perudo question

so yesterday my wife and I threw identical perudo rolls (4 aces and one 2. Obviously once I had rolled 4 aces and 1 2 the odds of her rolling the same were 5/7776 (7776 possible combinations of 5 dice, 5 of which will consist of 4 aces and 1 2). But it made me wonder what the odds were of us rolling the same thing before either of us roll. Rolls with 4 identical numbers are uncommon (30/7776), but doing the whole sum is eluding me. There's the problem that if I can roll all different, 1 pair, 2 pairs, 3 of a kind, a full house, 4 of a kind or Yahtzee all of which have different odds of occurring and being repeated. Anyone fancy having a stab?

I think it’s :

(6 x 1 + 150 x 5 + 300 x 10 + 1200 x 20 + 1800 x 30 + 3600 x 60 + 720 x 120) / ( 6 ^ 10) = 3,557 / 559,872

Or about 1 in 157.

Assuming you are asking what is the probability that all five of your dice match all five of your wife's, here's what I get:

Of the 7776 x 7776 = 60,466,176 possible pairs of rolls:
Five of a kind: 6 ways for you to roll it x 1 way for your wife to roll a matching set = 6
Four of a kind: 150 x 5 = 750
Full house: 300 x 10 = 3000
Three of a kind: 1200 x 20 = 24,000
Two pair: 1800 x 30 = 54,000
One pair: 3600 x 60 = 216,000
Five different numbers: 720 x 120 = 86,400
The probability is (6 + 750 + 3000 +24,000 + 54,000 + 216,000 + 86,400) / 60,466,176, which is 3557 / 559,872.

In other words, what Ace2 said.

Thanks folks. I’d got solutions for 5 of a kind, 4 of a kind, full house and all different, was close to 3 of a kind but 1 pair and 2 pairs was somewhat over stretching my cognition. I’m impressed with the speed of the solution!!!

Bw
Matt