Simplified poker

[slapdash]

Two people play. They each put an ante of $10 in the pot. Then they are each "dealt" a random number between 0 and 1. (Assume the numbers will always be different, so you don't need to worry about split pots.) Player A can then check, or bet another $10. If he checks, player B doesn't get a chance to bet. They just compare numbers and the higher number wins the pot of $20. If player A bets, then player B can fold or call (not raise). If he folds, then player A takes the $30 in the pot. If he calls, they compare numbers and the higher number takes the $40 in the pot. Player B's best strategy, assuming he knows what strategy player A is following, is easy to work out. When player A bets, he's giving player B pot odds of 3/1. So player B should call if he has at least a 25% chance of winning; i.e., if his number is bigger than at least a quarter of the numbers player A would have bet with. What about player A? Obviously he can break even on average by always checking. But can he do better? And how?

[GaF]

Re: Simplified poker Interesting........ Gut feel is player A should raise anything above average (0.5) ..... but I suspect the answer is a bit more complicated than that ...... Will give this some real thought later :ok

[Cloud]

Re: Simplified poker I'm thinking that he has to win over half of the pots where he raises. So if player B calls a raise with anything in the top 75% of the number range - player A would have to raise with anything in the top 37.5% of the range. So player A would raise with 0.625 or more for an optimum strategy. (This is where I find out I'm completely wrong :lol )

[Hedonist]

Re: Simplified poker Looks to me that player A would lose unless he bought a random element into his betting, betting the top 20% and the bottom 10% or 20%. How player B would work out what to bet with I do not know!

[slapdash]

Re: Simplified poker

Interesting........ Gut feel is player A should raise anything above average (0.5) ..... but I suspect the answer is a bit more complicated than that ...... Will give this some real thought later :ok

To save people some headaches, and because doing the calculation for a particular strategy is not really the point, here's how I found it's easiest to do the calculation: Just compare the outcome of your strategy with the outcome of the "always check" strategy for all possible deals. In many cases the outcome is the same. This will be the case when: (a) you check, or (b) you bet, he folds, and your number is higher. So, in GaF's "gut feel" strategy, you bet with greater than 0.5, and he calls with greater than 0.625. Call 0 to 0.5 "small", 0.5 to 0.625 "medium", and 0.625 to 1 "large. If you're small, that's (a), same as "always check". If you're large, he's small or medium, that's (b), same as "always check". If you're medium, he's small, that's also (b) But also, if you're both large, it's 50/50, as is "always check", although you're playing for double stakes, so on average this is the same as "always check" as well. That just leaves: (i) Both medium. This happens 1/64 of the time. In GaF's strategy, you win $10, but in "always check" you break even on average. This adds $10/64 to your expected winnings, compared to "always check". (ii) You're medium, he's large. This happens 3/64 of the time. GaF's strategy loses an extra $10. So this subtracts $30/64 from your expected winnings. So GaF's strategy loses $20/64 compared to "always check".

[slapdash]

Re: Simplified poker

How player B would work out what to bet with I do not know!

Player B is only allowed to call. And I'm assuming at the moment that he knows player A's strategy, so his calling strategy is simple: call if he has pot odds. I.e., call if he can beat 25% of the hands player A would bet with.

[GaF]

Re: Simplified poker

To save people some headaches' date=' [/quote'] Looks like I like headaches!!!! I just cannot accept that calling the bottom 50% of numbers and raising the top 50% of numbers has a negative ev!!! There are only 5 different possible outcomes, so ev should be easy enough to calculate - so I have :unsure 1) Your number is below 0.5 and you call and win (0.5 * 0.25 = 12.5% of the time) 50% of the time you call with an average number of 0.25. Your opponent has an average number of 0.5. Your number will be above your opponents and you will win $10 25% of the time. ev for this is therefore 0.5 * 0.25 * $10 = $1.25 2) Your number is below 0.5 and you call and lose(0.5 * 0.75 = 37.5% of the time) 50% of the time you call with an average number of 0.25. Your opponent has an average number of 0.5. Your number will be below your opponents and you will lose $10 75% of the time. ev for this is therefore 0.5 * 0.75 * -$10 = -$3.75 3) Your number is above 0.5, you are called and win (0.5 * 0.375 * 0.75 = 14.0625% of the time) 50% of the time you raise with an average number of 0.75. Your opponent will call with a number 0.625 or above. You will win $20 75% of the time. ev for this is therefore 0.5 * 0.375 * 0.75 * $20 = $2.81 4) Your number is above 0.5, you are called and lose (0.5 * 0.375 * 0.25 = 4.6875% of the time) 50% of the time you raise with an average number of 0.75. Your opponent will call with a number 0.625 or above. You will lose $20 25% of the time. ev for this is therefore 0.5 * 0.375 * 0.25 * -$20 = -$0.94 5) Your number is above 0.5, you are not called and win (0.5 * 0.625 = 31.25% of the time) 50% of the time you raise with an average number of 0.75. Your opponent will fold with a number below 0.625. You will win $10 100% of the time. ev for this is therefore 0.5 * 0.625 * 1 * $10 = $3.13 Overall ev is therefore $1.25-$3.75+2.81-$0.94+3.13=$2.50!! (positive ev!!) I'm pretty confident my calculations are right ....... but am sure I'll quickly be told if they're not!!!!

[slapdash]

Re: Simplified poker

3) Your number is above 0.5, you are called and win (0.5 * 0.375 * 0.75 = 14.0625% of the time) 50% of the time you raise with an average number of 0.75. Your opponent will call with a number 0.625 or above. You will win $20 75% of the time. ev for this is therefore 0.5 * 0.375 * 0.75 * $20 = $2.81

Why 75%? The times your number is between 0.5 and 1, and your opponent's is between 0.625 and 1, he's winning more often than you. I tried doing the calculations your way to begin with, and got at least five different answers! That's why I looked for the "easy" way! Your 1) and 2) look right to me, by the way, but I haven't checked 4) and 5).

[GaF]

Re: Simplified poker

Why 75%? The times your number is between 0.5 and 1, and your opponent's is between 0.625 and 1, he's winning more often than you.

Er....agreed - my number is wrong...... Lets see how big a headache I can get!!!! I still don't accept that the ev of this system can be less than the 0 ev (break even) of always calling......

[slapdash]

Re: Simplified poker If your number is bigger than 0.5 and his is bigger than 0.625, then you lose whenever yours is less than 0.625 (25% of the time) and the other 75% of the time, when you both have numbers bigger than 0.625, it's 50/50. So I think your 75% in 3) should be 37.5%, and your 25% in 4) should be 62.5%. If you make those changes, your answer agrees with mine.

[GaF]

Re: Simplified poker

So I think your 75% in 3) should be 37.5%, and your 25% in 4) should be 62.5%. If you make those changes, your answer agrees with mine.

Goddamn!!!!!

[Hedonist]

Re: Simplified poker If you bet with .8 to 1 and 0 to .2 then your average is .4 to .6. If player b calls with anything over .45 then you make a profit as far as I can calculate, but I am probably wrong.

[GaF]

Re: Simplified poker

If you bet with .8 to 1 and 0 to .2 then your average is .4 to .6. If player b calls with anything over .45 then you make a profit as far as I can calculate, but I am probably wrong.

I make that ev of -$0.25 (better than mine which turned out at ev of -$0.31). To put it into the "boxes" above.... 1) 40% chance of winning $10 2) 40% chance of losing $10 3) 5.625% chance of winning $20 4) 9.375% chance of losing $20 5) 5% chance of winning $10 Have it all formulated now though ...... so a goal seek should give me the optimal answer........

[GaF]

Re: Simplified poker

I make that ev of -$0.25 (better than mine which turned out at ev of -$0.31). To put it into the "boxes" above.... 1) 40% chance of winning $10 2) 40% chance of losing $10 3) 5.625% chance of winning $20 4) 9.375% chance of losing $20 5) 5% chance of winning $10 Have it all formulated now though ...... so a goal seek should give me the optimal answer........

Er forget this...... just tested it with the 50% raise and it gives a different answer - double checking ..... My head hurts :wall

[Hedonist]

Re: Simplified poker I worked it out that Player A will bet 40 times in 100, Player B will call 22 and fold 18. The 18 he folds he will lose 180 (18 * 10) The 22 he calls 11 will be against low hands so he will win 220 (11 * 20) The 11 he calls against high hands 7 will be with between .45 and .8 so he will lose 140 (7 * 20) The other 4 will be with both players between .8 and 1 so they can expect to win 2 each. both win 40. So Player A wins 360, Player B wins 260. The rest are even, Player B will win with hands .8 and above and lose with hands .2 and below. I am too worn out to calculate anything else.

[slapdash]

Re: Simplified poker

If you bet with .8 to 1 and 0 to .2 then your average is .4 to .6. If player b calls with anything over .45 then you make a profit as far as I can calculate, but I am probably wrong.

But if you bet with 0.8 to 1 and with 0 to 0.2, then player B will call with anything above 0.1, because then he's beating more than a quarter of the hands you bet with, so he has pot odds.

[Dodger]

Re: Simplified poker Has anybody incorporated the fact that player A's left eye twitches 20% of the time and he scratches his nose 10% of the time....and player B breaks out in a sweat when he has a high number??? The calculations above seem to centre on B knowing what hands A will play. If the game were to be played ''live'' A would raise with rubbish in the hope of getting B to fold, obviously, has this been considered? Also the inability of B to put in a single raise prevents him bluffing at all. What I am trying to say is that two bots against each other can be mathematically defined...chuck in the human element and all calculations go out of the window! If such a simplification of poker requires such enormous mathematical calculations to be considered, it's no wonder we are all so enthralled with the world of poker...it's so simple, yet so damned complex.

[Hedonist]

Re: Simplified poker

But if you bet with 0.8 to 1 and with 0 to 0.2, then player B will call with anything above 0.1, because then he's beating more than a quarter of the hands you bet with, so he has pot odds.

So Player A bets with 0.7 to 1 and 0 to 0.1 then B will call with anything above 0.1. Easy win for Player A

[slapdash]

Re: Simplified poker

So Player A bets with 0.7 to 1 and 0 to 0.1 then B will call with anything above 0.1. Easy win for Player A

:ok Exactly the right answer. Player A makes an average profit of $1. Actually, if player A does this it doesn't matter on average whether player B calls or folds with between 0.1 and 0.7, as he's getting exactly fair pot odds. But if he always folded these intermediate hands, then player A could exploit this by also betting some intermediate hands. And if he always called with the intermediate hands, player A could exploit this by bluffing less often. So in fact the "best" (Nash equilibrium) strategy for player A is to bet with 0 to 0.1 and 0.7 to 1. The "best" strategy for player B involves calling with 0.7 to 1, folding with 0 to 0.1, and using a randomized strategy with 0.1 to 0.7. I think it's a nice little model poker game, as it illustrates nicely what bluffing is for. Player A doesn't "bluff" with 0 to 0.1 to "fool" his opponent; it works even if he tells his opponent exactly which bad hands he'll bet. He does it to force his opponent to call with more hands, so he can make more profit with his good hands. It also illustrates nicely the principle that it's your very worst hands you should be bluffing with.