Not quite sure how one would calculate this, but here is the situation:


Two players are playing heads up. They will have 10 all-ins as they play over an unimportant number of hands (ie: these do not have to be 10 consecutive hands dealt).

Player 1 will choose 10 hands to shove and will do it whenever he has a top 75% hand.

Player 2 will call with a top 40% hand.


What are the odds that player 2 will have a dominating hand (70-30 or greater) all 10 hands in a row?

If that is impossible to determine, then what would be the odds of this happening if it were two random hands that were all-in each time.


Thanks in advance if there is a way of actually calculating this.

Not quite sure what your scenario is: player 1 shoves his first 10 top 75% hands, player 2 calls if and only if he has a top 40% hand, what are the odds he's 70/30+ 10 times in a row,

OR

Player 1 shoves all top 75% hands and you keep playing until player 2 has a top 40% hand to call with 10 times, etc.

either way, have to enumerate each qualifying hand combo for a deal and you can count the ones where player 2 is 70/30+, and raise that probability to the 10th.

Here is what I did.

Using shoving and calling hand ranges from the Kill Everyone book, I constructed a 75% shoving range and a 40% calling range as follows:

Player 1's 75% shoving range (actually this is a 74.962% range) = 22+,A2+,K2+,Q2+,J2+,T5o+,T2s+,96o+,93s+,86o+,84s+, 75o+,74s+,64s+,53s+

Player 2's 40% calling range (actually this is a 39.970% range) = 22+,A2+,K4o+,K2s+,Q8o+,Q6s+,J9+,T9s

I then dealt out 10,000,000 heads-up deals. I kept track of how often Player 1 (would) shove, how often Player 2 (would) call, how often Player 1 shoves and Player 2 calls, and how often Player 2's hand dominates Player 1's hand given that Player 1 shoved and Player 2 called.

For the purposes of this exercise, I considered any of the following cases to be a case of domination:

• Player 2 has a higher pair than Player 1 (KK>88)
• Player 2 has a pair & Player 1 does not have a pair & Player 1's highest card equals the rank of Player 2's pair (KK>K8)
• Neither player has a pair & they each have the same high card & Player 2's low card is higher than Player 1's low card (KJ>K8)
• Neither player has a pair & they each have the same low card & Player 2's high card is higher than Player 1's high card (K8>J8)

Here are the results of a simulation of 10 million total deals:

10,000,000 total deals
7,495,449 deals where Player 1 shoved (had a "top 75%" hand)
3,998,214 deals where Player 2 would have called (had a "top 40%" hand)
2,962,014 deals where Player 1 shoved and Player 2 called
333,754 deals where Player 1 shoved and Player 2 called & Player 2's hand dominated Player 1's hand

Note we would expect card removal effects to cause the probability Player 1 shoves and Player 2 calls to be slightly less than 0.74962*0.39970 = 0.29962, as confirmed by the numbers reported above.

The single-event probability that OP seeks (in my interpretation of OP) is then seen to be:

333,754 / 2,962,014 = 0.11268

As deals are independent, to find the probability that this event occurs in the next ten consecutive trials, we simply take the single-event probability to the 10th power.

(0.11268)^10 = 3.299*10^-10

Of course, the probability of a streak occurring anywhere in a sample of many trials is a different and higher probability.

As I always say in these circumstances, I tried to be careful in coding my simulation, but I would await confirmation from others before trusting these results.

Perfect, thanks!