Probability of (all-in) tied hands and effect of adding random caller

[dwr]

(My first post on 2+2 so please be gentle...)

This situation came up today. Satellite that paid top 5, 6 players left.

Seat 1: (1688 in chips)
Seat 3: (3146 in chips)
Seat 5: (4664 in chips)
Seat 6: (2140 in chips)
Seat 7: (1458 in chips)
Seat 101904 in chips)
Seat 10: posts big blind 600

Seat 6 folds
Seat 7 is all-in
Seat 10 folds
Seat 1 is all-in
Seat 3 calls

Now irrelevant of what the two short stacks hold, I was curious as to why the larger stack called. He said it was to reduce the probability of the 2 short stacks tie-ing and one not busting, and would call here with any 2.

While it seems sort of reasonable, I have the following questions:

1. How do you calculate (ie what is the maths) the likelihood of the 2 short stacks having the same hand, presuming you can put them on identical ranges?

2. What effect does the 3rd player have on these figures?

[adMIT defeat]

1. Probability P2 gets the same paired hand as P1 is: 2/50 * 1/49.
Probability P2 gets the same unpaired hand as P1 is: 6/50 * 3/49.
Determine the range of these players, and then multiply each hand in the range by one of these probabilities and add them together.

2. The third player has no effect on these figures.


You're missing the point of what the 3rd player was saying. The probability of the 2 short stacks tying is more than the probability of them having the same hand. This depends completely on the board. If they have 44 and 55, and the board comes QQKKA, they chop. The 3rd player being in the hand cuts down on the number of times this happens, because some of the time they chop, the 3rd player will have a better hand.

[Yepitis]

I did this once and they tied and still beat me, what are the chances of that?