No, your 1 in 3 choice is 1 in 3. Eliminate one choice and one goat and you are at 2 choices, one goat and one car, you still with me? You still get to choose and your odds are 50/50.
Ignore Bait: Highest IQ, Many Questions, Odds makers invited...
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That’s why you would lose 66% of the time if you didn’t swap.
Because you are apparently not able to grasp how this works.
Again, if you are correct, it’s a no brainer that you bring $50,000 and I bring $55,000 and we play for $50 and $55 per hand.
You can’t lose if it’s 50%.
That only works if you randomly eliminate one card.
The card that is turned over is not random at all. It’s always a goat. And it’s always *after* you picked a card randomly.
No, you aren't. He is throwing away one of his cards that you both know is a loser. Now you are left with two cards win and lose and it's your choice. 50/50
Please come take my money.
Your $50 vs my $55 per try. You can’t switch your hand.
We can play 100 times, or 1 million times. And I’ll never quit.
But the problem was the claim that you should switch. The only compelling thing about that is the 66% chance you picked a goat somebody mentioned back there. Still, requiring the switch now means a 50% chance of error.
The whole 66% stuff is smoke. So is your first choice, meaningless. He will always be left with at least one loser to discard. That leaves two choices a right and a wrong and it is always your pick. 50/50
Now don't just tell me I'm wrong, show where my reasoning is flawed.
Now don't just tell me I'm wrong, show where my reasoning is flawed.
This is why it’s such a great prop bet. Because even people who have been told and shown the answer don’t believe it.
And like flipping a coin, as long as you are betting more than you can win, you will lose.
So you can't tell me the flaw in my logic. Thanks.
Then please take my money.
I’m offering you better than 50% on what you say is 50%.
And I’m telling you I will *never* quit. The only stipulation is we play at least 100 times.
After 100x, you decide when we quit. I’ll keep playing you. I’ll bring $100,000 if you want.
The bet only goes once or the proposer is stupid. So takers only get the coin flip.This is why it’s such a great prop bet. Because even people who have been told and shown the answer don’t believe it.
You were a 66% loser when you picked your card.
He showed you 50% of that. With the knowledge that he can *NOT* flip the car over.
That’s where your logic is flawed.
You’re applying non conditional probability to a conditional probability scenario.
Your *unconditional* probability is 33% when you pick a card randomly.
By default his *unconditional* probability is 66%.
By looking at those two cards *AND* giving you a choice to swap, *after* he showed you 50% of the losing cards, he has now given away that 66% edge if you swap cards.
This is because it’s a conditional probability. You are no longer picking random cards.
You have more information (the conditional).
By default his *unconditional* probability is 66%.
By looking at those two cards *AND* giving you a choice to swap, *after* he showed you 50% of the losing cards, he has now given away that 66% edge if you swap cards.
This is because it’s a conditional probability. You are no longer picking random cards.
You have more information (the conditional).
You are only 66% loser until the third option is eliminated. Then there are precisely two options, win and lose, and it is entirely your choice. Nobody is going to pick 3 so eliminate it as an option. The conditional is eliminated when he eliminates the third option.
Looks like I need the 6-5 to make it close. Or maybe I need a little more like 3-2. There you go!
The flaw has been explained to you several times at this point.
The flaw is that you’re not understanding that if you do not change your mind, if you keep your original pick, then absolutely nothing has changed regardless of any subsequent knowledge that you may or may not have.
Your pick remains your pick.
And the odds of it being correct remain one in three. The only way you can change those odds, is to do something different. That means changing your selection.
I think if people took the time to read the article that was posted in post 10, they would understand it.
Here is a simple table showing why it's a win to switch.
Here is a simple table showing why it's a win to switch.
| You Pick | Prize Door | Don’t Switch | Switch |
| 1 | 1 | Win | Lose |
| 1 | 2 | Lose | Win |
| 1 | 3 | Lose | Win |
| 2 | 1 | Lose | Win |
| 2 | 2 | Win | Lose |
| 2 | 3 | Lose | Win |
| 3 | 1 | Lose | Win |
| 3 | 2 | Lose | Win |
| 3 | 3 | Win | Lose |
| 3 Wins (33%) | 6 Wins (66%) |
The Monty Hall Problem: A Statistical Illusion
The Monty Hall problem is extremely counter-intuitive because our statistical assumptions are incorrect. I explain how this happens and make sense of it.statisticsbyjim.com
It isn't that hard to understand when explained correctly.
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Here is the bogus chart from post #10:
Note that the third option remains as a choice and as the winner. That's a confusion factor since we aren't going to choose a known loser and it should never be listed in the Prize Door column, yet it erroneously is three times. The bogus graph shows Pick 3 as a win 2 out of three times when it shouldn't even be an option.
Their theorem only works when we accept data we know to be incorrect.
| You Pick | Prize Door | Don’t Switch | Switch |
| 1 | 1 | Win | Lose |
| 1 | 2 | Lose | Win |
| 1 | 3 | Lose | Win |
| 2 | 1 | Lose | Win |
| 2 | 2 | Win | Lose |
| 2 | 3 | Lose | Win |
| 3 | 1 | Lose | Win |
| 3 | 2 | Lose | Win |
| 3 | 3 | Win | Lose |
Their theorem only works when we accept data we know to be incorrect.