A really simple question.

Good old Mungo wins $4.20

Said, more in hope than expectation.

$4.20? Not in this airport, Buster! :-)

Your even a bigger crook than I thought, make that $8 profit. :wink:

Yup, $8 is right.

You call $8 being a bigger crook? What a sheltered life you must have lead! :-)

bm

Try

There are better ways to live your life than betting on those peculiar coins. Send $20 for my free booklet on ways for village lads to make it big.

Yup; I know I said free but the $20 is for postage and expenses. Would I lie to you?

Sent my solution by PM.

I'm sure it is correct, since it agrees with tryagain's winnings.

You gotta be kidding me, Try. How could the right answer have been a guess? Mark you, I have yet to see your reasoning.

Which brings me to Frank's reasoning. This is a strange problem in that the odds are calculated differently when in the bag from when a face has been shown. Your reasoning works, but does not take into account that a face has been shown, which does not effect the odds but does effect how they are arrived at. To illustrate; the game would be essentially unchanged if there were only two coins in the bag, one double headed and one 'normal'. If a tail shows, bets are doubled and the bag shaken again. If a head shows then I endeavor to guess as before.

There is now one chance in two of pulling out the double head. So do you think you would play against me on these rules?


BTW, remind me not to play games with Try and Frank in airports!

Try,

the notion "bm" is something you will run into in forums all the time, so you might as well know that often it means "bowel movement."

But here (and in most places) it means "bookmarking!"

That keeps the thread coming into my "My Posts" section -- and makes it easier to follow.

If that explanation does not make complete sense, ask me about it again. These forums are much more fun if you figure out how to monitor the threads you find interesting -- and that is the purpose of the "bm."

So; as both Try and Frank have exposed one of my best money makers I have seen the error of my ways and will now make amends. We will play dice instead!

Not ordinary dice, you understand, that is not my way. Out of a similar bag I pull four dice. They do not have the usual numbering. They have:

6, 6, 2, 2, 2, 2

5, 5, 5, 1, 1, 1

4, 4, 4, 4, 0, 0

3, 3, 3, 3, 3, 3

Just to show why I am still so widely known as 'Good Old Mungo' the bets will stay the same - I pay you $1 when you roll higher than me, you pay me 90 cents when I roll the highest. More than that, I will let you choose first each time which dice you will use. (We each roll one dice)

So now, after 30 rolls with the new reformed Mungo - if it follows the odds precisely, who will be ahead and by how much this time?

Can you both choose the same die?

Do you both throw at the same time -- or does one person go first?

Thank you Frank, I hope you get better soon.

Quote, "The new reformed Mungo" No change there then. Ha, ha. :wink:

Frank

Being a thoroughly nice person I will, of course, let my guest throw first unless he/she chooses otherwise. But he/she does get first pick of which dice because each chooses a dice before the throw.



P.S. You did not, nor did Try, say whether you felt the coin game would be the same if one of the double coins were removed from play.

Okay, let's take care of this oversight.


First of all -- let me give my solution to the original puzzle:

If everything goes "strictly according to the odds" -- in thirty tries, you will pull the double head coin from the bag 10 times; the double tail coin 10 times; and the head/tail coin 10 times.

In thirty tries, IF YOU ALWAYS GUESS THAT THE REVERSE IS THE SAME AS THE SIDE SHOWING -- you will lose 10 times (those ten occasions when the coin drawn was the heads/tails coin -- and win 20 times.

10 x $1 = $10 of losses
20 x $.90 = $18 of winnings

You win $8


Now let's take a look at the revision:

If you remove one of the double coins from the bag (let's say the two tails), you will be left with one coin with two heads and one coin with a head and a tail.

In thirty tries, if "everything goes strictly according to the odds", you will pull the two headed coin 15 times and the head/tails coin 15 times.

If you guess "heads" as the backside of the coin picked from the bag and placed on the table -- and if "everything goes strictly according to the odds" you will win all of the 15 times when the two headed coin is picked -- and half of the times that the heads/tails coin comes up. (Since you cannot have a half a win, we'll make the odd one a loss.)

You will have won 15 + 7 = 22 times x $.90 = $19.80

You will have lost 8 times x $1 = $8

You will have won - $11.80.


I'll get to your other puzzle in a bit.

QUESTION:

Can you pick the same die as your opponent -- or is that die only his and you must choose one of the others?

Frank

I'm sorry if I did not make that clear, so let me do so now.

Before each throw of the dice I bring out the bag, he/she selects a dice, I then select my dice from the three that remain. The bag is put away and we roll the dice with either one of us going first.

Frank

Regarding your explanation of how the odds would work for the revised version:

You say, "You will win half of the times when the head/tail coin comes up". I believe i said that, under the revised rules, if a tail was showing nobody wins, the bets are doubled, and we try again. So half of the time when the head/tail coin is selected no one wins. If my strategy is to say 'heads' every time then I do not win the other half either.

What does happen is that he/she wins one of four, one of four is retried, and two of four are, surprise surprise, won by me. I also have two-in-four chances of winning the retries with one in four chances of the retry being repeated. So, just as it was before, I would expect to be $8 up after 30 tries.

The point to remember is that the number of double coins is largely irrelevant. What matters is the faces. There are three ways that a head can be showing. If we assume that the heads of the double coin are heads A and B, and that of the head/tail coin is head C, of the three heads, if A comes up and I call 'heads' for the other side, I win. If head B comes up the same is true. Only when head C is showing will I lose. And the same is true of the tails. It does not matter if the double tail is removed and all appearance of the third tail are disregardced, there are still three heads that can be showing and I will win on any two of these if I call 'heads'.

In the real world the difficulty would be that a repeated calling of 'the same' would be quickly seen through. What would be interesting - but off-topic with this thread - would be to work out the best strategy that wins more than half but best disguises the 'same again' calling. I must have a think about that.