 25 chance of picking the $1,000,000 suitcase. when you get to the end and the 1 m suitcase has not been chosen, you have a 50/50 chance of choosing the 1 m suitcase if you switched as apposed to the 1 in 25 at the beginning. (I do not know how many suitcases they have, I never really watched it that much.)

 

[KenSmith]

However, if there are 25 cases, and you chose one, you have a 1 in 25 chance of picking the $1,000,000 suitcase. when you get to the end and the 1 m suitcase has not been chosen, you have a 50/50 chance of choosing the 1 m suitcase if you switched as apposed to the 1 in 25 at the beginning. (I do not know how many suitcases they have, I never really watched it that much.)

You get to the end and are offered the choice of switching...
If you switch, you have a 50% chance of winning the $1M.
If you don't switch, you have a 50% chance of winning the $1M.
Therefore, it doesn't matter if you switch or not.
This is different than the Monty Hall problem because there's no "Monty" using hidden information to eliminate non-winning cases.

 

[mdlbj]

The goat is better. baaaaa

You get to the end and are offered the choice of switching...
If you switch, you have a 50% chance of winning the $1M.
If you don't switch, you have a 50% chance of winning the $1M.
Therefore, it doesn't matter if you switch or not.
This is different than the Monty Hall problem because there's no "Monty" using hidden information to eliminate non-winning cases.

 

[callipygian]

If the producer of DoND is out there somewhere reading this, consider what it would be like to have a pro gambler on the show. It might give an interesting spin to the public's take on risk and reward. They could learn all about EV and Certainty Equivalent.

Certainty Equivalent as it applies to blackjack may not be completely appropriate because players only get to play the game once - long-term bankroll growth is not the goal. Given that you risked nothing to be on the show, you are (or should be) more likely to take a risk for a large prize than if you could play the game 10 times and accept a 10-game average winnings.

 

[ccibball50]

consider this though.

Your odds of you choosing the correct case do not change. You chose it at the begining so your chance is still 1 in 25. I know it is different from the 3 doors problem.

Say you have three doors and choose one to save for the end. You then open another door and it is a goat. Switching is a 50 percent chance of winning and your 33 percent just doesn't majically change unless you chose the goat first and opened it before choosing your second door.

I am still pondering the issue. If you have any math to show this it would be helpful.

 

[KenSmith]

Given that you risked nothing to be on the show, you are (or should be) more likely to take a risk for a large prize than if you could play the game 10 times and accept a 10-game average winnings.

It's certainly true that your behavior would be different if you had a 10-game schedule instead of one shot, but I think you have the risk backwards. A player who only gets to play once should be more conservative than one who will have multiple tries to win big.

While it is true that you pay nothing to get on the show, once you have a banker's offer, you're risking that money if you continue.

The concept of CE applies here once the offer is large enough to be "big money" in your view. Let's say you have three cases left $1M, $1000 and $1. The EV is $1001001/3 or $333367.
If the banker offers me $320K, I'm taking it. It's not worth an extra $13,367 of EV for me to risk $320,000 to continue.

In truth, the game is more complicated than that because in the times I've seen it, it appears that the banker offers can be MORE than EV late in the game. That means that I could be turning down more than an extra $13,367 in the example above because of a possible over-offer when only two cases remain. (Maybe they would offer $510,000 when only two cases remain for example.) If the details of how the offers are calculated is available anywhere that would have to be factored in.

 

[KenSmith]

Say you have three doors and choose one to save for the end. You then open another door and it is a goat. Switching is a 50 percent chance of winning and your 33 percent just doesn't majically change unless you chose the goat first and opened it before choosing your second door.

You've proved it right there. If switching means you have a 50% chance of winning, then NOT switching is obviously the OTHER 50% chance of winning. There is no advantage in switching.

 

[Kasi]

It's certainly true that your behavior would be different if you had a 10-game schedule instead of one shot, but I think you have the risk backwards. A player who only gets to play once should be more conservative than one who will have multiple tries to win big.

I was thinking if a player is neither risk-averse nor risk-prone, he would just continue to play until the offer exceeds EV. Maybe? lol?

If I play it a million times to the end, I'm only going to end up with the EV I had at the beginning anyway no matter what they offer?

Just asking - if i could play it a million times, I'd take the 2nd offer everytime no matter how much below EV it was lol. Maybe even the when I could only play it once lol.

My over/under side bet would be The Wise One would be outta there on the first offer lmao.

The concept of CE applies here once the offer is large enough to be "big money" in your view. Let's say you have three cases left $1M, $1000 and $1. The EV is $1001001/3 or $333367.
If the banker offers me $320K, I'm taking it. It's not worth an extra $13,367 of EV for me to risk $320,000 to continue.

In truth, the game is more complicated than that because in the times I've seen it, it appears that the banker offers can be MORE than EV late in the game. That means that I could be turning down more than an extra $13,367 in the example above because of a possible over-offer when only two cases remain. (Maybe they would offer $510,000 when only two cases remain for example.) If the details of how the offers are calculated is available anywhere that would have to be factored in.

This CE stuff always confuses me a little - would it be legit to assume I should take my total wealth into considerastion rather than just the money I might win on the show? If I'm a multi-millionaire going in, I guess I'd "gamble" more because the money would mean so little to me?

Not to mention I guess the reality of a higher tax bracket could logically make one more "risk-averse" than if taxes were not a consideration?

Anyway, what would you think could possibly make the banker of the show offer higher EV at the end? If he wants to make a player play as short as possible and win as little as possible, he wouldn't do it, would he?

If he's a risk-averse banker, scared sh*tless about possibly paying out $1MM, he offers more to cut his losses and settle out-of court so to speak lol.

It is assumed, unlike Monty, the player, the host and the banker have no idea how much is in any case at the beginning?

Maybe, it's like you suggest, the bank is sizing up the risk-level of the player along the way and making an offer accordingly, like maybe like poker when you are playing the player, not the house.

Like, if they read here, and knew FLASH was the next contestant, they'd know the calculations he can make in but a few seconds and offer him exactly EV so he'd be indifferent and take the offer.

They'd offer me $10K with an EV of $150K at the first chance guessing I might even possibly know that but still would take the offer knowing what a balless pathetic gambler I am and get over on me.

They'd offer Inplay way higher offers than EV at every step of the way but it wouldn't matter because he's going for broke no matter what they do :grin:

 

[sagefr0g]

...
My over/under side bet would be The Wise One would be outta there on the first offer lmao.
..

lmao. i can tell you that every time i watched that show, at some point i'd end up yelling at the boob tube, "take the money you #*#&&(#&&@ idiot!"
so me not understanding the game ev-wise or probability-wise or any way really if i managed to get the kind of offers i've seen a lot of the contestants get then yeah i'd be on my way with the money. just seems you can sort of tell once the money gets 'up there pretty good' and what amounts are left in cases and how many cases you have to choose 'blind', well at some point you know your risking a heck of a lot of money and could end up with much less or nothing. so too me there'd be a point where i'd likely figure it wouldn't be worth it to risk what i already had sort of thing. i guess i'd be mainly weighing the risk of losing what i had by how much against what i stood to gain sort of thing.
but yeah, i'd think the value one puts on some amount of money is gonna have a lot to do with how they would play the game. that utility function sort of stuff.

 

[tribute]

Just curious, there was a lecture scene where Micky gave Ben 3 "doors" to choose and then Ben chooses one. After that, Micky gave away the result of one door and now gives Ben another choice to choose the "winning door." Ben decides to choose a new door instead of staying with his original decision. What was the math reason behind that?



Thanks

In that movie scene, didn't Ben explain his reason to switch doors? I don't remember the lines, but his answer made a strong impression and helped land a spot on the team.

 

[callipygian]

In truth, the game is more complicated than that because in the times I've seen it, it appears that the banker offers can be MORE than EV late in the game.

Are you talking about Deal or No Deal?

I've never seen an offer greater than EV except when there are two cases remaining.

 

[KenSmith]

Are you talking about Deal or No Deal?

I've never seen an offer greater than EV except when there are two cases remaining.

Yes, I'm talking about DOND. And it is quite possible that the only +EV offers are at the end.