Discussing Probability & Odds: Does Flipping More Cards Increase Chances?

[caprirs302]

I am trying to settle a score on probability and odds, and I just wanted to see what everybody thought.

This springs from the fact that when you watch "Deal or No Deal" and cases are removed one at a time, Howie frequently announces that there is a 15% chance that the case holds a million dollars, then later it is a 25% chance, then 35% and so on, until (if it gets there) a 50% chance.

The question is, do the odds really improve?

My stance is that since the there were 26 cases present when the players case was chosen, that the odds will stay frozen at 1/26 up until the point that the player gets to "reguess" (if he gets that far) so essentially if there are two cases left, his case still has a 1/26 chance of being a winner, but if he is allowed to reguess with the new information it becomes 1/2.



Here is a better example.

I will randomly choose a card from a regular deck and proclaim that it is the Ace of Spades without showing it. The odds of me being right are of course 1/52. As far as I can see, there are only two ways of determining if I am right:

A: Flip over my card and check

B: Flip over all of the other cards, and if the Ace of Spades is not there, it must be my card

Do you agree so far?

Would either one of these give me a better chance at getting the Ace of Spades? NO!


Now, let's take a closer look at situation B. Let's say that instead of just flipping over all of the cards, I decide to flip them over one at a time, very slowly. This is ABSOLUTELY NO DIFFERENT FROM JUST FLIPPING THEM ALL OVER! My odds do not get any better as more cards are flipped over. My chances are still 1/52. If I happen to flip over the Ace, I know that I lost, but that has nothing to do with my odds.

Now, if somewhere in there you give me a chance to choose cards, then my odds become
1/however many cards remain.

 

[Office_Shredder]

Of course your odds improved. If the $1 box is opened, you now know that YOU don't have the $1 box. So your box is either the $2, $5, etc. boxes. Since there's one less value on the list of potential boxes, your odds of having any given one are greater.

 

[Whatever123]

Well then by that logic, when you have 1 case left and you know that it has to be the X dollar case, you still have a 1/26 chance of it being that case...

 

[caprirs302]

I think you are missing the point.

If I guess a case, it is 1/26. If I look at it and there is no million dollars, oh well, it was still a 1/26 chance. Right?

What is the difference between looking in my case, or determining my case through deduction by looking at the other cases.

NONE!


Think of it this way, if I chose a case and then walked out of the room, I would have a 1/26 chance of winning a million. Now that I am out of the room, I let Howie open the case for me to see if I got the million. I don't care how he determined that I had a million or not (either opening my case or opening all the rest) when he comes in the hallway he is either going to tell me that I got the million or I didn't. his methods for finding out what is in my case did not improve my odds.

 

[DavidSnider]

I think you are missing the point.

If I guess a case, it is 1/26. If I look at it and there is no million dollars, oh well, it was still a 1/26 chance. Right?

What is the difference between looking in my case, or determining my case through deduction by looking at the other cases.

NONE!Think of it this way, if I chose a case and then walked out of the room, I would have a 1/26 chance of winning a million. Now that I am out of the room, I let Howie open the case for me to see if I got the million. I don't care how he determined that I had a million or not (either opening my case or opening all the rest) when he comes in the hallway he is either going to tell me that I got the million or I didn't. his methods for finding out what is in my case did not improve my odds.

Yes, there is a 1/26 chance of getting the million in your case, but the point of the game is to try and maximize what the banker offers by weeding out the lower values.

 

[Whatever123]

I think you are missing the point.

If I guess a case, it is 1/26. If I look at it and there is no million dollars, oh well, it was still a 1/26 chance. Right?

What is the difference between looking in my case, or determining my case through deduction by looking at the other cases.

NONE!Think of it this way, if I chose a case and then walked out of the room, I would have a 1/26 chance of winning a million. Now that I am out of the room, I let Howie open the case for me to see if I got the million. I don't care how he determined that I had a million or not (either opening my case or opening all the rest) when he comes in the hallway he is either going to tell me that I got the million or I didn't. his methods for finding out what is in my case did not improve my odds.

No. This reminds me of an old problem called the "Monty Hall problem." In that problem, what you said holds true - there is a 1/3 chance at the end that the contestant will win with the door that they chose because the probability remains the same as in the beginning. But there is a fundamental difference in that problem and in this problem. In the Monty problem, the person choosing the doors (or cases) knows where the prize is at and is purposely not choosing it. In this problem, everything is completely random. Let's say that Howie knew where he million dollar case is and he was choosing the cases for you, purposely not choosing the million dollar case, then at the end, you would have a 1/26 chance of having the million dollar case and a 25/26 chance of not having it. Therefore, if you could, you would want to switch.

Let me put it another way. You're correct in saying that you have a 1/26 chance of having the million dollar case at the beginning... Just like you have a 1/26 chance of having the 1 dollar case, a 1/26 chance of having the 500 case, etc. So at the end, if it's down to just the 1 dollar case and the 1 million dollar case, you'll have a 50/50 shot no matter what because 1/26=1/26.

 

[Dertulm01]

I am trying to settle a score on probability and odds, and I just wanted to see what everybody thought.

This springs from the fact that when you watch "Deal or No Deal" and cases are removed one at a time, Howie frequently announces that there is a 15% chance that the case holds a million dollars, then later it is a 25% chance, then 35% and so on, until (if it gets there) a 50% chance.

The question is, do the odds really improve?

Sure they do. Howie may not be a mathematician, but a major network isn't going to get simple probability like this wrong and embarrass themselves.

I Googled "Deal or no deal" probability. The following came form the second hit:

Deal or no Deal

The probability that you are holding the $1,000,000 case is again 1/26, and again if you pick the $1,000,000 case, you will always end up with the million dollar choice. But the probability that you end up with the million dollar choice in general is only 2/26 in deal or no deal. The above formula shows that the probability that you are holding the $1,000,000 case given that you are choosing between two cases where one contains $1,000,000 to be 1/2.


It's clear that there is no advantage to switching in Deal or no Deal, because the odds that you are already holding the $1,000,000 case are 50:50.

http://en.wikibooks.org/wiki/Talk:Introduction_to_Game_Theory/Deal_Or_No_Deal

My stance is that since the there were 26 cases present when the players case was chosen, that the odds will stay frozen at 1/26 up until the point that the player gets to "reguess" (if he gets that far) so essentially if there are two cases left, his case still has a 1/26 chance of being a winner, but if he is allowed to reguess with the new information it becomes 1/2.

Huh? You're saying if he doesn't get a chance to switch the chances of him having the winning case is 1/26 but if he is given the choice his chances magically go to 1/2? When the contestant is down to two choices, his chances of winning $1,000,000 is 1/2 whether or not he is given a chance to switch.

Here is a better example.

I will randomly choose a card from a regular deck and proclaim that it is the Ace of Spades without showing it. The odds of me being right are of course 1/52. As far as I can see, there are only two ways of determining if I am right:

A: Flip over my card and check

B: Flip over all of the other cards, and if the Ace of Spades is not there, it must be my card

Do you agree so far?

Would either one of these give me a better chance at getting the Ace of Spades? NO!



Now, let's take a closer look at situation B. Let's say that instead of just flipping over all of the cards, I decide to flip them over one at a time, very slowly. This is ABSOLUTELY NO DIFFERENT FROM JUST FLIPPING THEM ALL OVER! My odds do not get any better as more cards are flipped over.

Of course they do. If you're lucky enough to never flip over the Ace of Spades when down to two cards, you have a 50/50 chance of having originally picked the Ace of Spades. Exactly what the wiki article I linked to above said about Deal or No Deal.

 

[Dertulm01]

Well then by that logic, when you have 1 case left and you know that it has to be the X dollar case, you still have a 1/26 chance of it being that case...

I think you are missing the point.

It's you that missed Whatever123's point. If your probability stays the same no matter how many cases are revealed, then by your own logic if it were to get narrowed down to one box, your probability of winning $1,000,000 is 1/26. You realize this is incorrect, right?

 

[caprirs302]

DavidSnider, thanks for seeing this the correct way. I understand the point of the game is to narrow down possibilities, but my question was based on the fact that the "odds" of having the million dollar case seem to chane every time a non million dollar case is chosen, and as you stated this is not the case.

For everyone else, my second example is a better one.

If you think that the odds change, please please please PLEASE explain to me how flipping over cards one at a time is any different than flipping them all over at once, and how either of these would improve my chances of picking the ace of spades over simply just picking a card and looking at it. There is no difference. There were 52 cards at the time I picked the card, and unless I somehow use the new information I gain to make a new decision (i.e. change cards in the middle of the flips, or at least get a chance to and decide to stay with my original) the odds do not change.

 

[Hurkyl]

It would help if you very precisely stated the example you want to consider.


But pay attention to this -- you have to analyze all cases if you want the odds to remain the same.

The odds that you flip over the ace of spades on the 52nd card is 1 in 52.

However, if you look only at those games where you haven't flipped over the ace of spades in the first 50 flips, then the probability of flipping over the ace of spades on the 52nd card is 1 in 2.

This is tempered by the fact that if you have already flipped over the ace of spades in the first 50 flips, the probability of flipping over the ace of spades on the 52nd card is zero, and this situation is much more likely.

 

[Office_Shredder]

DavidSnider, thanks for seeing this the correct way. I understand the point of the game is to narrow down possibilities, but my question was based on the fact that the "odds" of having the million dollar case seem to chane every time a non million dollar case is chosen, and as you stated this is not the case.

For everyone else, my second example is a better one.

If you think that the odds change, please please please PLEASE explain to me how flipping over cards one at a time is any different than flipping them all over at once, and how either of these would improve my chances of picking the ace of spades over simply just picking a card and looking at it. There is no difference. There were 52 cards at the time I picked the card, and unless I somehow use the new information I gain to make a new decision (i.e. change cards in the middle of the flips, or at least get a chance to and decide to stay with my original) the odds do not change.

Flipping over cards one at a time is only different than flipping them all over at the same time because, well, there's time in between. Which means instead of one instantaneous event that vastly changes the chances of having the 1,000,0000 dollars, there's a lot of events that change it slightly (unless it's revealed of course). If I have a deck of 52 cards and you pick one at random:

I flip over every card except for one left in the deck for you to look at. None of them are the ace of spades. There was a 1/52 chance that your card is the ace of spades, and a 1/52 chance the bottom card of the deck was the ace of spades. So since those are the only two cards left, there's a 1/2 chance of each card being the ace of spades. You seem to think you still only have a 1/52 chance of having the ace of spades. This means the odds of the bottom card of the deck being the ace of spades jumps to 51/52 by me revealing the other cards in the deck.

So let's play a new game. I give you a card. I flip over every card in the deck except for the bottom one. If the bottom card is then the ace of spades, you win. According to you, if the ace of spades does not show up in any of the cards that I flipped over, there's a 51/52 chance of you winning. But how is this game really any different from the one above? In one game you won if the top card was the ace of spades, in one game you won if the bottom card was the ace of spades. In both games I showed you every card besides the top and the bottom card

 

[caprirs302]

It is not more information that affects my decision on which card to choose, which was done at 1/52.

Think of it this way, if I were in a drawing with 51 other people, I have a 1/52 chance of winning. If a winner is randomly drawn, and then the LOSING names were called of one by one instead of simply calling the winner, and I happen to be one of the last two people, my odds are not 1/2 to win the drawing! they are still 1/52!

Just because names are being called as losers doesn't change the fact that the woinner has already been determined, and eliminating any amount of non wining people will not change the odds whatsoever unless a new winner is drawn after the fact.

How can you not see that?

 

[DoctorBinary]

The odds you PICKED the million dollar case stay the same -- 1/26. The odds you HAVE the million dollar case change -- they improve each time a non-million dollar case is exposed. "Picked" and "have" are different problems.

This applies to your cards example too.

 

[Hurkyl]

If a winner is randomly drawn, and then the LOSING names were called of one by one instead of simply calling the winner, and I happen to be one of the last two people, my odds are not 1/2 to win the drawing! they are still 1/52!

What if your name was already called as a loser? Do you still have a 1/52 chance of winning?

 

[caprirs302]

So let's play a new game. I give you a card. I flip over every card in the deck except for the bottom one. If the bottom card is then the ace of spades, you win. According to you, if the ace of spades does not show up in any of the cards that I flipped over, there's a 51/52 chance of you winning. But how is this game really any different from the one above? In one game you won if the top card was the ace of spades, in one game you won if the bottom card was the ace of spades. In both games I showed you every card besides the top and the bottom card

Now, here is the hole in your game. If we made a bet on that game at the very BEGINNING, the odds of me winning are one in 52. The act of you flipping over the cards simply delays us knowing the outcome of the game.

One the other hand, if you flip over a card and ask me to rewager, then you change the game completely because I am making a new decisin based on new information.

Again, I don't see how flipping over one card at a time gives me any more chances of winning than just shuffling the deck and looking at the bottom card. I am still only going to win this game with you 1/52 of the time regardless, giving me odds of 1/52.

 

[Dertulm01]

caprirs302, you are ignoring questions asked and points made and don't seem a bit interested in having your point of view changed. I'll play along a little longer and see if you're open to actually having a discussion and not just waving away any explanations that don't support your conclusion.

If you think that the odds change, please please please PLEASE explain to me how flipping over cards one at a time is any different than flipping them all over at once, and how either of these would improve my chances of picking the ace of spades over simply just picking a card and looking at it. There is no difference. There were 52 cards at the time I picked the card, and unless I somehow use the new information I gain to make a new decision (i.e. change cards in the middle of the flips, or at least get a chance to and decide to stay with my original) the odds do not change.

Lets do this with three cards instead of 52 for simplicity's sake. Take two random cards from a deck and the Ace of Spades. Mix up these three cards and throw them in a hat. Take one out. One out of every three times you do this (on average), you will have picked the Ace of Spades.

Now take the same cards, mix them up and throw them in a hat. Take one out and put it face down without ever seeing what card you've chosen. Remove another card from the hat and look at it. If it's not the Ace of Spades what are the chances that the card you have face down is the Ace of Spades. It's 1/2. Don't believe it? You can perform this experiment multiple times and you'll find that if every time you did not pick the the Ace of Spades the time you removed a card and peeked at it, your face down card will be the Ace of Spades about half the time.

You ask why? Because new information changes probability. Choose one of the 26 cases and you have 1/26 of having the $1,000,000 case. Peek at the insides of 24 of the remaining 25 cases and see that the $1,000,000 case is not in any of those and you have a 1/2 chance of having the $1,000,000 case. Peek inside the other case and see it isn't the $1,000,000 case and you believe your case still only has a 1/26 chance of containing $1,000,000? No, you can be very excited because you have a 1/1 chance of having the $1,000,000 case. You're a guaranteed winner.

 

[caprirs302]

caprirs302, you are ignoring questions asked and points made and don't seem a bit interested in having your point of view changed. I'll play along a little longer and see if you're open to actually having a discussion and not just waving away any explanations that don't support your conclusion.



Lets do this with three cards instead of 52 for simplicity's sake. Take two random cards from a deck and the Ace of Spades. Mix up these three cards and throw them in a hat. Take one out. One out of every three times you do this (on average), you will have picked the Ace of Spades.

Now take the same cards, mix them up and throw them in a hat. Take one out and put it face down without ever seeing what card you've chosen. Remove another card from the hat and look at it. If it's not the Ace of Spades what are the chances that the card you have face down is the Ace of Spades. It's 1/2. Don't believe it? You can perform this experiment multiple times and you'll find that if every time you did not pick the the Ace of Spades the time you removed a card and peeked at it, your face down card will be the Ace of Spades about half the time.

QUOTE]

I submit to you that that only works because you are eliminating the times that you would draw a card, reach into the hat and actually pull out the ace, thus proving that your card is not the ace. If you include these times, your odds are 1/3. And you have no reason to dismiss these situations since our goal is to simply determine if I picked the ace.

Your logic is true if you give me a chance to change my guess after we show a card, since I would now be picking 1 out of 2, but if I am given no options at all for changing my guess, the odds would forever be 1/3.

 

[Dertulm01]

Now, here is the hole in your game. If we made a bet on that game at the very BEGINNING, the odds of me winning are one in 52. The act of you flipping over the cards simply delays us knowing the outcome of the game.

It doesn't just simply delay the game; it provides new information. If we flip over the first card and it's the Ace of Spades, the game is over. If it's not, your chances of winning just got a little better. If we flip over 50 of them and none of them were the Ace of Spades, you wouldn't be happy because your odds of winning are significantly better than before you had information about the top 50 cards? You still think you only have a 1/52 chance of winning? No, if you perform this experiment enough times to get to the point where the top 50 cards weren't the Ace of Spades, you'll win 1/2 of the time. If you perform this experiment enough times to get to the point where the top 51 cards weren't the Ace of Spades, you'll win 100% of the time.

 

[caprirs302]

Again, by that logic I would have to not include all of the times where I was flipping cards and the ace did not come up in the last two (25/26) I cannot simply eliminate those from the set! They count too!

Yes, flipping cards over gives me new information, but if I do not get a chance to USE that information in making a decision it cannot affect my odds.

Again, going back to my original post:

If I simply pick a card and flip it over, the odds of it being an Ace of spades is 1/52

If I use deduction to see if I got the ace by flipping over the other 51 cards, my odds are still 1/52

If I use deduction by flipping over the other 51 cards one at a time you are telling me that i will get better odds of winning? This is awesome! I am going to go buy a lottery ticket right now, and have all of the losers called off one at a time, that way I have a better chance of winning!

 

[Dertulm01]

I submit to you that that only works because you are eliminating the times that you would draw a card, reach into the hat and actually pull out the ace, thus proving that your card is not the ace.

Exactly! Because the Ace didn't come out those times, your odds improve.

You said this in your OP:

This springs from the fact that when you watch "Deal or No Deal" and cases are removed one at a time, Howie frequently announces that there is a 15% chance that the case holds a million dollars, then later it is a 25% chance, then 35% and so on, until (if it gets there) a 50% chance.

The question is, do the odds really improve?

The odds really improved because Howie didn't reveal a million dollar case!

If you include these times, your odds are 1/3.

This is irrelevant to every scenario you posted. You brought up scenarios in what the probability would be if the winning case/card, etc. wasn't revealed and what the new probability would be.

And you have no reason to dismiss these situations since our goal is to simply determine if I picked the ace.

Yes, there is a reason to dismiss these situations. The reason is that those situations didn't play out. Once Howie removes 24 cases and none of them are the $1,000,000 case, we can dismiss them. There are two cases left and both have an equal chance of being the $1,000,000 case. The chances that it's the one you originally chose is 1/2.

Your logic is true if you give me a chance to change my guess after we show a card, since I would now be picking 1 out of 2, but if I am given no options at all for changing my guess, the odds would forever be 1/3.

How does changing your guess change the frequency in which your card will be the Ace of Spades? Explain how this works. The card will be the Ace of Spades about 1/2 the time whether or not you speak any words.

 

[caprirs302]

The last sentence that you just posted supports my argument whole heartedly, the chances of an ace coming up in your hat game will be 1/3, no matter how fance we get in determining what card I chose.

 

[Dertulm01]

Yes, flipping cards over gives me new information, but if I do not get a chance to USE that information in making a decision it cannot affect my odds.

You use that information to recalculate your odds.

If I simply pick a card and flip it over, the odds of it being an Ace of spades is 1/52

If I use deduction to see if I got the ace by flipping over the other 51 cards, my odds are still 1/52

Not if the other 51 cards are not the Ace of Spades. Then your probability of having the Ace of spades is 1/1.

If I use deduction by flipping over the other 51 cards one at a time you are telling me that i will get better odds of winning?

It depends on what you see after each flip. Every time it's not the Ace of Spades, your odds just got a little better. I already gave you an experiment you can try to prove it.

This is awesome! I am going to go buy a lottery ticket right now, and have all of the losers called off one at a time, that way I have a better chance of winning!

Every time one of the losers isn't you, your odds of being the winner just got a little better. Only thing is, the more people that get picked off, the more the chances of you being picked off increases.

Let's say there are 1,000,000 lottery tickets sold and one is the winner. You ask for a list of 999,998 losers not including yourself. What is your probability of winning? It's 1/1,000,000.

What if instead of asking for a list of losers, you ask for a random list of what 999,998 of the other people bought. You see that none of them won. What is your probability of winning? It's 1/2.

 

[Dertulm01]

The last sentence that you just posted supports my argument whole heartedly, the chances of an ace coming up in your hat game will be 1/3, no matter how fance we get in determining what card I chose.

Post edited. I meant to write "1/2", not "1/3".

Will you explain how changing your guess changes the frequency in which your card will be the Ace of Spades?

Every time one of the three cards is removed and it's not the Ace of Spades, your probability of having the Ace of Spades increased from 1/3 to 1/2. Try it and see.

 

[Dertulm01]

My stance is that since the there were 26 cases present when the players case was chosen, that the odds will stay frozen at 1/26 up until the point that the player gets to "reguess" (if he gets that far) so essentially if there are two cases left, his case still has a 1/26 chance of being a winner, but if he is allowed to reguess with the new information it becomes 1/2.

If he doesn't get a chance to switch, you're saying he will most likely lose because his probability is still 1/26. If he is given the choice to switch and keeps his case, he has an awesome chance of winning- 1/2? Please explain this.

 

[DavidSnider]

This isn't that complicated:

On the first turn there is a 1 in 26 chance that your case contains the million.

Each time you pick a case that is not the million the odds that your case contains the million are 1 in (26 - Turn). The odds improve because you have just eliminated one of the possibilities of what could be in your case.

If you pick the million your odds are zero that your case has it.

 

[Whatever123]

You just don't get it, do you? I already explained this once. You have a 1/52 chance at the beginning of it being an ace of spades, but you also have a 1/52 chance of it being an eight of hearts. So at the end, it's 50/50 because 1/52=1/52. Even basic logic disproves what you're saying because you can do the same for any card. You can say, I had a 1/52 chance of it being an ace of spades at the beginning, a 1/52 of an eight of hearts, and a 1/52 chance of a three of clubs. Now if there's three cards left (and let's say it happens to be those three), are they all still 1/52? What happened to the other 49/52 probability?

I'll explain this again... If somebody KNOWS where the ace of spades is and they're choosing cards to flip over and they're PURPOSELY NOT CHOOSING IT, then at the end, there's a 51/52 chance that the ace of spades is the card you didn't choose and a 1/52 chance that it's the card that you chose at the beginning.

 

[Dertulm01]

You just don't get it, do you? I already explained this once. You have a 1/52 chance at the beginning of it being an ace of spades, but you also have a 1/52 chance of it being an eight of hearts.

In case you don't understand what Whatever123 is saying, I'll break it down a little.

When you choose one card from a deck, you don't only have a 1/52 chance of it being the Ace of Spades, you also have a 1/52 chance of it being the Ace of Hearts, the Ace of Clubs, etc. If you remove one card randomly and it happens to be the King of Clubs, you have eliminated one of the possibilities. The chances that you now have the Ace of Spades is 1/51 because we take the King of Clubs out of the equation.

Now you may ask yourself, "So, I can just guarantee myself better odds by just removing a card?"

No. Because on average, one out of every fifty-two times you attempt that, the card you turn over to eliminate will be the Ace of Spades.

 

[DoctorBinary]

Let's say there are 1,000,000 lottery tickets sold and one is the winner. You ask for a list of 999,998 losers not including yourself. What is your probability of winning? It's 1/1,000,000.

What if instead of asking for a list of losers, you ask for a random list of what 999,998 of the other people bought. You see that none of them won. What is your probability of winning? It's 1/2.

And to drive the point home, let's ask: caprirs302, in the latter case, would you sell me your $1 lottery ticket for $1? DEAL!

 

[caprirs302]

And to drive the point home, let's ask: caprirs302, in the latter case, would you sell me your $1 lottery ticket for $1? DEAL!

Again, you are excluding the other 499,999 time that this scenario wouldn't happen.

Of course I would not sell you my lottery ticket, but that doesn't change the fact that I stil only have a 1/1,000,000 chance of having that winning ticket. By asking to sell you are giving me a new choice based on new information, I would be a fool to part with it.

I agree with whatever123 that at the end each card (or ticket) has equal odds of being the ace or a winner, but I am asking for my odds of winning, which cannot change from what they were when the final decision was made, or when the lottery ticket was sold.

Now, there may be some mathematical convention on odds calculating that you know of and I don't, but aside from that I cannot agree with you.

 

[Whatever123]

but aside from that I cannot agree with you.

Well... You're entitled to your opinion and you, of course, do not have to agree. But unfortunately for you, mathematics is not about opinion and just because you do not agree doesn't mean that you're any less wrong.

 

[Dertulm01]

caprirs302, you are again glossing over points made and questions asked. Are you the least bit interested in having your point of view changed or do you just not want to admit your wrong and have this score settled you mentioned in the OP?

At least answer to post # 24.

Again, you are excluding the other 499,999 time that this scenario wouldn't happen.

That's right. Because they didn't happen!

Of course I would not sell you my lottery ticket, but that doesn't change the fact that I stil only have a 1/1,000,000 chance of having that winning ticket. By asking to sell you are giving me a new choice based on new information, I would be a fool to part with it.

Yes, new information changes probability- not a new choice! How does a new choice change probability? You believe you only have a 1/1,000,000 chance of having the winning ticket, yet as soon as someone asks you to sell it, you have a lot to be excited about because your probability of winning is now 1/2? Explain.

I agree with whatever123 that at the end each card (or ticket) has equal odds of being the ace or a winner, but I am asking for my odds of winning, which cannot change from what they were when the final decision was made, or when the lottery ticket was sold.

Now, there may be some mathematical convention on odds calculating that you know of and I don't, but aside from that I cannot agree with you.

You don't need to know a lot about probability to figure this out. Simply try the experiment with three cards that I mentioned. You'll see that new information changes probability.

 

[Dertulm01]

Of course I would not sell you my lottery ticket, but that doesn't change the fact that I stil only have a 1/1,000,000 chance of having that winning ticket. By asking to sell you are giving me a new choice based on new information, I would be a fool to part with it.

Go over this a few times and see if it still makes sense to you. You're saying you'd be a fool to part with a lottery ticket that only has a 1/1,000,000 chance of being a winner.

 

[srijithju]

Again, you are excluding the other 499,999 time that this scenario wouldn't happen.

Of course I would not sell you my lottery ticket, but that doesn't change the fact that I stil only have a 1/1,000,000 chance of having that winning ticket. By asking to sell you are giving me a new choice based on new information, I would be a fool to part with it.

I agree with whatever123 that at the end each card (or ticket) has equal odds of being the ace or a winner, but I am asking for my odds of winning, which cannot change from what they were when the final decision was made, or when the lottery ticket was sold.

Now, there may be some mathematical convention on odds calculating that you know of and I don't, but aside from that I cannot agree with you.

Thank Goodness .. at least we are getting somewhere ..

So you have agreed that at the end each card (or ticket) has equal odds of being the ace or a winner.
Now let us see where that leads us ..
There are 2 cards - each has equal odds .. what does that mean --- each of these cards has a 1/2 a chance that it is the right card. So now what are your odds ? --- 1/2 !

 

[DoctorBinary]

Of course I would not sell you my lottery ticket, but that doesn't change the fact that I stil only have a 1/1,000,000 chance of having that winning ticket. By asking to sell you are giving me a new choice based on new information, I would be a fool to part with it.

"Making a new choice" doesn't have to play in here. Forget I offered to buy it. Wouldn't you feel good -- VERY good -- about your chances of winning at that point?

Back to Deal or No Deal, we'd all agree that the OVERALL odds of winning one million dollars is 1/26: before the show begins and after it ends; regardless of whether the million dollar case is exposed first, last or at any other time; regardless of the show's outcome. But we are analyzing individual instances of a show.

So which of these questions are you asking?

1) What are the odds of winning 1 million on Deal or No Deal? (Answer: 1/26)
2) What are the odds of winning 1 million on Deal or No Deal after 24 cases are exposed and 1 million isn't one of them? (Answer: 1/2)

You seem to be confusing the two.

 

[Office_Shredder]

caprirs, it's called conditional probability

You ask: What's the probability that event X occurs assuming that event Y has occurred

Can you see how that question allows you to ignore what the probability that event X occurs when Y doesn't?