The player places a wager of any size, let's use 10$ for example. The dealer now takes that 10$ ante and deals the cards. Texas Holdem 2 cards per player, one deck. The dealer goes all in blind, for ten times the amount of the ante, in this example the dealer takes the ten dollars from the player, and puts up 100$ on the line. The player looks at his cards and makes the decision to either call 100$ more, to win a total pot of 200$. Or fold his hand, and lose nothing but the cost of admission, which was 10$. If the player antes 5$ instead then the dealer only goes all in blind for 50$. If the player decides to call, then 5 community cards are dealt, and the best hand wins the pot. A tie is a push on the pot.

You lose the ante either way, would this be beatable?

Since you would only call the approximate top half of your hands you would have to win over 60% when you did call. Do you think you would?

If you could see the four other players cards could it be beatable?

eg j5o could be playable with an Ace and Q dead and no dead j or 5.

must be quite close

Surely this is beatable.

If you call top 30% of hands, you call 3 in every 10. So in 10 hands, you spend $400 and can win $300, you must win 57%.

Does the top 30% of hands beat an average hand 57% of the time? ive got no poker software.....

if you call top 20% you need to win 60%

if you call top 10% you need to win 66%

Is anyone aware of any graphs that show top x% of hands vs probability you beat an average hand?

Call 50% of the time

0.5(10x - 10(1-x) - 1 = 0
0.5(10x - 10 + 10x) - 1 = 0
5x - 5 + 5x - 1 = 0
10x = 6
x = 0.6
Have to win 60% of the time when calling to break even


Call 30% of the time

0.3(10x - 10(1-x) - 1 = 0
0.3(10x - 10 + 10x) - 1 = 0
3x - 3 + 3x - 1 = 0
6x = 4
x = 0.67
Have to win 67% of the time when calling to break even


Call 20% of the time

0.2(10x - 10(1-x) - 1 = 0
0.2(10x - 10 + 10x) - 1 = 0
2x - 2 + 2x - 1 = 0
4x = 3
x = 0.75
Have to win 75% of the time when calling to break even


Call 10% of the time

0.1(10x - 10(1-x) - 1 = 0
0.1(10x - 10 + 10x) - 1 = 0
1x - 1 + 1x - 1 = 0
2x = 2
x = 1
Have to win 100% of the time when calling to break even






EDIT:

What does the player in is losing the initial ante no matter what. Lose it when you fold. Lose it when you call and lose the hand. Lose it when you call and win the hand.