Monty Hall, we have a PROBLEM

[St. Clown]

Re: Monty Hall, we have a PROBLEM

Use scalar (Ts, Tc, whatever). Assume I bought in my town. or don't bother trying at all. whichever.

One of the tricky parts in the problem is estimating the odds of multiple winners....

And for those of us who ended their mathematics studies with applied calculus, how are we supposed to know to use scalar or what that even means now that we know it?

[FreshFish]

Re: Monty Hall, we have a PROBLEM

And for those of us who ended their mathematics studies with applied calculus, how are we supposed to know to use scalar or what that even means now that we know it?

sorry, I merely meant use a symbol to represent a number whose magnitude is unknown to you.

like c is the speed of light in a vacuum....nothing more than that. T is total tickets sold, Ts tickets in my state, Tt tickets in my town, whatever.

As I alluded, the vexation in the problem is "what are odds of splitting the prize" not so much, "once prize has been won (at least once), what are odds of holding winning ticket" though both are a bit gnarly.

Some of the math is reminiscent of our thread title: you get to pick one of three doors, they open a door that doesn't have the prize, so then do you switch your original choice, or keep it?

The odds before the drawing are x, but after the drawing is over and (at least) one winner is certain, how do the odds change?

This problem actually is probably too hard to solve as I posted it, because one also has to factor in winning something other than the jackpot as well.





at least I am not -100%....

[LynahFan]

Re: Monty Hall, we have a PROBLEM

Way too many unknowns - not just how many tickets sold, but how many unique guesses? And, if there was definitely a winner, how many people have already looked at a ticket with a unique losing combo? Every time that happens, your odds of winning go up.

Thought experiment extreme: 100M tickets sold, but all the other tickets guessed the same combo. As soon as a single other person checks his ticket and loses, your probability of winning becomes 100%. If the 100M were divided among 2 guesses (that are both different from yours), then when one loser checks his ticket, your probability is 50% - it's either your ticket or the 50M people who guessed the 3rd combo. Etc.

[FreshFish]

Re: Monty Hall, we have a PROBLEM

So like most offices we have a Super Bowl box pool. $20 for five squares (or $5 for one square). We have enough people to fill two boxes.

I am going to partner with someone to put in $40 total between us. Are we better off both in the same box, or in different boxes?

My intuition says that we are "probably" better off with both of us being in the same box, as that gives us 10 chances to win something in that particular box instead of five. (10% chance to win something in one box, 0% chance in other; compared to 5% chance to win in each box).

However, I'm not convinced that the edge is as great as it first appears to be, since it also reduces the chances we could be big winners in both boxes.

Payout is 10% for 1st quarter, 20% for halftime, 10% for 3rd quarter, 60% for final score (nothing for 4th quarter if there is OT). If we assume $400 for each box (everyone puts in $20).

If scores and distribution of boxes is random, hmm....

10% of 400 + 0% of 400 = 5% of 200 + 5% of 200 so I am having trouble reconciling my intuition with math here, I think the math is missing something though since we might have duplication of numbers if spread over 2 boxes...but on a net basis, is duplication bad? (both lose) or good? (both win).

Maybe both of us in one box has better chance of winning something while each of us in separate box has less likely chance but bigger potential payout (shorter wider distribution vs taller narrower distribution with same area under each).

Some days I wish I had gone beyond statistics 101....





PS I always thought it would be fun to have a secondary market after the random selection of boxes. Would getting some combination of 7, 3, 0 be worth more than others?

[unofan]

So like most offices we have a Super Bowl box pool. $20 for five squares (or $5 for one square). We have enough people to fill two boxes.

I am going to partner with someone to put in $40 total between us. Are we better off both in the same box, or in different boxes?

My intuition says that we are "probably" better off with both of us being in the same box, as that gives us 10 chances to win something in that particular box instead of five. (10% chance to win something in one box, 0% chance in other; compared to 5% chance to win in each box).

However, I'm not convinced that the edge is as great as it first appears to be, since it also reduces the chances we could be big winners in both boxes.

Payout is 10% for 1st quarter, 20% for halftime, 10% for 3rd quarter, 60% for final score (nothing for 4th quarter if there is OT). If we assume $400 for each box (everyone puts in $20).

If scores and distribution of boxes is random, hmm....

10% of 400 + 0% of 400 = 5% of 200 + 5% of 200 so I am having trouble reconciling my intuition with math here, I think the math is missing something though since we might have duplication of numbers if spread over 2 boxes...but on a net basis, is duplication bad? (both lose) or good? (both win).

Maybe both of us in one box has better chance of winning something while each of us in separate box has less likely chance but bigger potential payout (shorter wider distribution vs taller narrower distribution with same area under each).

Some days I wish I had gone beyond statistics 101....





PS I always thought it would be fun to have a secondary market after the random selection of boxes. Would getting some combination of 7, 3, 0 be worth more than others?

While I don't understand the specifics of your superb owl pool, it sounds like a variation of the "draw a number, and if it matches [total points, score differential, some condition X], you win" and there are enough participants to have two separate pools, of which you can contribute to one or both.

Your overall odds of winning, assuming all possibilities are equally likely to win, is simply the number of possibilities in your control divided by the total number of possibilities, regardless of how they're broken down.

At its simplest, think of flipping two coins, so there are four potential bets you could make (H1, T1, H2, T2). It costs you $1 to cover one of those. You decide to partner up with someone else to cover two possibilities. You could each take a different side of the same coin, guaranteeing you'll break even, but only break even. One of your bets must lose while the other must win, and you can't win anything from the other coin since you have no money on it.

If instead you each take one side on each coin, you will still be expected to break even. 1/2 the time that will actually happen, 1/4 of the time you'll lose both dollars, and 1/4 of the time you'll come out $2 ahead.

Or another way to think of it is like keno. Your choice is essentially like choosing between playing multiple tickets for a single drawing, or one ticket for multiple drawings. All else being equal, your overall expected value doesn't change.

[FreshFish]

Re: Monty Hall, we have a PROBLEM

While I don't understand the specifics of your superb owl pool [sic] ...

spell check can be a necessary but not a sufficient condition of accuracy sometimes....

[unofan]

spell check can be a necessary but not a sufficient condition of accuracy sometimes....

http://www.cc.com/video-clips/qs3r6w...the-super-bowl

That was not a typo.

[LynahFan]

So this obviously fabricated story is making the rounds. I do have a question that I didn't see brought up in any of the few dozen comments that I could stomach reading, and I would put it to this audience:

In decimal notation, doesn't "999" also include implied operators? That is, isn't 999 really just shorthand for 9x100 + 9x10 + 9?

(and now my head hurts from thinking about the number 10, which is really 1x10+0. But you can't use "10" in the definition of itself, so we should really break that down as 1x (1x10+0)+0, but....)

[unofan]

So this obviously fabricated story is making the rounds. I do have a question that I didn't see brought up in any of the few dozen comments that I could stomach reading, and I would put it to this audience:

In decimal notation, doesn't "999" also include implied operators? That is, isn't 999 really just shorthand for 9x100 + 9x10 + 9?

(and now my head hurts from thinking about the number 10, which is really 1x10+0. But you can't use "10" in the definition of itself, so we should really break that down as 1x (1x10+0)+0, but....)

I suppose, but you could break it out in an infinite number of ways (999 is also 1+1+1+1.... or 111 x3x3 or 10x10x9 + 9 x 10 x 9), so I'm not understanding why that matters or where you're going with it.

[unofan]

Re: Monty Hall, we have a PROBLEM

Btw, if anyone on here isn't reading the Riddler weekly on fivethirtyeight.com, do so.

[Twitch Boy]

Re: Monty Hall, we have a PROBLEM

In front of you are two boxes. Box A is transparent and you can see it contains $1,000. Box B is opaque and the host tells you it contains either $1 million or nothing. The host offers you a choice: you can either take both Box A and Box B, or just take Box B.

Here's the catch: the host is a mindreader. And a darn good one. As far as anyone can tell he is never wrong. If he predicts you'll take both Box A and Box B, he puts nothing in Box B. If he predicts you'll take just Box B, he puts the million inside. He has already loaded the boxes backstage and can't change them now.

Which is the best play?

Technically, either play can logically be the better play. If you just take Box B and he predicted it, bam, million dollars, end of story. However, if he predicted that and the boxes are already loaded, might as well take both and score an extra thousand as well.

[Handyman]

Re: Monty Hall, we have a PROBLEM

If there is no penalty for taking both boxes you should take both boxes. Seems pretty obvious...

[The Rube]

Re: Monty Hall, we have a PROBLEM

If there is no penalty for taking both boxes you should take both boxes. Seems pretty obvious...

That is what I was thinking. Glad to know I am not completely nuts.

[joecct]

That is what I was thinking. Glad to know I am not completely nuts.

But if we're all nuts, is the sane one nuts?

[The Rube]

Re: Monty Hall, we have a PROBLEM

But if we're all nuts, is the sane one nuts?

I'll go ask Schrodinger's cat.