- 08 Oct '07 02:48If 3 pennies are tossed and the order noted, there are 8 possible results. Say you choose one of those 8 in advance and I choose a different one, then we toss. There is only a 25% chance that the outcome will match either of our "predictions". So to get around this problem we toss a fourth coin, and use the last 3 as the "result", and if still no match then a fifth coin and so on until a result is obtained. We bet on each game and agree to play 100 times. The question is;;

Since you get first choice each time, and I MUST make a different selection, what odds should you give me in order to make it a fair bet? - 08 Oct '07 04:17 / 1 edit

1 in 2 because both options are equally probable.*Originally posted by luskin***If 3 pennies are tossed and the order noted, there are 8 possible results. Say you choose one of those 8 in advance and I choose a different one, then we toss. There is only a 25% chance that the outcome will match either of our "predictions". So to get around this problem we toss a fourth coin, and use the last 3 as the "result", and if still no match then a ...[text shortened]... UST make a different selection, what odds should you give me in order to make it a fair bet?**

Although, now that I think about it... there must be a case where you could bet at an advantage by having the last two of your coins match the first two of mine. That way, if I didn't win right away, I would have to risk having your set appear before mine.

Now I would say, less than 1 in 2, but I can't think of the exact probability. - 08 Oct '07 07:09 / 1 edit

There is no advantage to picking first, so 50/50 is fair odds.*Originally posted by luskin***If 3 pennies are tossed and the order noted, there are 8 possible results. Say you choose one of those 8 in advance and I choose a different one, then we toss. There is only a 25% chance that the outcome will match either of our \\\"predictions\\\". So to get around this problem we toss a fourth coin, and use the last 3 as the \\\"result\\\", and if still no ...[text shortened]... ifferent selection, what odds should you give me in order to make it a fair bet?**

[My interpretation of your wording is that you start with a bet and flip coins until a winning sequence occurs - no changing bets before that point. This is one \\\'game\\\'.] - 08 Oct '07 11:23

You're on the right track*Originally posted by Gastel***1 in 2 because both options are equally probable.**

Although, now that I think about it... there must be a case where you could bet at an advantage by having the last two of your coins match the first two of mine. That way, if I didn't win right away, I would have to risk having your set appear before mine.

Now I would say, less than 1 in 2, but I can't think of the exact probability. - 08 Oct '07 11:24

That interpretation is correct.*Originally posted by SwissGambit***There is no advantage to picking first, so 50/50 is fair odds.**

[My interpretation of your wording is that you start with a bet and flip coins until a winning sequence occurs - no changing bets before that point. This is one \\\'game\\\'.] - 08 Oct '07 14:45H=heads and T=tails (or however you spell it)

picking HHH would be pretty stupid, since you would pick THH and we only win 12,5% of the time, which is when the first three coins come heads, same goes for TTT.

How would you respond to any of the other choices?

Lets say we choose HTH, yuo would choose HHT (the last two of yours are the same as the first two of mine, therefore decreasing my chances of winning the second and so on throw, while your first two are different from my last 2). On the throw of the first three coins we have equal chances.

The second throw there are (8-2)x2=12 possibilities, of which 2 win for you and 1 wins for me.

The third throw there are (12-3)x2=18 possibilities, of which 3 win for you and 1 wins for me.

The fourth throw there are (18-4)x2=28 possibilities, of which 5 win for you and 2 win for me.

Now I don't really have the time to really see how this series is continuing, I'll come back to it later. - 09 Oct '07 01:39 / 2 edits

Darrie\'s post is enough to solve the problem.*Originally posted by Darrie***H=heads and T=tails (or however you spell it)**

picking HHH would be pretty stupid, since you would pick THH and we only win 12,5% of the time, which is when the first three coins come heads, same goes for TTT.

How would you respond to any of the other choices?

Lets say we choose HTH, yuo would choose HHT (the last two of yours are the same as the first two ...[text shortened]... the time to really see how this series is continuing, I\\\\\\\'ll come back to it later.

No matter what sequence the first player picks, the second player can best him. If we label player 1\'s sequence A-B-C, player 2 wins by picking A-A-B, except if it leads to all-tails or all-heads. In that case, he toggles the first coin in the sequence.

Darrie demonstrated that Player 2\'s chances increase with each coin toss. There is no limit on total coin tosses. Therefore, Player 2\'s chances increase infinitely, and only a sucker would take luskin\'s bet. The game is rigged in his favor. - 09 Oct '07 07:17 / 2 edits

The next line of your table is:*Originally posted by doodinthemood***This may be purely coincidental, but the probabilities appear to follow two fibonacci sequences:**

wins for him: wins for you:

2 1

3 1

5 2

8 3

The wins for you is always 2 steps behind. You should give your opponent odds somehow relating to the equation for a fibonacci sequence.

12 4

...which does not follow the Fibonacci sequence.