Odds and Even game - TOP Game Theory chapter

[teajay]

I was just thinking about the game described by skylansky in TOP chapter on bluffing and game theory.

I believe it was along the lines of playing odd-even with 101:100 payouts. The idea behind it was that if you were to use a coin flip to determine whether to show an odd or even number of fingers you could end up with a +0.5 EV, assuming all the conditions explained are met.

This made me wonder what would happen to your EV in a few different situations. Assume you are the opponent who is doing the guessing.

1) your opponent is not using a coin (or any external randomness) to make his decision. You are using a coin to decide. (This is just the reverse of what skylansky explains)

2) Both you and your opponent are using a coin make your decision. Since you have a 50/50 chance at heads or tails, I don't think it would matter if you use heads to decide odd and your opponent uses tails, or vice versa. Would it?

Thanks for any help you can provide... And sorry if i'm unclear on any of it. Just ask I may be able to explain better.

cheers,
travis

[k_squared]

I am not sure what your question is...

but the reason you use a coin to make your decision is precisely because that way your opponent cannot predict your decision, or find a pattern in your actions to take advantage of. In the case where you are faced with a bet that pays 101:100 you will always win if you randomize your decision because there is no way for him to predict what you will do and hence over the long run he will win half and you will win half. The person getting the best of the odds will win regardless of how the coin flips, or whether both people flip a coin. Once you you allow the element of randomness to enter the equation you can't take advantage of 'skill' which would otherwise allow you to be able to overcome the odds being layed in the other person's favor. If you take skill out of play and make it a game of random chance then the person getting the best odds will eventually win!

-K_squared

[pudley4]

[ QUOTE ]
I was just thinking about the game described by skylansky in TOP chapter on bluffing and game theory.

I believe it was along the lines of playing odd-even with 101:100 payouts. The idea behind it was that if you were to use a coin flip to determine whether to show an odd or even number of fingers you could end up with a +0.5 EV, assuming all the conditions explained are met.

This made me wonder what would happen to your EV in a few different situations. Assume you are the opponent who is doing the guessing.

1) your opponent is not using a coin (or any external randomness) to make his decision. You are using a coin to decide. (This is just the reverse of what skylansky explains)

2) Both you and your opponent are using a coin make your decision. Since you have a 50/50 chance at heads or tails, I don't think it would matter if you use heads to decide odd and your opponent uses tails, or vice versa. Would it?

Thanks for any help you can provide... And sorry if i'm unclear on any of it. Just ask I may be able to explain better.

cheers,
travis

[/ QUOTE ]

It doesn't matter. As long as at least one of you is generating a completely random result, you (and he) will both win 50% in the long run.

[teajay]

Figured it would be as simple as that... Thanks for your reply.

Cheers,
travis